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c++c++11rvalue-reference
thor
thor

Reputation: 22570

c++ syntax in understanding rvalue references

I was reading a blog about c++11 rvalue references by Thomas Becker, and the following syntax near the bottom of the page really confuses me.

  int& foo();
  foo() = 42; // ok, foo() is an lvalue

What exactly is foo? A function pointer that returns a int; an object?

If a function object, how can you assign a value to it, and why is foo a lvalue?

Upvotes: 0

Views: 119

Answers (3)

Pradhan
Pradhan

Reputation: 16777

Every function has at least one return expression. Let's look at function calling using the following incomplete, simplified description :

Assume your function signature is T f();, with one return expression return e;. For the sake simplicity, lets also assume that e and T have the same type after removing references and const-ness. In this situation, you can go to your function call location and replace the call f() with (T)e, to understand what's going on.

int& f(){....; return e;}

.
.
.

f() = 5;

becomes

(int&)e = 5;

Of course, if converting e to an lvalue reference(int&) was invalid, the compiler will throw an error.

Upvotes: 0

Praetorian
Praetorian

Reputation: 109289

int& foo(); declares a function foo that takes no arguments and returns a reference to an int. For example,

#include <iostream>

int a = 0;
int& foo();

int main()
{
    std::cout << a << std::endl;
    foo() = 42;
    std::cout << a << std::endl;
}

int& foo() { return a; }

Output:

0
42

Upvotes: 5

Brian Bi
Brian Bi

Reputation: 119641

foo is a function with return type int&. The line

int& foo();

simply declares the function, but does not define it.

When foo is called, the result is an lvalue of type int. Therefore you can assign to it: foo() = 42;

Upvotes: 3

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