右值引用参数和模板函数

Question

If I define a function which accepts an rvalue reference parameter:

template <typename T>
void fooT(T &&x) {}

I can call it, using GCC 4.5, with either a , ar , or arr :

int a, &ar = a, &&arr = 7;
fooT(a); fooT(ar); fooT(arr);

However, calling a similar, non-template function,

void fooInt(int &&x) {}

with any of those three arguments will fail. I was preparing to strengthen my knowledge of forward , but this has knocked me off course. Perhaps it's GCC 4.5; I was surprised to find that the first example from A Brief Introduction to Rvalue References also gives a compile error:

A a;
A&& a_ref2 = a;  // an rvalue reference

Answer 1

The behavior of deduction in template parameters is unique, and is the reason your template version works. I've explained exactly how this deduction works here , in the context of another question.

Summarized: when the argument is an lvalue, T is deduced to T& , and T& && collapses to T& . And with the parameter at T& , it is perfectly valid to supply an lvalue T to it. Otherwise, T remains T , and the parameter is T&& , which accepts rvalues arguments.

Contrarily, int&& is always int&& (no template deduction rules to coerce it to something else), and can only bind to rvalues.

Answer 2

In addition to GMan's correct answer A Brief Introduction to Rvalue References has an incorrect example because it was written prior to a language change which outlawed:

A a;
A&& a_ref2 = a;  // an rvalue reference (DISALLOWED in C++11)

Despite this change in the language, the main uses cases described in the article (move and forward) are still explained correctly in the article.

Update: Oh, and the same article was originally published here with (imho) slightly better formatting.