returning reference to local temporary object c++

Question

Overload the relational operator ==for the class stackTypethat returns trueiftwo stacks of the same type are the same; falseif otherwise.

my code:

template <class Type>
const stackType<Type>& stackType<Type>::operator ==
(const stackType<Type>& otherStack)
{
    if (this->stackTop != otherStack.stackTop)
        return false;
    for (int i = 0; i < stackTop; i++)
        if (this->list[i] != otherStack.list[i])
            return false;
    return true;
} //end operator==

I'm receiving a returning reference to local temporary object warning and from what i understand is that it's being destroyed once the function scope ends. Can anyone point me in the right direction to solving this warning?

To avoid the warning just don't do it ... You can safely return ref to attribute or allocated memory (new). — fievel, Oct 22 '17 at 20:24
@fievel but why allocate anything? The asker clearly needs to return bool and screwed up the return type. — user4581301, Oct 22 '17 at 20:41

score 2 · Answer 1 · answered Oct 22 '17 at 20:26

2

You are declaring your operator to return a const stackType<Type>&, bt then you return true or false. The compiler tries to convert true and false to const stackType<Type>&, and then return it. That’s probably not what you intended - rather declare it to return a bool.

Even if the conversion would work, the reference is to a local variable, which gives UB.

answered Oct 22 '17 at 20:26

Aganju

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Correct, it was meant to be a bool, but i still get a returning to local temp. How do I I not return to local temp? – noslov Oct 22 '17 at 20:50
1

Did you define it to return a `bool&`? Don't do that, never with elementary data types. Otherwise, the error must come from somewhere else in the code. – Aganju Oct 22 '17 at 21:17

returning reference to local temporary object c++

1 Answers1