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
- why is
foo()
an lvalue? is it becausefoo()
returnsint&
which is basically an lvalue? - why is
foobar()
an rvalue? is it becausefoobar()
returnsint
? - 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.
L-Values are locations, R-Values are actual values.
So:
- since
foo()
returns a reference(int&
), that makes it an lvalue itself. - Correct.
foobar()
is an rvalue becausefoobar()
returnsint
. - 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.
来源:https://stackoverflow.com/questions/13854518/c-function-lvalue-or-rvalue