Do we still need to write the empty angle brackets when using transparent std function objects?

up vote
14
down vote

favorite

With class template argument deduction we can write:

std::less Fn;

However, G++ 8.2 rejects this code:

#include <algorithm>
#include <vector>
#include <functional>

int main()

std::vector v= 1, 3, 2, 7, 5, 4 ;

std::sort(v.begin(),v.end(),std::greater());

emitting the following error:

error: cannot deduce template arguments for 'greater' from ()

Clang++ 7.0 and MSVC 15.8.0 compile it without warnings. Which compiler is right?

edited 36 mins ago

xskxzr

5,38081950

asked 1 hour ago

metalfox

1,344215

5

interesting gcc compiles with std::greater
â€“Â bolov
59 mins ago

@bolov I have added the missing header
â€“Â metalfox
57 mins ago

add a commentÂ |Â

up vote
14
down vote

favorite

With class template argument deduction we can write:

std::less Fn;

However, G++ 8.2 rejects this code:

#include <algorithm>
#include <vector>
#include <functional>

int main()

std::vector v= 1, 3, 2, 7, 5, 4 ;

std::sort(v.begin(),v.end(),std::greater());

emitting the following error:

error: cannot deduce template arguments for 'greater' from ()

Clang++ 7.0 and MSVC 15.8.0 compile it without warnings. Which compiler is right?

edited 36 mins ago

xskxzr

5,38081950

asked 1 hour ago

metalfox

1,344215

5

interesting gcc compiles with std::greater
â€“Â bolov
59 mins ago

@bolov I have added the missing header
â€“Â metalfox
57 mins ago

add a commentÂ |Â

up vote
14
down vote

favorite

With class template argument deduction we can write:

std::less Fn;

However, G++ 8.2 rejects this code:

#include <algorithm>
#include <vector>
#include <functional>

int main()

std::vector v= 1, 3, 2, 7, 5, 4 ;

std::sort(v.begin(),v.end(),std::greater());

emitting the following error:

error: cannot deduce template arguments for 'greater' from ()

Clang++ 7.0 and MSVC 15.8.0 compile it without warnings. Which compiler is right?

edited 36 mins ago

xskxzr

5,38081950

asked 1 hour ago

metalfox

1,344215

With class template argument deduction we can write:

std::less Fn;

However, G++ 8.2 rejects this code:

#include <algorithm>
#include <vector>
#include <functional>

int main()

std::vector v= 1, 3, 2, 7, 5, 4 ;

std::sort(v.begin(),v.end(),std::greater());

emitting the following error:

error: cannot deduce template arguments for 'greater' from ()

Clang++ 7.0 and MSVC 15.8.0 compile it without warnings. Which compiler is right?

c++ templates language-lawyer c++17 argument-deduction

edited 36 mins ago

xskxzr

5,38081950

asked 1 hour ago

metalfox

1,344215

edited 36 mins ago

xskxzr

5,38081950

asked 1 hour ago

metalfox

1,344215

edited 36 mins ago

xskxzr

5,38081950

edited 36 mins ago

xskxzr

5,38081950

edited 36 mins ago

xskxzr

5,38081950

asked 1 hour ago

metalfox

1,344215

asked 1 hour ago

metalfox

1,344215

asked 1 hour ago

metalfox

1,344215

5

interesting gcc compiles with std::greater
â€“Â bolov
59 mins ago

@bolov I have added the missing header
â€“Â metalfox
57 mins ago

add a commentÂ |Â

5

interesting gcc compiles with std::greater
â€“Â bolov
59 mins ago

@bolov I have added the missing header
â€“Â metalfox
57 mins ago

interesting gcc compiles with std::greater
â€“Â bolov
59 mins ago

@bolov I have added the missing header
â€“Â metalfox
57 mins ago

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
8
down vote

accepted

GCC is wrong. There is already a bug report.

[dcl.type.simple]/2 says:

A type-specifier of the form typename_optnested-name-specifier_opttemplate-name is a placeholder for a deduced class type ([dcl.type.class.deduct]).

And [dcl.type.class.deduct]/2 says:

A placeholder for a deduced class type can also be used in the type-specifier-seq in the new-type-id or type-id of a new-expression, as the simple-type-specifier in an explicit type conversion (functional notation) ([expr.type.conv]), or as the type-specifier in the parameter-declaration of a template-parameter. A placeholder for a deduced class type shall not appear in any other context.

Such use is allowed.

[temp.arg]/4 describes the syntax error that a template-id is required but there is no <>. However here std::greater is not resolved as a template-id so that paragraph does not apply.

edited 31 mins ago

answered 51 mins ago

xskxzr

5,38081950

Is this case an explicit type conversion (functional notation)?
â€“Â metalfox
43 mins ago

1

@metalfox Isn't it? [expr.type.conv]/1: "A simple-type-specifier or typename-specifier followed by a parenthesized optional expression-list or by a braced-init-list (the initializer) constructs a value of the specified type given the initializer. If the type is a placeholder for a deduced class type, it is replaced by the return type of the function selected by overload resolution for class template deduction for the remainder of this subclause."
â€“Â xskxzr
38 mins ago

I see. I didn't think of it as a type conversion
â€“Â metalfox
11 mins ago

add a commentÂ |Â

up vote
7
down vote

Clang and MSVC are correct. This should be well-formed because of the combination effect of implicitly-generated deduction guides (since C++17) and default template argument.

(emphasis mine)

When a function-style cast or declaration of a variable uses the name
of a primary class template C without an argument list as the type
specifier, deduction will proceed as follows:

If C is defined, for each constructor (or constructor template) Ci declared in the named primary template (if it is defined), a fictional
function template Fi, is constructed, such that

template parameters of Fi are the template parameters of C followed (if Ci is a constructor template) by the template parameters
of Ci (default template arguments are included too)

the function parameters of Fi are the constructor parameters

the return type of Fi is C followed by the template parameters of the class template enclosed in <>

If C is not defined or does not declare any constructors, an additional fictional function template is added, derived as above from
a hypothetical constructor C()

In any case, an additional fictional function template derived as above from a hypothetical constructor C(C) is added, called the copy
deduction candidate.

Template argument deduction and overload resolution is then performed
for initialization of a fictional object of hypothetical class type,
whose constructor signatures match the guides (except for return type)
for the purpose of forming an overload set, and the initializer is
provided by the context in which class template argument deduction was
performed, except that the first phase of list-initialization
(considering initializer-list constructors) is omitted if the
initializer list consists of a single expression of type (possibly
cv-qualified) U, where U is a specialization of C or a class derived
from a specialization of C.

These fictional constructors are public members of the hypothetical
class type. They are explicit if the guide was formed from an explicit
constructor. If overload resolution fails, the program is ill-formed.
Otherwise, the return type of the selected F template specialization
becomes the deduced class template specialization.

Given std::greater(), the implicitly-generated deduction guide is applied and the additional fictional function is selected at last. As the result of overload resolution the default argument void is applied, then the deduced type will be void. That means std::greater() should be same as writing std::greater<void>() or std::greater<>().

^{BTW: Gcc doesn't compile with std::greater(), but std::greater or std::greater g; are fine, it might be gcc's bug.}

edited 10 mins ago

answered 1 hour ago

songyuanyao

87.3k10169230

I'm not sure... std::less L compiles just fine. It is even shown as an example in cppreference
â€“Â metalfox
58 mins ago

@songyuanyao you are right. cppreference doesn't show neither a ctor, nor a deduction guide. Haven't checked the standard.
â€“Â bolov
54 mins ago

@metalfox Yes it's weird, std::greater g; is fine too. I'll check the standard further.
â€“Â songyuanyao
52 mins ago

@bolov I edited my answer, it seems your thought that it's relevent to deduction guide is correct.
â€“Â songyuanyao
19 mins ago

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
8
down vote

accepted

GCC is wrong. There is already a bug report.

[dcl.type.simple]/2 says:

A type-specifier of the form typename_optnested-name-specifier_opttemplate-name is a placeholder for a deduced class type ([dcl.type.class.deduct]).

And [dcl.type.class.deduct]/2 says:

A placeholder for a deduced class type can also be used in the type-specifier-seq in the new-type-id or type-id of a new-expression, as the simple-type-specifier in an explicit type conversion (functional notation) ([expr.type.conv]), or as the type-specifier in the parameter-declaration of a template-parameter. A placeholder for a deduced class type shall not appear in any other context.

Such use is allowed.

[temp.arg]/4 describes the syntax error that a template-id is required but there is no <>. However here std::greater is not resolved as a template-id so that paragraph does not apply.

edited 31 mins ago

answered 51 mins ago

xskxzr

5,38081950

Is this case an explicit type conversion (functional notation)?
â€“Â metalfox
43 mins ago

1

@metalfox Isn't it? [expr.type.conv]/1: "A simple-type-specifier or typename-specifier followed by a parenthesized optional expression-list or by a braced-init-list (the initializer) constructs a value of the specified type given the initializer. If the type is a placeholder for a deduced class type, it is replaced by the return type of the function selected by overload resolution for class template deduction for the remainder of this subclause."
â€“Â xskxzr
38 mins ago

I see. I didn't think of it as a type conversion
â€“Â metalfox
11 mins ago

add a commentÂ |Â

up vote
8
down vote

accepted

GCC is wrong. There is already a bug report.

[dcl.type.simple]/2 says:

A type-specifier of the form typename_optnested-name-specifier_opttemplate-name is a placeholder for a deduced class type ([dcl.type.class.deduct]).

And [dcl.type.class.deduct]/2 says:

A placeholder for a deduced class type can also be used in the type-specifier-seq in the new-type-id or type-id of a new-expression, as the simple-type-specifier in an explicit type conversion (functional notation) ([expr.type.conv]), or as the type-specifier in the parameter-declaration of a template-parameter. A placeholder for a deduced class type shall not appear in any other context.

Such use is allowed.

[temp.arg]/4 describes the syntax error that a template-id is required but there is no <>. However here std::greater is not resolved as a template-id so that paragraph does not apply.

edited 31 mins ago

answered 51 mins ago

xskxzr

5,38081950

Is this case an explicit type conversion (functional notation)?
â€“Â metalfox
43 mins ago

1

@metalfox Isn't it? [expr.type.conv]/1: "A simple-type-specifier or typename-specifier followed by a parenthesized optional expression-list or by a braced-init-list (the initializer) constructs a value of the specified type given the initializer. If the type is a placeholder for a deduced class type, it is replaced by the return type of the function selected by overload resolution for class template deduction for the remainder of this subclause."
â€“Â xskxzr
38 mins ago

I see. I didn't think of it as a type conversion
â€“Â metalfox
11 mins ago

add a commentÂ |Â

up vote
8
down vote

accepted

GCC is wrong. There is already a bug report.

[dcl.type.simple]/2 says:

A type-specifier of the form typename_optnested-name-specifier_opttemplate-name is a placeholder for a deduced class type ([dcl.type.class.deduct]).

And [dcl.type.class.deduct]/2 says:

A placeholder for a deduced class type can also be used in the type-specifier-seq in the new-type-id or type-id of a new-expression, as the simple-type-specifier in an explicit type conversion (functional notation) ([expr.type.conv]), or as the type-specifier in the parameter-declaration of a template-parameter. A placeholder for a deduced class type shall not appear in any other context.

Such use is allowed.

[temp.arg]/4 describes the syntax error that a template-id is required but there is no <>. However here std::greater is not resolved as a template-id so that paragraph does not apply.

edited 31 mins ago

answered 51 mins ago

xskxzr

5,38081950

GCC is wrong. There is already a bug report.

[dcl.type.simple]/2 says:

A type-specifier of the form typename_optnested-name-specifier_opttemplate-name is a placeholder for a deduced class type ([dcl.type.class.deduct]).

And [dcl.type.class.deduct]/2 says:

A placeholder for a deduced class type can also be used in the type-specifier-seq in the new-type-id or type-id of a new-expression, as the simple-type-specifier in an explicit type conversion (functional notation) ([expr.type.conv]), or as the type-specifier in the parameter-declaration of a template-parameter. A placeholder for a deduced class type shall not appear in any other context.

Such use is allowed.

[temp.arg]/4 describes the syntax error that a template-id is required but there is no <>. However here std::greater is not resolved as a template-id so that paragraph does not apply.

edited 31 mins ago

answered 51 mins ago

xskxzr

5,38081950

edited 31 mins ago

answered 51 mins ago

xskxzr

5,38081950

answered 51 mins ago

xskxzr

5,38081950

answered 51 mins ago

xskxzr

5,38081950

Is this case an explicit type conversion (functional notation)?
â€“Â metalfox
43 mins ago

1

@metalfox Isn't it? [expr.type.conv]/1: "A simple-type-specifier or typename-specifier followed by a parenthesized optional expression-list or by a braced-init-list (the initializer) constructs a value of the specified type given the initializer. If the type is a placeholder for a deduced class type, it is replaced by the return type of the function selected by overload resolution for class template deduction for the remainder of this subclause."
â€“Â xskxzr
38 mins ago

I see. I didn't think of it as a type conversion
â€“Â metalfox
11 mins ago

add a commentÂ |Â

Is this case an explicit type conversion (functional notation)?
â€“Â metalfox
43 mins ago

1

@metalfox Isn't it? [expr.type.conv]/1: "A simple-type-specifier or typename-specifier followed by a parenthesized optional expression-list or by a braced-init-list (the initializer) constructs a value of the specified type given the initializer. If the type is a placeholder for a deduced class type, it is replaced by the return type of the function selected by overload resolution for class template deduction for the remainder of this subclause."
â€“Â xskxzr
38 mins ago

I see. I didn't think of it as a type conversion
â€“Â metalfox
11 mins ago

Is this case an explicit type conversion (functional notation)?
â€“Â metalfox
43 mins ago

@metalfox Isn't it? [expr.type.conv]/1: "A simple-type-specifier or typename-specifier followed by a parenthesized optional expression-list or by a braced-init-list (the initializer) constructs a value of the specified type given the initializer. If the type is a placeholder for a deduced class type, it is replaced by the return type of the function selected by overload resolution for class template deduction for the remainder of this subclause."
â€“Â xskxzr
38 mins ago

I see. I didn't think of it as a type conversion
â€“Â metalfox
11 mins ago

add a commentÂ |Â

up vote
7
down vote

Clang and MSVC are correct. This should be well-formed because of the combination effect of implicitly-generated deduction guides (since C++17) and default template argument.

(emphasis mine)

When a function-style cast or declaration of a variable uses the name
of a primary class template C without an argument list as the type
specifier, deduction will proceed as follows:

If C is defined, for each constructor (or constructor template) Ci declared in the named primary template (if it is defined), a fictional
function template Fi, is constructed, such that

template parameters of Fi are the template parameters of C followed (if Ci is a constructor template) by the template parameters
of Ci (default template arguments are included too)

the function parameters of Fi are the constructor parameters

the return type of Fi is C followed by the template parameters of the class template enclosed in <>

If C is not defined or does not declare any constructors, an additional fictional function template is added, derived as above from
a hypothetical constructor C()

In any case, an additional fictional function template derived as above from a hypothetical constructor C(C) is added, called the copy
deduction candidate.

Template argument deduction and overload resolution is then performed
for initialization of a fictional object of hypothetical class type,
whose constructor signatures match the guides (except for return type)
for the purpose of forming an overload set, and the initializer is
provided by the context in which class template argument deduction was
performed, except that the first phase of list-initialization
(considering initializer-list constructors) is omitted if the
initializer list consists of a single expression of type (possibly
cv-qualified) U, where U is a specialization of C or a class derived
from a specialization of C.

These fictional constructors are public members of the hypothetical
class type. They are explicit if the guide was formed from an explicit
constructor. If overload resolution fails, the program is ill-formed.
Otherwise, the return type of the selected F template specialization
becomes the deduced class template specialization.

^{BTW: Gcc doesn't compile with std::greater(), but std::greater or std::greater g; are fine, it might be gcc's bug.}

edited 10 mins ago

answered 1 hour ago

songyuanyao

87.3k10169230

I'm not sure... std::less L compiles just fine. It is even shown as an example in cppreference
â€“Â metalfox
58 mins ago

@songyuanyao you are right. cppreference doesn't show neither a ctor, nor a deduction guide. Haven't checked the standard.
â€“Â bolov
54 mins ago

@metalfox Yes it's weird, std::greater g; is fine too. I'll check the standard further.
â€“Â songyuanyao
52 mins ago

@bolov I edited my answer, it seems your thought that it's relevent to deduction guide is correct.
â€“Â songyuanyao
19 mins ago

add a commentÂ |Â

up vote
7
down vote

Clang and MSVC are correct. This should be well-formed because of the combination effect of implicitly-generated deduction guides (since C++17) and default template argument.

(emphasis mine)

When a function-style cast or declaration of a variable uses the name
of a primary class template C without an argument list as the type
specifier, deduction will proceed as follows:

If C is defined, for each constructor (or constructor template) Ci declared in the named primary template (if it is defined), a fictional
function template Fi, is constructed, such that

template parameters of Fi are the template parameters of C followed (if Ci is a constructor template) by the template parameters
of Ci (default template arguments are included too)

the function parameters of Fi are the constructor parameters

the return type of Fi is C followed by the template parameters of the class template enclosed in <>

If C is not defined or does not declare any constructors, an additional fictional function template is added, derived as above from
a hypothetical constructor C()

In any case, an additional fictional function template derived as above from a hypothetical constructor C(C) is added, called the copy
deduction candidate.

Template argument deduction and overload resolution is then performed
for initialization of a fictional object of hypothetical class type,
whose constructor signatures match the guides (except for return type)
for the purpose of forming an overload set, and the initializer is
provided by the context in which class template argument deduction was
performed, except that the first phase of list-initialization
(considering initializer-list constructors) is omitted if the
initializer list consists of a single expression of type (possibly
cv-qualified) U, where U is a specialization of C or a class derived
from a specialization of C.

These fictional constructors are public members of the hypothetical
class type. They are explicit if the guide was formed from an explicit
constructor. If overload resolution fails, the program is ill-formed.
Otherwise, the return type of the selected F template specialization
becomes the deduced class template specialization.

^{BTW: Gcc doesn't compile with std::greater(), but std::greater or std::greater g; are fine, it might be gcc's bug.}

edited 10 mins ago

answered 1 hour ago

songyuanyao

87.3k10169230

I'm not sure... std::less L compiles just fine. It is even shown as an example in cppreference
â€“Â metalfox
58 mins ago

@songyuanyao you are right. cppreference doesn't show neither a ctor, nor a deduction guide. Haven't checked the standard.
â€“Â bolov
54 mins ago

@metalfox Yes it's weird, std::greater g; is fine too. I'll check the standard further.
â€“Â songyuanyao
52 mins ago

@bolov I edited my answer, it seems your thought that it's relevent to deduction guide is correct.
â€“Â songyuanyao
19 mins ago

add a commentÂ |Â

up vote
7
down vote

Clang and MSVC are correct. This should be well-formed because of the combination effect of implicitly-generated deduction guides (since C++17) and default template argument.

(emphasis mine)

When a function-style cast or declaration of a variable uses the name
of a primary class template C without an argument list as the type
specifier, deduction will proceed as follows:

If C is defined, for each constructor (or constructor template) Ci declared in the named primary template (if it is defined), a fictional
function template Fi, is constructed, such that

template parameters of Fi are the template parameters of C followed (if Ci is a constructor template) by the template parameters
of Ci (default template arguments are included too)

the function parameters of Fi are the constructor parameters

the return type of Fi is C followed by the template parameters of the class template enclosed in <>

If C is not defined or does not declare any constructors, an additional fictional function template is added, derived as above from
a hypothetical constructor C()

In any case, an additional fictional function template derived as above from a hypothetical constructor C(C) is added, called the copy
deduction candidate.

Template argument deduction and overload resolution is then performed
for initialization of a fictional object of hypothetical class type,
whose constructor signatures match the guides (except for return type)
for the purpose of forming an overload set, and the initializer is
provided by the context in which class template argument deduction was
performed, except that the first phase of list-initialization
(considering initializer-list constructors) is omitted if the
initializer list consists of a single expression of type (possibly
cv-qualified) U, where U is a specialization of C or a class derived
from a specialization of C.

These fictional constructors are public members of the hypothetical
class type. They are explicit if the guide was formed from an explicit
constructor. If overload resolution fails, the program is ill-formed.
Otherwise, the return type of the selected F template specialization
becomes the deduced class template specialization.

^{BTW: Gcc doesn't compile with std::greater(), but std::greater or std::greater g; are fine, it might be gcc's bug.}

edited 10 mins ago

answered 1 hour ago

songyuanyao

87.3k10169230

Clang and MSVC are correct. This should be well-formed because of the combination effect of implicitly-generated deduction guides (since C++17) and default template argument.

(emphasis mine)

When a function-style cast or declaration of a variable uses the name
of a primary class template C without an argument list as the type
specifier, deduction will proceed as follows:

If C is defined, for each constructor (or constructor template) Ci declared in the named primary template (if it is defined), a fictional
function template Fi, is constructed, such that

template parameters of Fi are the template parameters of C followed (if Ci is a constructor template) by the template parameters
of Ci (default template arguments are included too)

the function parameters of Fi are the constructor parameters

the return type of Fi is C followed by the template parameters of the class template enclosed in <>

If C is not defined or does not declare any constructors, an additional fictional function template is added, derived as above from
a hypothetical constructor C()

In any case, an additional fictional function template derived as above from a hypothetical constructor C(C) is added, called the copy
deduction candidate.

Template argument deduction and overload resolution is then performed
for initialization of a fictional object of hypothetical class type,
whose constructor signatures match the guides (except for return type)
for the purpose of forming an overload set, and the initializer is
provided by the context in which class template argument deduction was
performed, except that the first phase of list-initialization
(considering initializer-list constructors) is omitted if the
initializer list consists of a single expression of type (possibly
cv-qualified) U, where U is a specialization of C or a class derived
from a specialization of C.

These fictional constructors are public members of the hypothetical
class type. They are explicit if the guide was formed from an explicit
constructor. If overload resolution fails, the program is ill-formed.
Otherwise, the return type of the selected F template specialization
becomes the deduced class template specialization.

^{BTW: Gcc doesn't compile with std::greater(), but std::greater or std::greater g; are fine, it might be gcc's bug.}

edited 10 mins ago

answered 1 hour ago

songyuanyao

87.3k10169230

edited 10 mins ago

answered 1 hour ago

songyuanyao

87.3k10169230

answered 1 hour ago

songyuanyao

87.3k10169230

answered 1 hour ago

songyuanyao

87.3k10169230

I'm not sure... std::less L compiles just fine. It is even shown as an example in cppreference
â€“Â metalfox
58 mins ago

@songyuanyao you are right. cppreference doesn't show neither a ctor, nor a deduction guide. Haven't checked the standard.
â€“Â bolov
54 mins ago

@metalfox Yes it's weird, std::greater g; is fine too. I'll check the standard further.
â€“Â songyuanyao
52 mins ago

@bolov I edited my answer, it seems your thought that it's relevent to deduction guide is correct.
â€“Â songyuanyao
19 mins ago

add a commentÂ |Â

I'm not sure... std::less L compiles just fine. It is even shown as an example in cppreference
â€“Â metalfox
58 mins ago

@songyuanyao you are right. cppreference doesn't show neither a ctor, nor a deduction guide. Haven't checked the standard.
â€“Â bolov
54 mins ago

@metalfox Yes it's weird, std::greater g; is fine too. I'll check the standard further.
â€“Â songyuanyao
52 mins ago

@bolov I edited my answer, it seems your thought that it's relevent to deduction guide is correct.
â€“Â songyuanyao
19 mins ago

I'm not sure... std::less L compiles just fine. It is even shown as an example in cppreference
â€“Â metalfox
58 mins ago

@songyuanyao you are right. cppreference doesn't show neither a ctor, nor a deduction guide. Haven't checked the standard.
â€“Â bolov
54 mins ago

@metalfox Yes it's weird, std::greater g; is fine too. I'll check the standard further.
â€“Â songyuanyao
52 mins ago

@bolov I edited my answer, it seems your thought that it's relevent to deduction guide is correct.
â€“Â songyuanyao
19 mins ago

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