Why the temporary object can be returned in this example?

Question

char*[] object can be passed to function version2() & version 3() by the const reference, which means s2 is a temporary object equals to "###" or "@@@".

But why the temporary object s2 can be returned back, while the temporary object temp cannot.

int main()
{
    result = version2(input, "###");
    cout << "Your string enhanced: " << result << endl;

    result = version3(input, "@@@");
    cout << "Your string enhanced: " << result << endl;
    return 0;
}

// No Error
const string & version2(string & s1, const string & s2)
{
    return s2; 
}

// Error
const string & version3(string & s1, const string & s2)
{
    string temp = s2 + s1 + s2;
    return temp;
}

Answer 1

The point is object life time.

For the 1st case, as you said, a temporary std::string is created and passed to version2 (being bound to the parameter s2). The temporary will be destroyed after the full-expression, i.e. the whole result = version2(input, "###");. That means the reference returned by version2 remains valid, it's fine to use it for assigning to result.

All temporary objects are destroyed as the last step in evaluating the full-expression that (lexically) contains the point where they were created,

For the 2nd case, a local object temp is created inside version3, it will be destroyed when get out of version3. That means version3 will always return a dangled reference. And using it to assign to result leads to UB. The compiler might give a diagnosis for it; but doesn't have to do.