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Home/ Questions/Q 8852243
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T13:21:02+00:00 2026-06-14T13:21:02+00:00

Consider the following source code in C++ vector <char *> myFunction() { vector <char

  • 0

Consider the following source code in C++

vector <char *> myFunction()
{
    vector <char *> vRetVal;
    char *szSomething = new char[7];

    strcpy(szSomething,"Hello!");
    vRetVal.push_back(szSomething); // here vRetVal[0] address == &szSomething

    delete[] szSomething; // delete[]ing szSomething will "corrupt" vRetVal[0]
    szSomething = NULL;

    return vRetVal; // here i return a "corrupted" vRetVal
}

Any idea on how to use push_back to make a copy of the parameter I pass instead of taking it by reference? Any other idea is also accepted and appreciated.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-14T13:21:03+00:00Added an answer on June 14, 2026 at 1:21 pm

    The object whose pointer you’ve pushed to the vector is destroyed by delete statement in your code. That means, the item (which is pointer) in the vector is pointing to a deleted object. I’m sure you don’t want that.

    Use std::string:

    std::vector<std::string> myFunction()
    {
        std::vector<std::string> v;
        v.push_back("Hello"); 
        v.push_back("World");
        return v;
    }
    

    In C++11, you could just write this:

    std::vector<std::string> myFunction()
    {
       std::vector<std::string> v{"Hello", "World"};
       return v;
    }
    

    Or this,

    std::vector<std::string> myFunction()
    {
       return {"Hello", "World"};
    }
    
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