在结构中使用指针更改数据

Question

I have a program with one structnamed sample, it contains 2 int members and one char *. when creating 2 objects called a and b , I try assign a new dynamic string to a with the pointer and then copy all the values to b . so b = a . But later on when try to make changes to a like this : a.ptr[1] = 'X'; the pointer in b also changes. I want to know why, and how can I solve this.

struct Sample{
    int one;
    int two;
    char* sPtr = nullptr;
};
int _tmain(int argc, _TCHAR* argv[])
{
    Sample a;
    Sample b;
    char *s = "Hello, World";
    a.sPtr = new char[strlen(s) + 1];
    strcpy_s(a.sPtr, strlen(s) + 1, s);
    a.one = 1;
    a.two = 2;
    b.one = b.two = 9999;



    b = a;
    cout << "After assigning a to b:" << endl;
    cout << "b=(" << b.one << "," << b.two << "," << b.sPtr << ")" << endl << endl;

    a.sPtr[1] = 'X' ;
    cout << "After changing sPtr[1] with 'x', b also changed value : " << endl;
    cout << "a=(" << a.one << "," << a.two << "," << a.sPtr << ")" << endl;
    cout << "b=(" << b.one << "," << b.two << "," << b.sPtr << ")" << endl;

    cout << endl << "testing adresses for a and b: " << &a.sPtr << " & b is: " << &b.sPtr << endl;



    return 0;
}

Answer 1

Your struct contains a char* . When you assign all values in a to b, the pointer is also copied.

This means that a and b now point to the same char array. Therefore changing a value in this char array changes it for both structs.

If you do not want this, make a new char array for b and use strcpy .

Answer 2

You are copying the pointer not the value. To solve this you could override your assignment operator in the structure:

struct Sample{
    int one;
    int two;
    char* sPtr = nullptr;
    Sample& operator=(const Sample& inputSample)
    {
        one = inputSample.one;
        two = inputSample.two;
        sPtr = new char[strlen(inputSample.sPtr) + 1];
        strcpy (sPtr, inputSample.sPtr);
        return *this;
    }
};