How do we print out the value_type of a C++ STL container?

Clash Royale CLAN TAG#URR8PPP

up vote
11
down vote

favorite

2

I know that STL containers have a value_type parameter and I've seen how it can be used to declare a type of a variable like:

vector<int>::value_type foo;

But can we just print this value_type to a console?

I want to see "string" in this example:

int main() 
 vector<string> v = "apple", "facebook", "microsoft", "google";
 cout << v << endl;
 cout << v.value_type << endl;
 return 0;

c++ stl

share|improve this question

edited Sep 3 at 19:56

Peter Mortensen

12.9k1983111

asked Sep 3 at 9:06

Niakrais

14610

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The symbol value_type is not a member variable, it's a scoped type-name. Would you expect to be able to print the value of the keyword int?
  – Some programmer dude
  Sep 3 at 9:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  it's just a type, so you want to get type name as string?
  – apple apple
  Sep 3 at 9:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Quite possible, but you'll need to add some code.
  – bipll
  Sep 3 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That would be equivalent to cout << string;, which is not the same as cout << "string";.
  – molbdnilo
  Sep 3 at 9:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yeah @molbdnilo I want a type as human readable string
  – Niakrais
  Sep 3 at 9:26
  

  
  
  
  

 | 
show 1 more comment

up vote
11
down vote

favorite

2

I know that STL containers have a value_type parameter and I've seen how it can be used to declare a type of a variable like:

vector<int>::value_type foo;

But can we just print this value_type to a console?

I want to see "string" in this example:

int main() 
 vector<string> v = "apple", "facebook", "microsoft", "google";
 cout << v << endl;
 cout << v.value_type << endl;
 return 0;

c++ stl

share|improve this question

edited Sep 3 at 19:56

Peter Mortensen

12.9k1983111

asked Sep 3 at 9:06

Niakrais

14610

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The symbol value_type is not a member variable, it's a scoped type-name. Would you expect to be able to print the value of the keyword int?
  – Some programmer dude
  Sep 3 at 9:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  it's just a type, so you want to get type name as string?
  – apple apple
  Sep 3 at 9:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Quite possible, but you'll need to add some code.
  – bipll
  Sep 3 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That would be equivalent to cout << string;, which is not the same as cout << "string";.
  – molbdnilo
  Sep 3 at 9:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yeah @molbdnilo I want a type as human readable string
  – Niakrais
  Sep 3 at 9:26
  

  
  
  
  

 | 
show 1 more comment

up vote
11
down vote

favorite

2

up vote
11
down vote

favorite

2

2

I know that STL containers have a value_type parameter and I've seen how it can be used to declare a type of a variable like:

vector<int>::value_type foo;

But can we just print this value_type to a console?

I want to see "string" in this example:

int main() 
 vector<string> v = "apple", "facebook", "microsoft", "google";
 cout << v << endl;
 cout << v.value_type << endl;
 return 0;

c++ stl

share|improve this question

edited Sep 3 at 19:56

Peter Mortensen

12.9k1983111

asked Sep 3 at 9:06

Niakrais

14610

I know that STL containers have a value_type parameter and I've seen how it can be used to declare a type of a variable like:

vector<int>::value_type foo;

But can we just print this value_type to a console?

I want to see "string" in this example:

int main() 
 vector<string> v = "apple", "facebook", "microsoft", "google";
 cout << v << endl;
 cout << v.value_type << endl;
 return 0;

c++ stl

share|improve this question

edited Sep 3 at 19:56

Peter Mortensen

12.9k1983111

asked Sep 3 at 9:06

Niakrais

14610

share|improve this question

share|improve this question

edited Sep 3 at 19:56

Peter Mortensen

12.9k1983111

edited Sep 3 at 19:56

Peter Mortensen

12.9k1983111

edited Sep 3 at 19:56

Peter Mortensen

12.9k1983111

12.9k1983111

asked Sep 3 at 9:06

Niakrais

14610

asked Sep 3 at 9:06

Niakrais

14610

asked Sep 3 at 9:06

Niakrais

14610

14610

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The symbol value_type is not a member variable, it's a scoped type-name. Would you expect to be able to print the value of the keyword int?
  – Some programmer dude
  Sep 3 at 9:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  it's just a type, so you want to get type name as string?
  – apple apple
  Sep 3 at 9:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Quite possible, but you'll need to add some code.
  – bipll
  Sep 3 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That would be equivalent to cout << string;, which is not the same as cout << "string";.
  – molbdnilo
  Sep 3 at 9:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yeah @molbdnilo I want a type as human readable string
  – Niakrais
  Sep 3 at 9:26
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The symbol value_type is not a member variable, it's a scoped type-name. Would you expect to be able to print the value of the keyword int?
  – Some programmer dude
  Sep 3 at 9:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  it's just a type, so you want to get type name as string?
  – apple apple
  Sep 3 at 9:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Quite possible, but you'll need to add some code.
  – bipll
  Sep 3 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That would be equivalent to cout << string;, which is not the same as cout << "string";.
  – molbdnilo
  Sep 3 at 9:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  yeah @molbdnilo I want a type as human readable string
  – Niakrais
  Sep 3 at 9:26
  

  
  
  
  

The symbol value_type is not a member variable, it's a scoped type-name. Would you expect to be able to print the value of the keyword int?
– Some programmer dude
Sep 3 at 9:08

The symbol value_type is not a member variable, it's a scoped type-name. Would you expect to be able to print the value of the keyword int?
– Some programmer dude
Sep 3 at 9:08

it's just a type, so you want to get type name as string?
– apple apple
Sep 3 at 9:08

it's just a type, so you want to get type name as string?
– apple apple
Sep 3 at 9:08

Quite possible, but you'll need to add some code.
– bipll
Sep 3 at 9:09

Quite possible, but you'll need to add some code.
– bipll
Sep 3 at 9:09

That would be equivalent to cout << string;, which is not the same as cout << "string";.
– molbdnilo
Sep 3 at 9:14

That would be equivalent to cout << string;, which is not the same as cout << "string";.
– molbdnilo
Sep 3 at 9:14

yeah @molbdnilo I want a type as human readable string
– Niakrais
Sep 3 at 9:26

yeah @molbdnilo I want a type as human readable string
– Niakrais
Sep 3 at 9:26

 | 
show 1 more comment

5 Answers
5

active

oldest

votes

up vote
12
down vote



accepted

X::value_type is no different from any other type alias (aka typedef) in this regard — C++ has no native way of converting a type to its source code string representation (not the least because that could be ambiguous). What you can do is use std::type_info::name:

cout << typeid(decltype(v)::value_type).name() << endl;

The resulting text is compiler dependent (and not even guaranteed to be easily human readable). It will be consistent within the same build of a program, but you cannot expect it to be the same across different compilers. In other words, it's useful for debugging, but cannot reliably be used in a persistent fashion.

share|improve this answer

edited Sep 3 at 9:12

answered Sep 3 at 9:09

Angew

127k11236333

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you, on my machine it gives me NSt3__112basic_stringIcNS_11char_traitsIcEENS_9allocatorIcEEEE
  – Niakrais
  Sep 3 at 9:24
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @You can use __cxa_demangle to get a readable type name: stackoverflow.com/a/23266701/1968
  – Konrad Rudolph
  Sep 3 at 9:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Isn't it only for gcc?
  – Niakrais
  Sep 3 at 9:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Niakrais: Not quite, it's for all compilers/platforms combinations using the Itanium ABI. The Itanium ABI is used on all Linux/MacOS platforms by all mainstream C++ compilers; the only widespread platform not using it is Windows, and thus MSVC and Clang in MSVC-compatible mode will not use it. Even on Windows, using Cygwin or MinGW means using the Itanium ABI.
  – Matthieu M.
  Sep 3 at 11:14
  

  
  
  
  

add a comment | 

up vote
5
down vote

In addition to the typeid-based answer, I see one further possibility using Boost like this:

#include <boost/type_index.hpp>

cout << boost::typeindex::type_id_with_cvr<decltype(v)::value_type>().pretty_name() << "n";

The advantage is a portable, human-readable name that includes const/volatile qualifications and is consistent across major compilers and systems. The output on my machine is

// when compiled with gcc:
std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >
// ... and with clang:
std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> >

Decide for yourself whether this output qualifies as "human-readable", but also compare it to the typeid(...).name() approach, which on my machine gives:

NSt3__112basic_stringIcNS_11char_traitsIcEENS_9allocatorIcEEEE

share|improve this answer

edited Sep 3 at 19:57

Peter Mortensen

12.9k1983111

answered Sep 3 at 9:12

lubgr

6,0701339

add a comment | 

up vote
1
down vote

It's probably safe to assume that you need this type info for debug purposes(?).
If it is, don't forget gdb's built in whatis and ptype commands:

$ cat main.cpp
#include <iostream>
#include <vector>
using namespace std;
int main() 
 vector<string> v = "apple", "facebook", "microsoft", "google" ;
 cout << v[2] << endl;
 // cout << v.value_type << endl;
 return 0;

$ cat gdbCommands.txt
break main
run
next
next
whatis v
quit
$ g++ -ggdb -o main main.cpp
$ gdb -x gdbCommands.txt ./main
GNU gdb (Ubuntu 8.0.1-0ubuntu1) 8.0.1
// bla bla bla ...
type = std::vector<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >>

share|improve this answer

answered Sep 3 at 10:13

OrenIshShalom

845520

add a comment | 

up vote
1
down vote

I had a rather similar question some time ago. The answer there shown a function able to translate a type into a human readable string:

template <typename T> std::string TypeName() 
 auto name = typeid(T()).name(); // function type, not a constructor call!
 int status = 0;

 std::unique_ptr<char, void(*)(void*)> res 
 abi::__cxa_demangle(name, NULL, NULL, &status),
 std::free
 ;

 std::string ret((status == 0) ? res.get() : name);
 if (ret.substr(ret.size() - 3) == " ()") ret.resize(ret.size() - 3);
 return ret;

Once you have this TypeName function you can achieve your goal:

int main() 
 vector<string> v = "apple", "facebook", "microsoft", "google" ;
 cout << v << endl;
 cout << TypeName<v.value_type>() << endl;
 return 0;

Which should output (GCC HEAD 9 201809):

std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >

share|improve this answer

answered Sep 3 at 13:55

Paula_plus_plus

6,58442758

add a comment | 

up vote
1
down vote

If it's for debug purpose, then this solution would work without any dependencies:

#include <regex>

template <typename T>
struct TypeToNameT

 static std::string getTypeFromName() 
#ifdef _MSC_VER
 std::regex re("<(.*)>::.*");
 std::string templateType(__FUNCTION__);
#else
 std::regex re(" = (.*);");
 std::string templateType(__PRETTY_FUNCTION__);
#endif
 std::smatch match;
 try
 
 if (std::regex_search(templateType, match, re) && match.size() > 1)
 return match.str(1);
 
 catch (std::regex_error& e) 
 // Syntax error in the regular expression
 
 return "";
 
;
/** Get the templated type name. This does not depends on RTTI, but on the preprocessor, so it should be quite safe to use even on old compilers */
template <typename T>
std::string getTypeName() return TypeToNameT<T>::getTypeFromName(); 


int main() 
 std::vector<std::string> v = "apple", "facebook", "microsoft", "google" ;
 std::cout << getTypeName<decltype(v)::value_type>() << std::endl; // Returns: "std::__cxx11::basic_string<char>"
 return 0;

Unlike the typeid based answer, this use the preprocessor and as such is not subject to name mangling (so the name is... readable).

Please notice that you can't use v.value_type as a type directly, it's not part of the instance v but of the type of v: std::vector<std::string>

The regex part is because the std::string class is too dumb and you need such oddities for a simple search/split code. If you have a decent string class, you'll not need regex here for extracting the part that's after the "=" string and before the ";".

share|improve this answer

answered Sep 3 at 14:14

xryl669

1,4301327

• 
  
  
  

  
  

  
  
  
  
  
  

  
  __FUNCTION__ prints just the function name, without any additional information. You want __FUNCSIG__.
  – HolyBlackCat
  Sep 3 at 16:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Just tried it here: rextester.com/FDZT51977 and it seems to work with FUNCTION. With FUNCSIG, the regex must be modified since the signature is more complex. In all cases, thanks, I did not know about FUNCSIG before.
  – xryl669
  Sep 3 at 16:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nevermind, you're right. __FUNCTION__ doesn't contain template parameters of the function itself, so when I used this trick before, I resorted to use __FUNCSIG__. But it missed me that it contains the template parameters of enclosing classes. That's a nice idea.
  – HolyBlackCat
  Sep 3 at 16:33
  
  

  
  
  
  

add a comment | 

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
12
down vote



accepted

X::value_type is no different from any other type alias (aka typedef) in this regard — C++ has no native way of converting a type to its source code string representation (not the least because that could be ambiguous). What you can do is use std::type_info::name:

cout << typeid(decltype(v)::value_type).name() << endl;

The resulting text is compiler dependent (and not even guaranteed to be easily human readable). It will be consistent within the same build of a program, but you cannot expect it to be the same across different compilers. In other words, it's useful for debugging, but cannot reliably be used in a persistent fashion.

share|improve this answer

edited Sep 3 at 9:12

answered Sep 3 at 9:09

Angew

127k11236333

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you, on my machine it gives me NSt3__112basic_stringIcNS_11char_traitsIcEENS_9allocatorIcEEEE
  – Niakrais
  Sep 3 at 9:24
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @You can use __cxa_demangle to get a readable type name: stackoverflow.com/a/23266701/1968
  – Konrad Rudolph
  Sep 3 at 9:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Isn't it only for gcc?
  – Niakrais
  Sep 3 at 9:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Niakrais: Not quite, it's for all compilers/platforms combinations using the Itanium ABI. The Itanium ABI is used on all Linux/MacOS platforms by all mainstream C++ compilers; the only widespread platform not using it is Windows, and thus MSVC and Clang in MSVC-compatible mode will not use it. Even on Windows, using Cygwin or MinGW means using the Itanium ABI.
  – Matthieu M.
  Sep 3 at 11:14
  

  
  
  
  

add a comment | 

up vote
12
down vote



accepted

X::value_type is no different from any other type alias (aka typedef) in this regard — C++ has no native way of converting a type to its source code string representation (not the least because that could be ambiguous). What you can do is use std::type_info::name:

cout << typeid(decltype(v)::value_type).name() << endl;

The resulting text is compiler dependent (and not even guaranteed to be easily human readable). It will be consistent within the same build of a program, but you cannot expect it to be the same across different compilers. In other words, it's useful for debugging, but cannot reliably be used in a persistent fashion.

share|improve this answer

edited Sep 3 at 9:12

answered Sep 3 at 9:09

Angew

127k11236333

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you, on my machine it gives me NSt3__112basic_stringIcNS_11char_traitsIcEENS_9allocatorIcEEEE
  – Niakrais
  Sep 3 at 9:24
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @You can use __cxa_demangle to get a readable type name: stackoverflow.com/a/23266701/1968
  – Konrad Rudolph
  Sep 3 at 9:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Isn't it only for gcc?
  – Niakrais
  Sep 3 at 9:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Niakrais: Not quite, it's for all compilers/platforms combinations using the Itanium ABI. The Itanium ABI is used on all Linux/MacOS platforms by all mainstream C++ compilers; the only widespread platform not using it is Windows, and thus MSVC and Clang in MSVC-compatible mode will not use it. Even on Windows, using Cygwin or MinGW means using the Itanium ABI.
  – Matthieu M.
  Sep 3 at 11:14
  

  
  
  
  

add a comment | 

up vote
12
down vote



accepted

up vote
12
down vote



accepted

X::value_type is no different from any other type alias (aka typedef) in this regard — C++ has no native way of converting a type to its source code string representation (not the least because that could be ambiguous). What you can do is use std::type_info::name:

cout << typeid(decltype(v)::value_type).name() << endl;

The resulting text is compiler dependent (and not even guaranteed to be easily human readable). It will be consistent within the same build of a program, but you cannot expect it to be the same across different compilers. In other words, it's useful for debugging, but cannot reliably be used in a persistent fashion.

share|improve this answer

edited Sep 3 at 9:12

answered Sep 3 at 9:09

Angew

127k11236333

X::value_type is no different from any other type alias (aka typedef) in this regard — C++ has no native way of converting a type to its source code string representation (not the least because that could be ambiguous). What you can do is use std::type_info::name:

cout << typeid(decltype(v)::value_type).name() << endl;

The resulting text is compiler dependent (and not even guaranteed to be easily human readable). It will be consistent within the same build of a program, but you cannot expect it to be the same across different compilers. In other words, it's useful for debugging, but cannot reliably be used in a persistent fashion.

share|improve this answer

edited Sep 3 at 9:12

answered Sep 3 at 9:09

Angew

127k11236333

share|improve this answer

share|improve this answer

edited Sep 3 at 9:12

edited Sep 3 at 9:12

edited Sep 3 at 9:12

answered Sep 3 at 9:09

Angew

127k11236333

answered Sep 3 at 9:09

Angew

127k11236333

answered Sep 3 at 9:09

Angew

127k11236333

127k11236333

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you, on my machine it gives me NSt3__112basic_stringIcNS_11char_traitsIcEENS_9allocatorIcEEEE
  – Niakrais
  Sep 3 at 9:24
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @You can use __cxa_demangle to get a readable type name: stackoverflow.com/a/23266701/1968
  – Konrad Rudolph
  Sep 3 at 9:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Isn't it only for gcc?
  – Niakrais
  Sep 3 at 9:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Niakrais: Not quite, it's for all compilers/platforms combinations using the Itanium ABI. The Itanium ABI is used on all Linux/MacOS platforms by all mainstream C++ compilers; the only widespread platform not using it is Windows, and thus MSVC and Clang in MSVC-compatible mode will not use it. Even on Windows, using Cygwin or MinGW means using the Itanium ABI.
  – Matthieu M.
  Sep 3 at 11:14
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you, on my machine it gives me NSt3__112basic_stringIcNS_11char_traitsIcEENS_9allocatorIcEEEE
  – Niakrais
  Sep 3 at 9:24
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @You can use __cxa_demangle to get a readable type name: stackoverflow.com/a/23266701/1968
  – Konrad Rudolph
  Sep 3 at 9:26
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Isn't it only for gcc?
  – Niakrais
  Sep 3 at 9:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Niakrais: Not quite, it's for all compilers/platforms combinations using the Itanium ABI. The Itanium ABI is used on all Linux/MacOS platforms by all mainstream C++ compilers; the only widespread platform not using it is Windows, and thus MSVC and Clang in MSVC-compatible mode will not use it. Even on Windows, using Cygwin or MinGW means using the Itanium ABI.
  – Matthieu M.
  Sep 3 at 11:14
  

  
  
  
  

Thank you, on my machine it gives me NSt3__112basic_stringIcNS_11char_traitsIcEENS_9allocatorIcEEEE
– Niakrais
Sep 3 at 9:24

Thank you, on my machine it gives me NSt3__112basic_stringIcNS_11char_traitsIcEENS_9allocatorIcEEEE
– Niakrais
Sep 3 at 9:24

1

1

@You can use __cxa_demangle to get a readable type name: stackoverflow.com/a/23266701/1968
– Konrad Rudolph
Sep 3 at 9:26

@You can use __cxa_demangle to get a readable type name: stackoverflow.com/a/23266701/1968
– Konrad Rudolph
Sep 3 at 9:26

Isn't it only for gcc?
– Niakrais
Sep 3 at 9:30

Isn't it only for gcc?
– Niakrais
Sep 3 at 9:30

@Niakrais: Not quite, it's for all compilers/platforms combinations using the Itanium ABI. The Itanium ABI is used on all Linux/MacOS platforms by all mainstream C++ compilers; the only widespread platform not using it is Windows, and thus MSVC and Clang in MSVC-compatible mode will not use it. Even on Windows, using Cygwin or MinGW means using the Itanium ABI.
– Matthieu M.
Sep 3 at 11:14

@Niakrais: Not quite, it's for all compilers/platforms combinations using the Itanium ABI. The Itanium ABI is used on all Linux/MacOS platforms by all mainstream C++ compilers; the only widespread platform not using it is Windows, and thus MSVC and Clang in MSVC-compatible mode will not use it. Even on Windows, using Cygwin or MinGW means using the Itanium ABI.
– Matthieu M.
Sep 3 at 11:14

add a comment | 

up vote
5
down vote

In addition to the typeid-based answer, I see one further possibility using Boost like this:

#include <boost/type_index.hpp>

cout << boost::typeindex::type_id_with_cvr<decltype(v)::value_type>().pretty_name() << "n";

The advantage is a portable, human-readable name that includes const/volatile qualifications and is consistent across major compilers and systems. The output on my machine is

// when compiled with gcc:
std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >
// ... and with clang:
std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> >

Decide for yourself whether this output qualifies as "human-readable", but also compare it to the typeid(...).name() approach, which on my machine gives:

NSt3__112basic_stringIcNS_11char_traitsIcEENS_9allocatorIcEEEE

share|improve this answer

edited Sep 3 at 19:57

Peter Mortensen

12.9k1983111

answered Sep 3 at 9:12

lubgr

6,0701339

add a comment | 

up vote
5
down vote

In addition to the typeid-based answer, I see one further possibility using Boost like this:

#include <boost/type_index.hpp>

cout << boost::typeindex::type_id_with_cvr<decltype(v)::value_type>().pretty_name() << "n";

The advantage is a portable, human-readable name that includes const/volatile qualifications and is consistent across major compilers and systems. The output on my machine is

// when compiled with gcc:
std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >
// ... and with clang:
std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> >

Decide for yourself whether this output qualifies as "human-readable", but also compare it to the typeid(...).name() approach, which on my machine gives:

NSt3__112basic_stringIcNS_11char_traitsIcEENS_9allocatorIcEEEE

share|improve this answer

edited Sep 3 at 19:57

Peter Mortensen

12.9k1983111

answered Sep 3 at 9:12

lubgr

6,0701339

add a comment | 

up vote
5
down vote

up vote
5
down vote

In addition to the typeid-based answer, I see one further possibility using Boost like this:

#include <boost/type_index.hpp>

cout << boost::typeindex::type_id_with_cvr<decltype(v)::value_type>().pretty_name() << "n";

The advantage is a portable, human-readable name that includes const/volatile qualifications and is consistent across major compilers and systems. The output on my machine is

// when compiled with gcc:
std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >
// ... and with clang:
std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> >

Decide for yourself whether this output qualifies as "human-readable", but also compare it to the typeid(...).name() approach, which on my machine gives:

NSt3__112basic_stringIcNS_11char_traitsIcEENS_9allocatorIcEEEE

share|improve this answer

edited Sep 3 at 19:57

Peter Mortensen

12.9k1983111

answered Sep 3 at 9:12

lubgr

6,0701339

In addition to the typeid-based answer, I see one further possibility using Boost like this:

#include <boost/type_index.hpp>

cout << boost::typeindex::type_id_with_cvr<decltype(v)::value_type>().pretty_name() << "n";

The advantage is a portable, human-readable name that includes const/volatile qualifications and is consistent across major compilers and systems. The output on my machine is

// when compiled with gcc:
std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >
// ... and with clang:
std::__1::basic_string<char, std::__1::char_traits<char>, std::__1::allocator<char> >

Decide for yourself whether this output qualifies as "human-readable", but also compare it to the typeid(...).name() approach, which on my machine gives:

NSt3__112basic_stringIcNS_11char_traitsIcEENS_9allocatorIcEEEE

share|improve this answer

edited Sep 3 at 19:57

Peter Mortensen

12.9k1983111

answered Sep 3 at 9:12

lubgr

6,0701339

share|improve this answer

share|improve this answer

edited Sep 3 at 19:57

Peter Mortensen

12.9k1983111

edited Sep 3 at 19:57

Peter Mortensen

12.9k1983111

edited Sep 3 at 19:57

Peter Mortensen

12.9k1983111

12.9k1983111

answered Sep 3 at 9:12

lubgr

6,0701339

answered Sep 3 at 9:12

lubgr

6,0701339

answered Sep 3 at 9:12

lubgr

6,0701339

6,0701339

add a comment | 

add a comment | 

up vote
1
down vote

It's probably safe to assume that you need this type info for debug purposes(?).
If it is, don't forget gdb's built in whatis and ptype commands:

$ cat main.cpp
#include <iostream>
#include <vector>
using namespace std;
int main() 
 vector<string> v = "apple", "facebook", "microsoft", "google" ;
 cout << v[2] << endl;
 // cout << v.value_type << endl;
 return 0;

$ cat gdbCommands.txt
break main
run
next
next
whatis v
quit
$ g++ -ggdb -o main main.cpp
$ gdb -x gdbCommands.txt ./main
GNU gdb (Ubuntu 8.0.1-0ubuntu1) 8.0.1
// bla bla bla ...
type = std::vector<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >>

share|improve this answer

answered Sep 3 at 10:13

OrenIshShalom

845520

add a comment | 

up vote
1
down vote

It's probably safe to assume that you need this type info for debug purposes(?).
If it is, don't forget gdb's built in whatis and ptype commands:

$ cat main.cpp
#include <iostream>
#include <vector>
using namespace std;
int main() 
 vector<string> v = "apple", "facebook", "microsoft", "google" ;
 cout << v[2] << endl;
 // cout << v.value_type << endl;
 return 0;

$ cat gdbCommands.txt
break main
run
next
next
whatis v
quit
$ g++ -ggdb -o main main.cpp
$ gdb -x gdbCommands.txt ./main
GNU gdb (Ubuntu 8.0.1-0ubuntu1) 8.0.1
// bla bla bla ...
type = std::vector<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >>

share|improve this answer

answered Sep 3 at 10:13

OrenIshShalom

845520

add a comment | 

up vote
1
down vote

up vote
1
down vote

It's probably safe to assume that you need this type info for debug purposes(?).
If it is, don't forget gdb's built in whatis and ptype commands:

$ cat main.cpp
#include <iostream>
#include <vector>
using namespace std;
int main() 
 vector<string> v = "apple", "facebook", "microsoft", "google" ;
 cout << v[2] << endl;
 // cout << v.value_type << endl;
 return 0;

$ cat gdbCommands.txt
break main
run
next
next
whatis v
quit
$ g++ -ggdb -o main main.cpp
$ gdb -x gdbCommands.txt ./main
GNU gdb (Ubuntu 8.0.1-0ubuntu1) 8.0.1
// bla bla bla ...
type = std::vector<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >>

share|improve this answer

answered Sep 3 at 10:13

OrenIshShalom

845520

It's probably safe to assume that you need this type info for debug purposes(?).
If it is, don't forget gdb's built in whatis and ptype commands:

$ cat main.cpp
#include <iostream>
#include <vector>
using namespace std;
int main() 
 vector<string> v = "apple", "facebook", "microsoft", "google" ;
 cout << v[2] << endl;
 // cout << v.value_type << endl;
 return 0;

$ cat gdbCommands.txt
break main
run
next
next
whatis v
quit
$ g++ -ggdb -o main main.cpp
$ gdb -x gdbCommands.txt ./main
GNU gdb (Ubuntu 8.0.1-0ubuntu1) 8.0.1
// bla bla bla ...
type = std::vector<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >>

share|improve this answer

answered Sep 3 at 10:13

OrenIshShalom

845520

share|improve this answer

share|improve this answer

answered Sep 3 at 10:13

OrenIshShalom

845520

answered Sep 3 at 10:13

OrenIshShalom

845520

answered Sep 3 at 10:13

OrenIshShalom

845520

845520

add a comment | 

add a comment | 

up vote
1
down vote

I had a rather similar question some time ago. The answer there shown a function able to translate a type into a human readable string:

template <typename T> std::string TypeName() 
 auto name = typeid(T()).name(); // function type, not a constructor call!
 int status = 0;

 std::unique_ptr<char, void(*)(void*)> res 
 abi::__cxa_demangle(name, NULL, NULL, &status),
 std::free
 ;

 std::string ret((status == 0) ? res.get() : name);
 if (ret.substr(ret.size() - 3) == " ()") ret.resize(ret.size() - 3);
 return ret;

Once you have this TypeName function you can achieve your goal:

int main() 
 vector<string> v = "apple", "facebook", "microsoft", "google" ;
 cout << v << endl;
 cout << TypeName<v.value_type>() << endl;
 return 0;

Which should output (GCC HEAD 9 201809):

std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >

share|improve this answer

answered Sep 3 at 13:55

Paula_plus_plus

6,58442758

add a comment | 

up vote
1
down vote

I had a rather similar question some time ago. The answer there shown a function able to translate a type into a human readable string:

template <typename T> std::string TypeName() 
 auto name = typeid(T()).name(); // function type, not a constructor call!
 int status = 0;

 std::unique_ptr<char, void(*)(void*)> res 
 abi::__cxa_demangle(name, NULL, NULL, &status),
 std::free
 ;

 std::string ret((status == 0) ? res.get() : name);
 if (ret.substr(ret.size() - 3) == " ()") ret.resize(ret.size() - 3);
 return ret;

Once you have this TypeName function you can achieve your goal:

int main() 
 vector<string> v = "apple", "facebook", "microsoft", "google" ;
 cout << v << endl;
 cout << TypeName<v.value_type>() << endl;
 return 0;

Which should output (GCC HEAD 9 201809):

std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >

share|improve this answer

answered Sep 3 at 13:55

Paula_plus_plus

6,58442758

add a comment | 

up vote
1
down vote

up vote
1
down vote

I had a rather similar question some time ago. The answer there shown a function able to translate a type into a human readable string:

template <typename T> std::string TypeName() 
 auto name = typeid(T()).name(); // function type, not a constructor call!
 int status = 0;

 std::unique_ptr<char, void(*)(void*)> res 
 abi::__cxa_demangle(name, NULL, NULL, &status),
 std::free
 ;

 std::string ret((status == 0) ? res.get() : name);
 if (ret.substr(ret.size() - 3) == " ()") ret.resize(ret.size() - 3);
 return ret;

Once you have this TypeName function you can achieve your goal:

int main() 
 vector<string> v = "apple", "facebook", "microsoft", "google" ;
 cout << v << endl;
 cout << TypeName<v.value_type>() << endl;
 return 0;

Which should output (GCC HEAD 9 201809):

std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >

share|improve this answer

answered Sep 3 at 13:55

Paula_plus_plus

6,58442758

I had a rather similar question some time ago. The answer there shown a function able to translate a type into a human readable string:

template <typename T> std::string TypeName() 
 auto name = typeid(T()).name(); // function type, not a constructor call!
 int status = 0;

 std::unique_ptr<char, void(*)(void*)> res 
 abi::__cxa_demangle(name, NULL, NULL, &status),
 std::free
 ;

 std::string ret((status == 0) ? res.get() : name);
 if (ret.substr(ret.size() - 3) == " ()") ret.resize(ret.size() - 3);
 return ret;

Once you have this TypeName function you can achieve your goal:

int main() 
 vector<string> v = "apple", "facebook", "microsoft", "google" ;
 cout << v << endl;
 cout << TypeName<v.value_type>() << endl;
 return 0;

Which should output (GCC HEAD 9 201809):

std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >

share|improve this answer

answered Sep 3 at 13:55

Paula_plus_plus

6,58442758

share|improve this answer

share|improve this answer

answered Sep 3 at 13:55

Paula_plus_plus

6,58442758

answered Sep 3 at 13:55

Paula_plus_plus

6,58442758

answered Sep 3 at 13:55

Paula_plus_plus

6,58442758

6,58442758

add a comment | 

add a comment | 

up vote
1
down vote

If it's for debug purpose, then this solution would work without any dependencies:

#include <regex>

template <typename T>
struct TypeToNameT

 static std::string getTypeFromName() 
#ifdef _MSC_VER
 std::regex re("<(.*)>::.*");
 std::string templateType(__FUNCTION__);
#else
 std::regex re(" = (.*);");
 std::string templateType(__PRETTY_FUNCTION__);
#endif
 std::smatch match;
 try
 
 if (std::regex_search(templateType, match, re) && match.size() > 1)
 return match.str(1);
 
 catch (std::regex_error& e) 
 // Syntax error in the regular expression
 
 return "";
 
;
/** Get the templated type name. This does not depends on RTTI, but on the preprocessor, so it should be quite safe to use even on old compilers */
template <typename T>
std::string getTypeName() return TypeToNameT<T>::getTypeFromName(); 


int main() 
 std::vector<std::string> v = "apple", "facebook", "microsoft", "google" ;
 std::cout << getTypeName<decltype(v)::value_type>() << std::endl; // Returns: "std::__cxx11::basic_string<char>"
 return 0;

Unlike the typeid based answer, this use the preprocessor and as such is not subject to name mangling (so the name is... readable).

Please notice that you can't use v.value_type as a type directly, it's not part of the instance v but of the type of v: std::vector<std::string>

The regex part is because the std::string class is too dumb and you need such oddities for a simple search/split code. If you have a decent string class, you'll not need regex here for extracting the part that's after the "=" string and before the ";".

share|improve this answer

answered Sep 3 at 14:14

xryl669

1,4301327

• 
  
  
  

  
  

  
  
  
  
  
  

  
  __FUNCTION__ prints just the function name, without any additional information. You want __FUNCSIG__.
  – HolyBlackCat
  Sep 3 at 16:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Just tried it here: rextester.com/FDZT51977 and it seems to work with FUNCTION. With FUNCSIG, the regex must be modified since the signature is more complex. In all cases, thanks, I did not know about FUNCSIG before.
  – xryl669
  Sep 3 at 16:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nevermind, you're right. __FUNCTION__ doesn't contain template parameters of the function itself, so when I used this trick before, I resorted to use __FUNCSIG__. But it missed me that it contains the template parameters of enclosing classes. That's a nice idea.
  – HolyBlackCat
  Sep 3 at 16:33
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

If it's for debug purpose, then this solution would work without any dependencies:

#include <regex>

template <typename T>
struct TypeToNameT

 static std::string getTypeFromName() 
#ifdef _MSC_VER
 std::regex re("<(.*)>::.*");
 std::string templateType(__FUNCTION__);
#else
 std::regex re(" = (.*);");
 std::string templateType(__PRETTY_FUNCTION__);
#endif
 std::smatch match;
 try
 
 if (std::regex_search(templateType, match, re) && match.size() > 1)
 return match.str(1);
 
 catch (std::regex_error& e) 
 // Syntax error in the regular expression
 
 return "";
 
;
/** Get the templated type name. This does not depends on RTTI, but on the preprocessor, so it should be quite safe to use even on old compilers */
template <typename T>
std::string getTypeName() return TypeToNameT<T>::getTypeFromName(); 


int main() 
 std::vector<std::string> v = "apple", "facebook", "microsoft", "google" ;
 std::cout << getTypeName<decltype(v)::value_type>() << std::endl; // Returns: "std::__cxx11::basic_string<char>"
 return 0;

Unlike the typeid based answer, this use the preprocessor and as such is not subject to name mangling (so the name is... readable).

Please notice that you can't use v.value_type as a type directly, it's not part of the instance v but of the type of v: std::vector<std::string>

The regex part is because the std::string class is too dumb and you need such oddities for a simple search/split code. If you have a decent string class, you'll not need regex here for extracting the part that's after the "=" string and before the ";".

share|improve this answer

answered Sep 3 at 14:14

xryl669

1,4301327

• 
  
  
  

  
  

  
  
  
  
  
  

  
  __FUNCTION__ prints just the function name, without any additional information. You want __FUNCSIG__.
  – HolyBlackCat
  Sep 3 at 16:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Just tried it here: rextester.com/FDZT51977 and it seems to work with FUNCTION. With FUNCSIG, the regex must be modified since the signature is more complex. In all cases, thanks, I did not know about FUNCSIG before.
  – xryl669
  Sep 3 at 16:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nevermind, you're right. __FUNCTION__ doesn't contain template parameters of the function itself, so when I used this trick before, I resorted to use __FUNCSIG__. But it missed me that it contains the template parameters of enclosing classes. That's a nice idea.
  – HolyBlackCat
  Sep 3 at 16:33
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

If it's for debug purpose, then this solution would work without any dependencies:

#include <regex>

template <typename T>
struct TypeToNameT

 static std::string getTypeFromName() 
#ifdef _MSC_VER
 std::regex re("<(.*)>::.*");
 std::string templateType(__FUNCTION__);
#else
 std::regex re(" = (.*);");
 std::string templateType(__PRETTY_FUNCTION__);
#endif
 std::smatch match;
 try
 
 if (std::regex_search(templateType, match, re) && match.size() > 1)
 return match.str(1);
 
 catch (std::regex_error& e) 
 // Syntax error in the regular expression
 
 return "";
 
;
/** Get the templated type name. This does not depends on RTTI, but on the preprocessor, so it should be quite safe to use even on old compilers */
template <typename T>
std::string getTypeName() return TypeToNameT<T>::getTypeFromName(); 


int main() 
 std::vector<std::string> v = "apple", "facebook", "microsoft", "google" ;
 std::cout << getTypeName<decltype(v)::value_type>() << std::endl; // Returns: "std::__cxx11::basic_string<char>"
 return 0;

Unlike the typeid based answer, this use the preprocessor and as such is not subject to name mangling (so the name is... readable).

Please notice that you can't use v.value_type as a type directly, it's not part of the instance v but of the type of v: std::vector<std::string>

The regex part is because the std::string class is too dumb and you need such oddities for a simple search/split code. If you have a decent string class, you'll not need regex here for extracting the part that's after the "=" string and before the ";".

share|improve this answer

answered Sep 3 at 14:14

xryl669

1,4301327

If it's for debug purpose, then this solution would work without any dependencies:

#include <regex>

template <typename T>
struct TypeToNameT

 static std::string getTypeFromName() 
#ifdef _MSC_VER
 std::regex re("<(.*)>::.*");
 std::string templateType(__FUNCTION__);
#else
 std::regex re(" = (.*);");
 std::string templateType(__PRETTY_FUNCTION__);
#endif
 std::smatch match;
 try
 
 if (std::regex_search(templateType, match, re) && match.size() > 1)
 return match.str(1);
 
 catch (std::regex_error& e) 
 // Syntax error in the regular expression
 
 return "";
 
;
/** Get the templated type name. This does not depends on RTTI, but on the preprocessor, so it should be quite safe to use even on old compilers */
template <typename T>
std::string getTypeName() return TypeToNameT<T>::getTypeFromName(); 


int main() 
 std::vector<std::string> v = "apple", "facebook", "microsoft", "google" ;
 std::cout << getTypeName<decltype(v)::value_type>() << std::endl; // Returns: "std::__cxx11::basic_string<char>"
 return 0;

Unlike the typeid based answer, this use the preprocessor and as such is not subject to name mangling (so the name is... readable).

Please notice that you can't use v.value_type as a type directly, it's not part of the instance v but of the type of v: std::vector<std::string>

The regex part is because the std::string class is too dumb and you need such oddities for a simple search/split code. If you have a decent string class, you'll not need regex here for extracting the part that's after the "=" string and before the ";".

share|improve this answer

answered Sep 3 at 14:14

xryl669

1,4301327

share|improve this answer

share|improve this answer

answered Sep 3 at 14:14

xryl669

1,4301327

answered Sep 3 at 14:14

xryl669

1,4301327

answered Sep 3 at 14:14

xryl669

1,4301327

1,4301327

• 
  
  
  

  
  

  
  
  
  
  
  

  
  __FUNCTION__ prints just the function name, without any additional information. You want __FUNCSIG__.
  – HolyBlackCat
  Sep 3 at 16:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Just tried it here: rextester.com/FDZT51977 and it seems to work with FUNCTION. With FUNCSIG, the regex must be modified since the signature is more complex. In all cases, thanks, I did not know about FUNCSIG before.
  – xryl669
  Sep 3 at 16:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nevermind, you're right. __FUNCTION__ doesn't contain template parameters of the function itself, so when I used this trick before, I resorted to use __FUNCSIG__. But it missed me that it contains the template parameters of enclosing classes. That's a nice idea.
  – HolyBlackCat
  Sep 3 at 16:33
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  __FUNCTION__ prints just the function name, without any additional information. You want __FUNCSIG__.
  – HolyBlackCat
  Sep 3 at 16:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Just tried it here: rextester.com/FDZT51977 and it seems to work with FUNCTION. With FUNCSIG, the regex must be modified since the signature is more complex. In all cases, thanks, I did not know about FUNCSIG before.
  – xryl669
  Sep 3 at 16:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Nevermind, you're right. __FUNCTION__ doesn't contain template parameters of the function itself, so when I used this trick before, I resorted to use __FUNCSIG__. But it missed me that it contains the template parameters of enclosing classes. That's a nice idea.
  – HolyBlackCat
  Sep 3 at 16:33
  
  

  
  
  
  

__FUNCTION__ prints just the function name, without any additional information. You want __FUNCSIG__.
– HolyBlackCat
Sep 3 at 16:25

__FUNCTION__ prints just the function name, without any additional information. You want __FUNCSIG__.
– HolyBlackCat
Sep 3 at 16:25

Just tried it here: rextester.com/FDZT51977 and it seems to work with FUNCTION. With FUNCSIG, the regex must be modified since the signature is more complex. In all cases, thanks, I did not know about FUNCSIG before.
– xryl669
Sep 3 at 16:30

Just tried it here: rextester.com/FDZT51977 and it seems to work with FUNCTION. With FUNCSIG, the regex must be modified since the signature is more complex. In all cases, thanks, I did not know about FUNCSIG before.
– xryl669
Sep 3 at 16:30

Nevermind, you're right. __FUNCTION__ doesn't contain template parameters of the function itself, so when I used this trick before, I resorted to use __FUNCSIG__. But it missed me that it contains the template parameters of enclosing classes. That's a nice idea.
– HolyBlackCat
Sep 3 at 16:33

Nevermind, you're right. __FUNCTION__ doesn't contain template parameters of the function itself, so when I used this trick before, I resorted to use __FUNCSIG__. But it missed me that it contains the template parameters of enclosing classes. That's a nice idea.
– HolyBlackCat
Sep 3 at 16:33

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