lvalue c++ question

Thanks for your response.

From your remark, I got the impression that function returned values are temporary values which are also lvalues. Could you please show me in the C++ standard where this is and why it is necessary? Is it so that we can have statements like:   foo() = some-value; ?

There appears to be a difference between temporary values generated from User Defined Types and Built-In Types, as noted in this program:

template <class T>
class Lvalues {
public:
  Lvalues(T i) : val(i) {}
  T sumT(T arg) { (arg+val) = val; return (arg+val); }
  T val;
};

int main() {
  Lvalues<int> b1(4), b2(5), b3(6);
  Lvalues<string> a1("4"), a2("5"), a3("6");
  a1 = a2.sumT("12");  // OK, no compiler error
  cout << a1.val;
//  b1 = b2.sumT(12);
}

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The line b1 = b2.sumT(12); has the compiler error: lvalue required as left operand of assignment. When commented out, the program gives the expected result: 125.

Why should it work for UDT, but not for Built-In-Type?

Ok, thanks for the comments. I will accept that you are representing the standard reasonably well. Thanks again, and Happy Holidays to you.

I hope I am (not 100% sure, but 90% should be OK - focusing on that very "temporary object"), but trying my best, and my hoilday wishs go back in your direction ;o)

Loking at that after a few days, let me make one correction:

I wrote

But every operation that returns an object, that object is assumed to be assignable (which I assume to be "OO-101"), thus being fit for being taken as an L-value - and that's where the conundrum starts.

and that should have been

every operation that returns a non-const object

- maybe 'constness' could be the key to that very conundrum here?

>> every operation that returns a non-const object
Right, I figured that was what you meant. Thanks for clarifying.
Happy New Year!