Is returning uniform initialized reference valid?

Question

Is this code sample valid?

using ref = char&;

ref foo(ref x) {
  return ref{x};
}

int main() {
  char a;
  foo(a);
  return 0;
}

seems that:

clang 3.5 says YES

gcc 4.9 says NO

main.cpp: In function 'char& foo(ref)':
main.cpp:4:15: error: invalid cast of an rvalue expression of type 'char' to type 'ref {aka char&}'
   return ref{x};
               ^

http://coliru.stacked-crooked.com/a/cb6604b81083393f

So which compiler is right? or is it unspecified?

It very easy so overcome gcc build error by:

using parenthesis instead of braces
```
ref foo(ref x) {
  return ref(x);
}
```

by naming returned value

ref foo(ref x) {
  ref ret{x};
  return ret;
}

option 1. breaks uniform initialization, option 2. adds useless line of code.

Similar question was already aked here: Why can't I initialize a reference in an initializer list with uniform initialization?

But mentioned pr50025 is fixed in gcc 4.9.

I know that above code sample is pretty useless, but I oversimplified it intentionally to point out the issue. In real life code problem can be hidden in a generic function like:

#include <utility>
template <typename Tp, typename... Us>
Tp bar(Us&&... us) {
  return Tp{std::forward<Us>(us)...};
}

Unknown Unknown · Accepted Answer · 2015-01-28T20:27:16

This seems like an omission in the standard, where GCC is implementing exactly what the standard requires, and clang is going for what's probably intended.

From C++11 (emphasis mine):

5.2.3 Explicit type conversion (functional notation) [expr.type.conv]

1 A simple-type-specifier (7.1.6.2) or typename-specifier (14.6) followed by a parenthesized expression-list constructs a value of the specified type given the expression list. If the expression list is a single expression, the type conversion expression is equivalent (in definedness, and if defined in meaning) to the corresponding cast expression (5.4). [...]

[...]

3 Similarly, a simple-type-specifier or typename-specifier followed by a braced-init-list creates a temporary object of the specified type direct-list-initialized (8.5.4) with the specified braced-init-list, and its value is that temporary object as a prvalue.

For the braced-init-list case, the standard doesn't specify that that this works just like a C-style cast. And it doesn't:

typedef char *cp;
int main() {
  int i;
  (cp(&i)); // okay: C-style casts can be used as reinterpret_cast
  (cp{&i}); // error: no implicit conversion from int * to char *
}

Unfortunately, T(expr) being equivalent to (T)expr is also the one exception in which a functional cast doesn't necessarily produce a prvalue. The standard fails to specify a similar exception for a functional cast using a braced-init-list to a reference type. As a result, in your example, ref{x} constructs a temporary of type ref, direct-list-initialised from {x}. That temporary is then treated as a prvalue, because that's what the standard says the behaviour should be, and that prvalue cannot be used for binding to an lvalue reference.

I strongly suspect that if this were brought up to the ISO C++ committee, the standard would be changed to require clang's behaviour, but based on the current wording of the standard, I think it's GCC that's correct, at least for your specific example.

Instead of adding a variable, or switching to parentheses, you can omit ref (Tp) to avoid the problem:

template <typename Tp, typename... Us>
Tp bar(Us&&... us) {
  return {std::forward<Us>(us)...};
}

Is returning uniform initialized reference valid?

1 Answers