Re: returning lvalue in C vs C++

Ben C wrote:

On 2006-03-29, Ben C <spamspam@xxxxxxxxx> wrote:
On 2006-03-29, Phlip <phlip2005@xxxxxxxxx> wrote:
Ben C wrote:

struct thing
{
int x;
};

struct thing f(void)
{
struct thing ret = {100};
return ret;
}

int g(void)
{
return 10;
}

int main(void)
{
#if 0
g() = 4; /* "invalid lvalue" in C and C++ */
#endif
f().x = 200; /* "invalid lvalue" in C, but not in C++ apparently
*/

Consider the general scheme of C++, and specifically its overloaded
operators.

x.operator=() (figuratively) could have a side-effect that your code
depends on. So regardless where 'thing' appears, thing.x is accessible
as an lvalue.

It would be thing.operator=() but I see the point.

Sorry, no it wouldn't, you were right. It's x's operator= we'd be
calling.

But x is an int, so if f().x = 200 works on that basis, why not g() = 4?

Yup: why not? I don't know. But for classes, it sure works:


#include <iostream>

struct dummy {

dummy & operator= ( dummy const & other ) {
std::cout << "hello world!\n";
return ( *this );
}

};

dummy g ( void ) {
dummy tmp;
return ( tmp );
}

int main ( void ) {
g() = g();
}


Best

Kai-Uwe Bux
.

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