c ++：函数左值或右值

Question

I just started learning about rvalue references in c++11 by reading this page , but I got stuck into the very first page. Here is the code I took from that page.

  int& foo();
  foo() = 42; // ok, foo() is an lvalue
  int* p1 = &foo(); // ok, foo() is an lvalue

  int foobar();
  j = foobar(); // ok, foobar() is an rvalue
  int* p2 = &foobar(); // error, cannot take the address of an rvalue

1. why is foo() an lvalue? is it because foo() returns int& which is basically an lvalue?
2. why is foobar() an rvalue? is it because foobar() returns int ?
3. In general, why would you care if a function is an rvalue or not? I think if I read the rest of that article, I'll get my answer to this.

Answer 1

L-Values are locations, R-Values are actual values.

So:

1. since foo() returns a reference( int& ), that makes it an lvalue itself.
2. Correct. foobar() is an rvalue because foobar() returns int .
3. We don't care that much if a function is an R-Value or not. What we are getting excited about is R-Value references.

The article you pointed to is interesting and I had not considered forwarding or the use in factories before. The reason I was excited about R-Value references was the move semantics, such as this:

BigClass my_function (const int& val, const OtherClass & valb);

BigClass x;
x = my_function(5, other_class_instance);

In that example, x is destroyed, then the return of my_function is copied into x with a copy constructor. To get around that historically, you would write:

void my_function (BigClass *ret, const int& val, const OtherClass & valb);

BigClass x;
my_function(&x, 5, other_class_instance);

which means that now my_function has side effects, plus it isn't as plain to read. Now, with C++11, we can instead write:

BigClass & my_function (const int& val, const OtherClass & valb);

BigClass x;
x = my_function(5, other_class_instance);

And have it operate as efficiently as the second example.