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Why can't I pass an rvalue-reference as it is to another function in C++11?

I have a code:

void f(int&& i) {
  auto lambda = [](int&& j) { (void)j; }
  lambda(i);
}

int main() {
  f(5);
}

Clang++ gives an error: no known conversion from 'int' to 'int &&' for 1st argument

Why the i changes its type to int when being passed to the lambda()?

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abyss.7 Avatar asked Feb 21 '14 17:02

abyss.7


2 Answers

i is of type int&&, that is, it's of type "rvalue reference to int." However, note that i itself is an lvalue (because it has a name). And as an lvalue, it cannot bind to a "reference to rvalue."

To bind it, you must turn it back to an rvalue, using std::move() or std::forward().

To expand a bit: the type of an expression and its value category are (largely) independent concepts. The type of i is int&&. The value category of i is lvalue.

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Angew is no longer proud of SO Avatar answered Nov 01 '22 12:11

Angew is no longer proud of SO


There are two elements at work here:

  • Type: The parameter i has type int&&, or "rvalue reference to int", where "rvalue reference" is the name for the && feature, allowing binding rvalues to a reference;

  • Value category: This is the key. i has a name and so the expression naming it is an lvalue, regardless of its type or what the standard committee decided to call that type. :)

(Note that the expression that looks like i has type int, not int&&, because reasons. The rvalue ref is lost when you start to use the parameter, unless you use something like std::move to get it back.)

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Lightness Races in Orbit Avatar answered Nov 01 '22 14:11

Lightness Races in Orbit