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Can tr1::function swallow return values?

The boost::function FAQ item 3 specifically addresses the scenario I am interested in:

Why are there workarounds for void returns? C++ allows them! Void returns are permitted by the C++ standard, as in this code snippet:

void f();
void g() { return f(); }

This is a valid usage of boost::function because void returns are not used. With void returns, we would attempting to compile ill-formed code similar to:

int f();
void g() { return f(); }

In essence, not using void returns allows boost::function to swallow a return value. This is consistent with allowing the user to assign and invoke functions and function objects with parameters that don't exactly match.

Unfortunately, this doesn't work in VS2008:

int Foo();
std::tr1::function<void()> Bar = Foo;

This produces errors starting with:

c:\Program Files\Microsoft Visual Studio 9.0\VC\include\xxcallfun(7) : error C2562: 'std::tr1::_Callable_fun<_Ty>::_ApplyX' : 'void' function returning a value

Is this a failing of the VS2008 TR1 implementation? Does this work in VS2010? Does TR1 address this capability? How about C++0x?

like image 689
Chris Avatar asked Jul 08 '11 17:07

Chris


1 Answers

I believe tr1 addresses this issue. N1836 (the latest tr1 draft) says:

A function object f of type F is Callable for argument types T1, T2, ..., TN and a return type R, if, given lvalues t1, t2, ..., tNoftypesT1, T2, ..., TN,respectively,INVOKE(f, t1, t2, ..., tN)is well-formed([3.3]) and, if R is not void, convertible to R.

In your example R is void, and so the last part of the requirements for Callable (convertible to R) is ignored.

However it looks like C++0x (C++11) changes the rules. In C++11 Callable is defined as INVOKE(f, t1, t2, ..., tN, R) which is defined in [func.require] as requiring INVOKE(f, t1, t2, ..., tN) to be implicitly convertible to R, with no exception for when R is void. So in C++11, your example should fail.

like image 125
Howard Hinnant Avatar answered Nov 11 '22 07:11

Howard Hinnant