Why isn't this call to a template ambiguous?

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I declare two templates, the first converts the argument x from type T to type U and the second from type U to type T. If I call cast with 10, the compiler does not complain. I think both meet the requirements to be used and therefore there should be ambiguity, is that true? This code prints 10.

#include <iostream>

template<typename T, typename U>
U cast(T x) 
 return static_cast<U>(x);


template<typename T, typename U>
T cast(U x) 
 return static_cast<T>(x);


int main() 
 std::cout << cast<int,float>(10) << 'n';

edited 12 mins ago

Deduplicator

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wic

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If you change your code to cast<int,float>(10.0f) the second overload would be called. And if you changed it to cast<int,float>(10.0) the call would be ambiguous.
â€“Â Praetorian
5 hours ago

add a commentÂ |Â

up vote
12
down vote

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#include <iostream>

template<typename T, typename U>
U cast(T x) 
 return static_cast<U>(x);


template<typename T, typename U>
T cast(U x) 
 return static_cast<T>(x);


int main() 
 std::cout << cast<int,float>(10) << 'n';

edited 12 mins ago

Deduplicator

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asked 6 hours ago

wic

1087

New contributor

If you change your code to cast<int,float>(10.0f) the second overload would be called. And if you changed it to cast<int,float>(10.0) the call would be ambiguous.
â€“Â Praetorian
5 hours ago

add a commentÂ |Â

up vote
12
down vote

favorite

#include <iostream>

template<typename T, typename U>
U cast(T x) 
 return static_cast<U>(x);


template<typename T, typename U>
T cast(U x) 
 return static_cast<T>(x);


int main() 
 std::cout << cast<int,float>(10) << 'n';

edited 12 mins ago

Deduplicator

33.2k64786

asked 6 hours ago

wic

1087

New contributor

#include <iostream>

template<typename T, typename U>
U cast(T x) 
 return static_cast<U>(x);


template<typename T, typename U>
T cast(U x) 
 return static_cast<T>(x);


int main() 
 std::cout << cast<int,float>(10) << 'n';

c++ templates overload-resolution template-deduction

edited 12 mins ago

Deduplicator

33.2k64786

asked 6 hours ago

wic

1087

New contributor

edited 12 mins ago

Deduplicator

33.2k64786

asked 6 hours ago

wic

1087

New contributor

edited 12 mins ago

Deduplicator

33.2k64786

edited 12 mins ago

Deduplicator

33.2k64786

edited 12 mins ago

Deduplicator

33.2k64786

asked 6 hours ago

wic

1087

New contributor

asked 6 hours ago

wic

1087

asked 6 hours ago

wic

1087

New contributor

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If you change your code to cast<int,float>(10.0f) the second overload would be called. And if you changed it to cast<int,float>(10.0) the call would be ambiguous.
â€“Â Praetorian
5 hours ago

add a commentÂ |Â

If you change your code to cast<int,float>(10.0f) the second overload would be called. And if you changed it to cast<int,float>(10.0) the call would be ambiguous.
â€“Â Praetorian
5 hours ago

If you change your code to cast<int,float>(10.0f) the second overload would be called. And if you changed it to cast<int,float>(10.0) the call would be ambiguous.
â€“Â Praetorian
5 hours ago

add a commentÂ |Â

2 Answers
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When you use cast<int, float>, both templates are considered.

template<typename T=int,typename U=float>
U cast(T x);
template<typename T=int,typename U=float>
T cast(U x);

we then substitute:

template<typename T=int,typename U=float>
float cast(int x);
template<typename T=int,typename U=float>
int cast(float x);

at this point, there are no types to deduce. So we go to overload resolution.

In one case, we can take an int and convert to float when calling cast, and in the other we take a int and convert to int when calling cast. Note I'm not looking at all into the body of cast; the body is not relevant to overload resolution.

The second non-conversion (at the point of call) is a better match, so that overload is chosen.

If you did this:

std::cout << cast<int>(10) << "n";

things get more interesting:

template<typename T=int,typename U=?>
U cast(T x);
template<typename T=int,typename U=?>
T cast(U x);

for the first one, we cannot deduce U. For the second one we can.

template<typename T=int,typename U=?>
U cast(int x);
template<typename T=int,typename U=int>
int cast(int x);

so here we have one viable overload, and it is used.

edited 5 hours ago

answered 5 hours ago

Yakk - Adam Nevraumont

172k18175354

what OP's code actually does is convert from int to float (verified in assembler), so what non-conversion?
â€“Â jakub_d
5 hours ago

1

@jakub_d In the overload resolution of the argument. The body of a function does not change which function is called.
â€“Â Yakk - Adam Nevraumont
5 hours ago

add a commentÂ |Â

up vote
2
down vote

The instantiation is not ambiguous because the function argument exactly matches the first template parameter: the literal 10 is an int, which is also explicitly specified for the first template type.

You can make the instantiation ambiguous when the argument type does not match the explicitly specified types (thus, a conversion is necessary), e.g.

// error: call of overloaded 'cast<int, float>(unsigned int)' is ambiguous
std::cout << cast<int,float>(10u) << "n";

answered 5 hours ago

lubgr

7,64521442

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
21
down vote

accepted

When you use cast<int, float>, both templates are considered.

template<typename T=int,typename U=float>
U cast(T x);
template<typename T=int,typename U=float>
T cast(U x);

we then substitute:

template<typename T=int,typename U=float>
float cast(int x);
template<typename T=int,typename U=float>
int cast(float x);

at this point, there are no types to deduce. So we go to overload resolution.

The second non-conversion (at the point of call) is a better match, so that overload is chosen.

If you did this:

std::cout << cast<int>(10) << "n";

things get more interesting:

template<typename T=int,typename U=?>
U cast(T x);
template<typename T=int,typename U=?>
T cast(U x);

for the first one, we cannot deduce U. For the second one we can.

template<typename T=int,typename U=?>
U cast(int x);
template<typename T=int,typename U=int>
int cast(int x);

so here we have one viable overload, and it is used.

edited 5 hours ago

answered 5 hours ago

Yakk - Adam Nevraumont

172k18175354

what OP's code actually does is convert from int to float (verified in assembler), so what non-conversion?
â€“Â jakub_d
5 hours ago

1

@jakub_d In the overload resolution of the argument. The body of a function does not change which function is called.
â€“Â Yakk - Adam Nevraumont
5 hours ago

add a commentÂ |Â

up vote
21
down vote

accepted

When you use cast<int, float>, both templates are considered.

template<typename T=int,typename U=float>
U cast(T x);
template<typename T=int,typename U=float>
T cast(U x);

we then substitute:

template<typename T=int,typename U=float>
float cast(int x);
template<typename T=int,typename U=float>
int cast(float x);

at this point, there are no types to deduce. So we go to overload resolution.

The second non-conversion (at the point of call) is a better match, so that overload is chosen.

If you did this:

std::cout << cast<int>(10) << "n";

things get more interesting:

template<typename T=int,typename U=?>
U cast(T x);
template<typename T=int,typename U=?>
T cast(U x);

for the first one, we cannot deduce U. For the second one we can.

template<typename T=int,typename U=?>
U cast(int x);
template<typename T=int,typename U=int>
int cast(int x);

so here we have one viable overload, and it is used.

edited 5 hours ago

answered 5 hours ago

Yakk - Adam Nevraumont

172k18175354

what OP's code actually does is convert from int to float (verified in assembler), so what non-conversion?
â€“Â jakub_d
5 hours ago

1

@jakub_d In the overload resolution of the argument. The body of a function does not change which function is called.
â€“Â Yakk - Adam Nevraumont
5 hours ago

add a commentÂ |Â

up vote
21
down vote

accepted

When you use cast<int, float>, both templates are considered.

template<typename T=int,typename U=float>
U cast(T x);
template<typename T=int,typename U=float>
T cast(U x);

we then substitute:

template<typename T=int,typename U=float>
float cast(int x);
template<typename T=int,typename U=float>
int cast(float x);

at this point, there are no types to deduce. So we go to overload resolution.

The second non-conversion (at the point of call) is a better match, so that overload is chosen.

If you did this:

std::cout << cast<int>(10) << "n";

things get more interesting:

template<typename T=int,typename U=?>
U cast(T x);
template<typename T=int,typename U=?>
T cast(U x);

for the first one, we cannot deduce U. For the second one we can.

template<typename T=int,typename U=?>
U cast(int x);
template<typename T=int,typename U=int>
int cast(int x);

so here we have one viable overload, and it is used.

edited 5 hours ago

answered 5 hours ago

Yakk - Adam Nevraumont

172k18175354

When you use cast<int, float>, both templates are considered.

template<typename T=int,typename U=float>
U cast(T x);
template<typename T=int,typename U=float>
T cast(U x);

we then substitute:

template<typename T=int,typename U=float>
float cast(int x);
template<typename T=int,typename U=float>
int cast(float x);

at this point, there are no types to deduce. So we go to overload resolution.

The second non-conversion (at the point of call) is a better match, so that overload is chosen.

If you did this:

std::cout << cast<int>(10) << "n";

things get more interesting:

template<typename T=int,typename U=?>
U cast(T x);
template<typename T=int,typename U=?>
T cast(U x);

for the first one, we cannot deduce U. For the second one we can.

template<typename T=int,typename U=?>
U cast(int x);
template<typename T=int,typename U=int>
int cast(int x);

so here we have one viable overload, and it is used.

edited 5 hours ago

answered 5 hours ago

Yakk - Adam Nevraumont

172k18175354

edited 5 hours ago

answered 5 hours ago

Yakk - Adam Nevraumont

172k18175354

answered 5 hours ago

Yakk - Adam Nevraumont

172k18175354

answered 5 hours ago

Yakk - Adam Nevraumont

172k18175354

what OP's code actually does is convert from int to float (verified in assembler), so what non-conversion?
â€“Â jakub_d
5 hours ago

1

@jakub_d In the overload resolution of the argument. The body of a function does not change which function is called.
â€“Â Yakk - Adam Nevraumont
5 hours ago

add a commentÂ |Â

what OP's code actually does is convert from int to float (verified in assembler), so what non-conversion?
â€“Â jakub_d
5 hours ago

1

@jakub_d In the overload resolution of the argument. The body of a function does not change which function is called.
â€“Â Yakk - Adam Nevraumont
5 hours ago

what OP's code actually does is convert from int to float (verified in assembler), so what non-conversion?
â€“Â jakub_d
5 hours ago

@jakub_d In the overload resolution of the argument. The body of a function does not change which function is called.
â€“Â Yakk - Adam Nevraumont
5 hours ago

add a commentÂ |Â

up vote
2
down vote

You can make the instantiation ambiguous when the argument type does not match the explicitly specified types (thus, a conversion is necessary), e.g.

// error: call of overloaded 'cast<int, float>(unsigned int)' is ambiguous
std::cout << cast<int,float>(10u) << "n";

answered 5 hours ago

lubgr

7,64521442

add a commentÂ |Â

up vote
2
down vote

You can make the instantiation ambiguous when the argument type does not match the explicitly specified types (thus, a conversion is necessary), e.g.

// error: call of overloaded 'cast<int, float>(unsigned int)' is ambiguous
std::cout << cast<int,float>(10u) << "n";

answered 5 hours ago

lubgr

7,64521442

add a commentÂ |Â

up vote
2
down vote

You can make the instantiation ambiguous when the argument type does not match the explicitly specified types (thus, a conversion is necessary), e.g.

// error: call of overloaded 'cast<int, float>(unsigned int)' is ambiguous
std::cout << cast<int,float>(10u) << "n";

answered 5 hours ago

lubgr

7,64521442

You can make the instantiation ambiguous when the argument type does not match the explicitly specified types (thus, a conversion is necessary), e.g.

// error: call of overloaded 'cast<int, float>(unsigned int)' is ambiguous
std::cout << cast<int,float>(10u) << "n";

answered 5 hours ago

lubgr

7,64521442

answered 5 hours ago

lubgr

7,64521442

answered 5 hours ago

lubgr

7,64521442

answered 5 hours ago

lubgr

7,64521442

add a commentÂ |Â

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