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

[weaknessforcats]

This is silly code.

X() creates an X object by calling the default constructor. Therefore:

X() = X()

Creates two X objects and assigns the one on the right to the one on the left. That left object is bound as the non-const reference used on the call to f().

The code above uis the same as:

X obj1;
X obj2;
obj1 = obj2;
f(obj1);
 

[George2]

Thanks weaknessforcats,


Do you think my original sample breaks the rule that non-const reference can not binded to a rvalue (temporary object return by X())?

This is silly code.

X() creates an X object by calling the default constructor. Therefore:

X() = X()

Creates two X objects and assigns the one on the right to the one on the left. That left object is bound as the non-const reference used on the call to f().

The code above uis the same as:

X obj1;
X obj2;
obj1 = obj2;
f(obj1);
 

regards,
George

[weaknessforcats]

Do you think my original sample breaks the rule that non-const reference can not binded to a rvalue (temporary object return by X())?

Where did you hear this??

Syntatically, you can always return a non-const pointer or reference to a local object. However, the the pointer or reference will become garbage once the temporary object disappears. On my compiler doing this generates a warning that I am doing is a bad thing.

[George2]

Thanks for your advice, weaknessforcats!

Where did you hear this??

Syntatically, you can always return a non-const pointer or reference to a local object. However, the the pointer or reference will become garbage once the temporary object disappears. On my compiler doing this generates a warning that I am doing is a bad thing.

regards,
George