CODEWITHSUNDEEP
c++stringpointerschar-pointer
MrMiserMeister
MrMiserMeister

Reputation: 71

Two different char pointers end up with the same value

I'm assuming this is undefined behavior, but I'm not exactly sure why?

When I compile the above code with g++ 4.4.7 and execute it:

#include <iostream>
#include <string>

using namespace std;

int main()
{
    string A = "-A";
    string B = "_B";
    string common = "Common";
    const char* commonA = (common + A).c_str();
    const char* commonB = (common + B).c_str();
    
    cout << commonA << endl;
    cout << commonB << endl;
    
    return 0;
}

The resulting prints show that both commonA and commonB evaluate to "Common_B".

I noticed that when I put the same code into an online c++ ide I get the expected result where commonA is "Common-A" and commonB is "Common_B". Hence why I think what I'm doing is undefined behavior.

I was able to get the original snippet of code to give me the expected behavior in g++ 4.4.7 by separating the string concatenation and c_str() call into two steps. i.e.:

    string commonA_str = common + A;
    string commonB_str = common + B;

    const char* commonA = commonA_str.c_str();
    const char* commonB = commonB_str.c_str();

What exactly is going on here, and what is a safe way to do this?

Thanks!

Upvotes: 0

Views: 54

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