Passing Lvalue argument to Rvalue referece parameter of template function?

Question

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here the template function definition

C++

template<typename T>
void foo(T&& t){}

call of foo

C++

int i = 11;
foo(i);//compile and run!

In the previous code ,are we binding an rvalue reference to a lvalue? why is that possible ?
because we know that if we define foo as :

C++

void foo(int&& t){}

we'll get an error from the call of foo(i).

appreciate for reading.

Posted 12-Oct-13 2:11am

Wang Xin USTC

Updated 12-Oct-13 4:51am

v5

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Comments

Richard MacCutchan 12-Oct-13 9:00am

Build the program with a bit more code and step through with your debugger and you will find out.

Wang Xin USTC 12-Oct-13 10:09am

yes,I know from debugging the code that T is deducted as int& ,but I'd like to know what happened behind the scenes : why here we are allowed to bind a rvalue reference to a lvalue?

2 solutions

Solution 1

It compiles because it is a feature in C++. T& is an lvalue reference and T&& is an rvalue reference.

C++

T a;
T& l_ref_to_a = a;  // an lvalue reference
T&& r_ref_to_a = a;  // an rvalue reference

"An rvalue reference to A is created with the syntax A&&. To distinguish it from the existing reference (A&), the existing reference is now termed an lvalue reference. The new rvalue reference behaves just like the existing lvalue reference except where noted...."

Read the rest :
http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2004/n1690.html[^]

Posted 12-Oct-13 4:31am

Captain Price

Comments

Wang Xin USTC 12-Oct-13 10:44am

well ,thank you.
what I intended to ask is :

void foo(int&& i);
int j = 11;
foo(j); // error because try to bind a rvalue ref to an lvalue.

but when it comes to template function, as in the initial question ,things changed. that is my point.

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Wang Xin USTC · Accepted Answer · 2013-10-12T05:42:00

"If P is an rvalue reference to a cvunqualified
template parameter and the argument is an lvalue, the type “lvalue reference to A” is used in
place of A for type deduction." [ Example:

C++

template <class T> int f(T&&);
template <class T> int g(const T&&);
int i;
int n1 = f(i); // calls f<int&>(int&)
int n2 = f(0); // calls f<int>(int&&)
int n3 = g(i); // error: would call g<int>(const int&&), which
// would bind an rvalue reference to an lvalue

----- ISO (N3690) 14.8.2.1
So,in this question ,T is deducted as int&,so the function is of argument int& && ,which is equivalent of int& (if you take a reference to a reference to a type ,you got a reference to that type)
modified version of codes in the initial question can demonstrate that :

C++

template<typename t="">
void foo(T&& rr){
	rr++;
}
int main(){	
	int i = 11;	
	foo(i);
	cout << i << endl; // output is : 12,because foo modified the local variable
}
</typename>