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RVO (Return Value Optimization)

Lately I was recalling and practicing C++ more. Not fully that I want to preach about it, but it's a realization that industry accepts it, and it'd be hard to find work without knowing it for my current situation.

This post talks about RVO or (Return Value Optimization).

The gotcha came during the time I implemented a postfix conversion function, then checked address of returned value against the outside (calling site) variable that receives the result. Surprisingly, it turns out to be the same!

See the actual code below

std::string ConvertToPostfix(const std::string& infix) {
    std::stringstream ss;
    Stack<char> ts;

    for (auto it = infix.begin(); it != infix.end(); ++it) {
        char c = *it;

        if (c == '+' ||
            c == '-' ||
            c == '*') {
           ts.Push(c); 
        } 
        else if (c >= '0' && c <= '9') {
            ss << c << " ";
        }
        else if (c == ')') {
            ss << ts.Pop() << " ";

            if (it == infix.end() - 1) {
                while (!ts.IsEmpty()) {
                    ss << ts.Pop();
                }
            }
        }
    }

    return ss.str();
}

No need to pay attention much to the actual implementation, I just want to be clear adding it there so in case you want to see what happens with std::stringstream ss so you can check it.

std::stringstream::str() function will return a temporary std::string every time we call it. Below is what cppreference has a say about it.

Notes The copy of the underlying string returned by str is a temporary object that will be destructed at the end of the expression, so directly calling c_str() on the result of str() (for example in auto *ptr = out.str().c_str();) results in a dangling pointer.

So taking that into account, a decision to make above function return by value should be safe and correct. Then I'll do some validations.

Change returning line into the following

...
std::string tmpStr = ss.str();
std::cout << "Before: " << std::addressof(tmpStr) << "\n";
return tmpStr;

On the call site, we'd have

std::string postfix = ConvertToPostfix("5 * (((9+8) * (4*6)) + 7)");
std::cout << "After: " << std::addressof(postfix) << std::endl;

Now what do you think about the result? Probably first thought would be, both addresses definitely be different. But it turns out to be opposite.

Before: 0x7ffec387ec90
After: 0x7ffec387ec90

So what's is going on? After research, it's due to compiler optimization specifically it's called RVO or Return Value Optimization. In short, compiler eliminates the temporary object which is needed to be created to hold a return value from function. So now it acts like such return value is declared on stack at same time of that receiving end at the call site.

How to turn RVO off?

Specify -fno-elide-constructors as compile flags to gcc. Now the result...

Before: 0x7ffc5a823450
After: 0x7ffc5a823690



First published on July, 21, 2019






Written by Wasin Thonkaew
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