Why does the implicit type conversion not work in template deduction?

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In the following code, I want to call a template function by implicitly converting an int to a Scalar<int> object.

#include<iostream>
using namespace std;

template<typename Dtype>
class Scalar
public:
 Scalar(Dtype v) : value_(v)
private:
 Dtype value_;
;

template<typename Dtype>
void func(int a, Scalar<Dtype> b) 
 cout << "ok" <<endl;


int main()
 int a = 1;
 func(a, 2); 
 //int b = 2;
 //func(a, b);
 return 0;

Why does the template argument deduction/substitution fail? And the commented-codes are also wrong.

test.cpp: In function ‘int main()’:
test.cpp:19:12: error: no matching function for call to ‘func(int&, int)’
 func(a, 2);
 ^
test.cpp:19:12: note: candidate is:
test.cpp:13:6: note: template<class Dtype> void func(int, Scalar<Dtype>)
 void func(int a, Scalar<Dtype> b){
 ^
test.cpp:13:6: note: template argument deduction/substitution failed:
test.cpp:19:12: note: mismatched types ‘Scalar<Dtype>’ and ‘int’
 func(a, 2);

edited 1 hour ago

YSC

18.5k34388

asked 1 hour ago

Yulong Ao

441617

3

Possible duplicate of C++ implicit type conversion with template
– jwismar
1 hour ago

add a comment |

up vote
10
down vote

favorite

In the following code, I want to call a template function by implicitly converting an int to a Scalar<int> object.

#include<iostream>
using namespace std;

template<typename Dtype>
class Scalar
public:
 Scalar(Dtype v) : value_(v)
private:
 Dtype value_;
;

template<typename Dtype>
void func(int a, Scalar<Dtype> b) 
 cout << "ok" <<endl;


int main()
 int a = 1;
 func(a, 2); 
 //int b = 2;
 //func(a, b);
 return 0;

Why does the template argument deduction/substitution fail? And the commented-codes are also wrong.

test.cpp: In function ‘int main()’:
test.cpp:19:12: error: no matching function for call to ‘func(int&, int)’
 func(a, 2);
 ^
test.cpp:19:12: note: candidate is:
test.cpp:13:6: note: template<class Dtype> void func(int, Scalar<Dtype>)
 void func(int a, Scalar<Dtype> b){
 ^
test.cpp:13:6: note: template argument deduction/substitution failed:
test.cpp:19:12: note: mismatched types ‘Scalar<Dtype>’ and ‘int’
 func(a, 2);

edited 1 hour ago

YSC

18.5k34388

asked 1 hour ago

Yulong Ao

441617

3

Possible duplicate of C++ implicit type conversion with template
– jwismar
1 hour ago

add a comment |

up vote
10
down vote

favorite

In the following code, I want to call a template function by implicitly converting an int to a Scalar<int> object.

#include<iostream>
using namespace std;

template<typename Dtype>
class Scalar
public:
 Scalar(Dtype v) : value_(v)
private:
 Dtype value_;
;

template<typename Dtype>
void func(int a, Scalar<Dtype> b) 
 cout << "ok" <<endl;


int main()
 int a = 1;
 func(a, 2); 
 //int b = 2;
 //func(a, b);
 return 0;

Why does the template argument deduction/substitution fail? And the commented-codes are also wrong.

test.cpp: In function ‘int main()’:
test.cpp:19:12: error: no matching function for call to ‘func(int&, int)’
 func(a, 2);
 ^
test.cpp:19:12: note: candidate is:
test.cpp:13:6: note: template<class Dtype> void func(int, Scalar<Dtype>)
 void func(int a, Scalar<Dtype> b){
 ^
test.cpp:13:6: note: template argument deduction/substitution failed:
test.cpp:19:12: note: mismatched types ‘Scalar<Dtype>’ and ‘int’
 func(a, 2);

edited 1 hour ago

YSC

18.5k34388

asked 1 hour ago

Yulong Ao

441617

In the following code, I want to call a template function by implicitly converting an int to a Scalar<int> object.

#include<iostream>
using namespace std;

template<typename Dtype>
class Scalar
public:
 Scalar(Dtype v) : value_(v)
private:
 Dtype value_;
;

template<typename Dtype>
void func(int a, Scalar<Dtype> b) 
 cout << "ok" <<endl;


int main()
 int a = 1;
 func(a, 2); 
 //int b = 2;
 //func(a, b);
 return 0;

Why does the template argument deduction/substitution fail? And the commented-codes are also wrong.

test.cpp: In function ‘int main()’:
test.cpp:19:12: error: no matching function for call to ‘func(int&, int)’
 func(a, 2);
 ^
test.cpp:19:12: note: candidate is:
test.cpp:13:6: note: template<class Dtype> void func(int, Scalar<Dtype>)
 void func(int a, Scalar<Dtype> b){
 ^
test.cpp:13:6: note: template argument deduction/substitution failed:
test.cpp:19:12: note: mismatched types ‘Scalar<Dtype>’ and ‘int’
 func(a, 2);

c++ c++11 templates

edited 1 hour ago

YSC

18.5k34388

asked 1 hour ago

Yulong Ao

441617

edited 1 hour ago

YSC

18.5k34388

asked 1 hour ago

Yulong Ao

441617

edited 1 hour ago

YSC

18.5k34388

edited 1 hour ago

YSC

18.5k34388

edited 1 hour ago

YSC

18.5k34388

asked 1 hour ago

Yulong Ao

441617

asked 1 hour ago

Yulong Ao

441617

asked 1 hour ago

Yulong Ao

441617

3

Possible duplicate of C++ implicit type conversion with template
– jwismar
1 hour ago

add a comment |

3

Possible duplicate of C++ implicit type conversion with template
– jwismar
1 hour ago

Possible duplicate of C++ implicit type conversion with template
– jwismar
1 hour ago

add a comment |

2 Answers
2

active

oldest

votes

up vote
10
down vote

Because template argument deduction is not that smart: it does not (by design) consider user-defined conversions. And int -> Scalar<int> is a user-defined conversion.

If you want to use TAD, you need to convert your argument at the caller site:

func(a, Scalar<int>2);

or define a deduction guide¹ for Scalar and call f:

func(a, Scalar2); // C++17 only

Alternatively, you can explicitly instantiate f:

func<int>(a, 2);

¹⁾ The default deduction guide is sufficient: demo.

edited 1 hour ago

answered 1 hour ago

YSC

18.5k34388

I've tested all three of your solutions and they will work as is; however if the OP declared Scalar() as explicit the third one will not compile. Not sure if this is something worth mentioning or not, but might be something important for the OP to be aware of.
– Francis Cugler
1 hour ago

1

@FrancisCugler This is true ... but ... If OP did X, according to X, all, some, or none of my solutions would break.
– YSC
1 hour ago

add a comment |

up vote
1
down vote

template<typename Dtype>
void func(int a, Scalar<Dtype> b) 
 cout << "ok" <<endl;

template<typename Dtype>
void func(int a, Dtype b) 
 func(a, Scalar<Dtype>(std::move(b)));

template argument deduction is pattern matching, and it only matches the types or their base types exactly. It does no conversion.

Conversion is done later, at overload resolution & function call time.

Here, we add another overload that explicitly forwards to the one you want.

answered 55 mins ago

Yakk - Adam Nevraumont

176k19179361

add a comment |

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2 Answers
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active

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2 Answers
2

active

oldest

votes

up vote
10
down vote

Because template argument deduction is not that smart: it does not (by design) consider user-defined conversions. And int -> Scalar<int> is a user-defined conversion.

If you want to use TAD, you need to convert your argument at the caller site:

func(a, Scalar<int>2);

or define a deduction guide¹ for Scalar and call f:

func(a, Scalar2); // C++17 only

Alternatively, you can explicitly instantiate f:

func<int>(a, 2);

¹⁾ The default deduction guide is sufficient: demo.

edited 1 hour ago

answered 1 hour ago

YSC

18.5k34388

I've tested all three of your solutions and they will work as is; however if the OP declared Scalar() as explicit the third one will not compile. Not sure if this is something worth mentioning or not, but might be something important for the OP to be aware of.
– Francis Cugler
1 hour ago

1

@FrancisCugler This is true ... but ... If OP did X, according to X, all, some, or none of my solutions would break.
– YSC
1 hour ago

add a comment |

up vote
10
down vote

Because template argument deduction is not that smart: it does not (by design) consider user-defined conversions. And int -> Scalar<int> is a user-defined conversion.

If you want to use TAD, you need to convert your argument at the caller site:

func(a, Scalar<int>2);

or define a deduction guide¹ for Scalar and call f:

func(a, Scalar2); // C++17 only

Alternatively, you can explicitly instantiate f:

func<int>(a, 2);

¹⁾ The default deduction guide is sufficient: demo.

edited 1 hour ago

answered 1 hour ago

YSC

18.5k34388

I've tested all three of your solutions and they will work as is; however if the OP declared Scalar() as explicit the third one will not compile. Not sure if this is something worth mentioning or not, but might be something important for the OP to be aware of.
– Francis Cugler
1 hour ago

1

@FrancisCugler This is true ... but ... If OP did X, according to X, all, some, or none of my solutions would break.
– YSC
1 hour ago

add a comment |

up vote
10
down vote

Because template argument deduction is not that smart: it does not (by design) consider user-defined conversions. And int -> Scalar<int> is a user-defined conversion.

If you want to use TAD, you need to convert your argument at the caller site:

func(a, Scalar<int>2);

or define a deduction guide¹ for Scalar and call f:

func(a, Scalar2); // C++17 only

Alternatively, you can explicitly instantiate f:

func<int>(a, 2);

¹⁾ The default deduction guide is sufficient: demo.

edited 1 hour ago

answered 1 hour ago

YSC

18.5k34388

Because template argument deduction is not that smart: it does not (by design) consider user-defined conversions. And int -> Scalar<int> is a user-defined conversion.

If you want to use TAD, you need to convert your argument at the caller site:

func(a, Scalar<int>2);

or define a deduction guide¹ for Scalar and call f:

func(a, Scalar2); // C++17 only

Alternatively, you can explicitly instantiate f:

func<int>(a, 2);

¹⁾ The default deduction guide is sufficient: demo.

edited 1 hour ago

answered 1 hour ago

YSC

18.5k34388

edited 1 hour ago

answered 1 hour ago

YSC

18.5k34388

answered 1 hour ago

YSC

18.5k34388

answered 1 hour ago

YSC

18.5k34388

I've tested all three of your solutions and they will work as is; however if the OP declared Scalar() as explicit the third one will not compile. Not sure if this is something worth mentioning or not, but might be something important for the OP to be aware of.
– Francis Cugler
1 hour ago

1

@FrancisCugler This is true ... but ... If OP did X, according to X, all, some, or none of my solutions would break.
– YSC
1 hour ago

add a comment |

I've tested all three of your solutions and they will work as is; however if the OP declared Scalar() as explicit the third one will not compile. Not sure if this is something worth mentioning or not, but might be something important for the OP to be aware of.
– Francis Cugler
1 hour ago

1

@FrancisCugler This is true ... but ... If OP did X, according to X, all, some, or none of my solutions would break.
– YSC
1 hour ago

I've tested all three of your solutions and they will work as is; however if the OP declared Scalar() as explicit the third one will not compile. Not sure if this is something worth mentioning or not, but might be something important for the OP to be aware of.
– Francis Cugler
1 hour ago

@FrancisCugler This is true ... but ... If OP did X, according to X, all, some, or none of my solutions would break.
– YSC
1 hour ago

add a comment |

up vote
1
down vote

template<typename Dtype>
void func(int a, Scalar<Dtype> b) 
 cout << "ok" <<endl;

template<typename Dtype>
void func(int a, Dtype b) 
 func(a, Scalar<Dtype>(std::move(b)));

template argument deduction is pattern matching, and it only matches the types or their base types exactly. It does no conversion.

Conversion is done later, at overload resolution & function call time.

Here, we add another overload that explicitly forwards to the one you want.

answered 55 mins ago

Yakk - Adam Nevraumont

176k19179361

add a comment |

up vote
1
down vote

template<typename Dtype>
void func(int a, Scalar<Dtype> b) 
 cout << "ok" <<endl;

template<typename Dtype>
void func(int a, Dtype b) 
 func(a, Scalar<Dtype>(std::move(b)));

template argument deduction is pattern matching, and it only matches the types or their base types exactly. It does no conversion.

Conversion is done later, at overload resolution & function call time.

Here, we add another overload that explicitly forwards to the one you want.

answered 55 mins ago

Yakk - Adam Nevraumont

176k19179361

add a comment |

up vote
1
down vote

template<typename Dtype>
void func(int a, Scalar<Dtype> b) 
 cout << "ok" <<endl;

template<typename Dtype>
void func(int a, Dtype b) 
 func(a, Scalar<Dtype>(std::move(b)));

template argument deduction is pattern matching, and it only matches the types or their base types exactly. It does no conversion.

Conversion is done later, at overload resolution & function call time.

Here, we add another overload that explicitly forwards to the one you want.

answered 55 mins ago

Yakk - Adam Nevraumont

176k19179361

template<typename Dtype>
void func(int a, Scalar<Dtype> b) 
 cout << "ok" <<endl;

template<typename Dtype>
void func(int a, Dtype b) 
 func(a, Scalar<Dtype>(std::move(b)));

template argument deduction is pattern matching, and it only matches the types or their base types exactly. It does no conversion.

Conversion is done later, at overload resolution & function call time.

Here, we add another overload that explicitly forwards to the one you want.

answered 55 mins ago

Yakk - Adam Nevraumont

176k19179361

answered 55 mins ago

Yakk - Adam Nevraumont

176k19179361

answered 55 mins ago

Yakk - Adam Nevraumont

176k19179361

answered 55 mins ago

Yakk - Adam Nevraumont

176k19179361

add a comment |

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