Explicit cast to r-value reference does not clear ambiguity

Asked Nov 29 '16 at 15:25

Active Nov 29 '16 at 15:26

Viewed 78 times

Regarding the following code:

void foo( int in )
{
  std::cout << "no ref" << std::endl;
}

void foo( int&& in )
{
  std::cout << "ref&&" << std::endl;
}


int main()
{ 
  foo( static_cast<int&&>(1) );

  return 0;
}

I wonder why the call to foo( static_cast< int&& >(1) ) is ambiguous:

error: call to 'foo' is ambiguous
  foo( static_cast<int&&>(1) );
  ^~~

Because of the explicit cast to int&&, I expected that void foo( int&& in ) will be called.

edited Nov 29 '16 at 15:26

Angew is no longer proud of SO

167,307
17
350
455

asked Nov 29 '16 at 15:25

abraham_hilbert

2,221
1
13
30

2

The expression `1` already *is* an rvalue. – Some programmer dude Nov 29 '16 at 15:27
Probably the same reason `void foo(const int& in)` would be ambiguous - both functions can take the same type without needing to convert it. – Mark Ransom Nov 29 '16 at 15:31
@MarkRansom `void foo( int&& in )` is preferred over `void foo(const int& in)` if it is an rvalue. – NathanOliver Nov 29 '16 at 15:32
@NathanOliver there are *so* many corner cases in C++, I learned something new today. Thank you! – Mark Ransom Nov 29 '16 at 15:35
@MarkRansom No problem. I find it hard to keep them all straight myself. – NathanOliver Nov 29 '16 at 15:35
2

I didn't even know you could call `static_cast` to an r-value. I figured that's what `std::move` is for... – Guy Kogus Nov 29 '16 at 15:41
2

@GuyKogus `std::move` is implemented precisely as `std::static_cast` – Angew is no longer proud of SO Nov 29 '16 at 15:51

Explicit cast to r-value reference does not clear ambiguity

0 Answers0