Get lvalue through rvalue

[George2]

Hello everyone,
I do not know how in the following code, rvalue -- return of X(),
could result in a lvalue finally and binded to a non-const reference
input parameter of function f.

Any ideas?

struct X {
 
 
};
 
void f (X& x) {}
 
int main()
{
f (X() = X());
 
return 0;
}
 

thanks in advance,
George

[Victor Bazarov]

George2 wrote:

I do not know how in the following code, rvalue -- return of X(),
could result in a lvalue finally and binded to a non-const reference
input parameter of function f.

You're confusing the return type of 'X()' and the full expression
that you can see inside the parentheses where 'f' is called.

>
Any ideas?

struct X {


};

void f (X& x) {}

int main()
{
f (X() = X());

return 0;
}
 

The trick here is based on the fact that every temporary of a class
type is a separate object, for which you can call member functions.
The operator= is a member function. What is its usual signature?
Rewrite the expression

f( X() = X() )

in terms of member function calls, and you will hopefully see how
an r-value can be turned into an l-value.

V
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