Question

This question arose from being unable to use uniform-initialisation syntax with the auto keyword because it treats it as a std::initializer_list<T> (explanation in the comments here).

Take the following code example:

class X { };
int x( X() ); // function prototype (1)
auto x( X() );  // copy/move construction of an X, function prototype or compile-time error?

What does the compiler do with auto x?

Reasoning for each possibility:

Copy/move construction: I can see this being the proper behaviour due to (1) being seen as a kind of defect.

Function prototype: Seems unlikely as there is no trailing return type.

Compile-time error: If the compiler does parse this as a function prototype, it may cause a compile-time error due to lacking a trailing return type.

What does the C++0x standard say this should be interpreted as?

Answer 1

I get

error: 'x' function uses 'auto' type specifier without late return type

The compiler is expecting something like

auto x( X() ) -> int;

which would be equivalent to line 2.