Why is there a difference between using c++ string+string temporary object by different compilers?

Question

Please look at this code.

#include <iostream>
#include <string>
using namespace std;

int main() {
    string hello = "Hello"
         , world = "World";

    const char *p = (hello+world).c_str();
    cout << "STRING: " << p <<endl;

    return 0;
}

I have no reputation, can't post images so that I will write results by hand.

= Visual Studio 2013 ver.12.0.30110.00

STRING:

= Dev-C++ ver.4.9.9.2

STRING: HelloWorld

The first following is execution result that compiled by Visual Studio.

Second is compiled by Dev-C++.

I wonder what makes this difference.

I will be looking forward to your reply. Thanks :)

It is *undefined behaviour*. That is a reason you can observe a difference. — juanchopanza, Apr 25 '14 at 08:25
Thanks @juanchopanza, I think it is important that avoid using undefined behavior... — Leed, Apr 25 '14 at 08:33

score 3 · Accepted Answer · answered Apr 25 '14 at 08:26

3

(hello+world).c_str() is only valid until the trailing ;. Accessing the memory afterwards is undefined behavior.

Visual studio probably actually clears the memory, Dev-C++ doesn't bother. Try building a release version with Visual studio (optimizations on) and you'll probably see the same behavior.

answered Apr 25 '14 at 08:26

Luchian Grigore

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It could also depend on how the `<<` operator uses memory. If `std::string` uses SSO, it will also depend on whether the temporary string is small enough to fit; if so, `p` will point to an on stack location, which will be overwritten when `operator<<` is called. And so on: you really can't make any general statement. – James Kanze Apr 25 '14 at 08:31
Thanks @Luchian Grigore I think Visual Studio bother with optimizations :) – Leed Apr 25 '14 at 08:34

Why is there a difference between using c++ string+string temporary object by different compilers?

1 Answers1