Why I can't change a value via passing pointer using new-operator in cpp? [duplicate]

Question

I wanted to changed a variable with passing the method. I used traditional C way. I wrote this code in Visual Studio 2010 with Visual C++. However it does not give expected result.

Code have been a purpose but I changed it for easy understandability.

#include<cstdio>
using namespace std;
void exampleMethod(char *array) {
    array = new char[6];

    array[0] = 'h';
    array[1] = 'e';
    array[2] = 'l';
    array[3] = 'l';
    array[4] = 'o';
    array[5] = 0; // for terminating

    fprintf(stdout, "exampleMethod(): myArray is:%s.\n", array);
}
void main() {
    char* myArray = 0;

    exampleMethod(myArray);

    fprintf(stdout,"main(): myArray is:%s.\n", myArray);

    getchar(); // to hold the console.
}

Output of this code is:

exampleMethod(): myArray is:hello.
main(): myArray is:(null).

I don't understand why pointer value was not changed in main(). I know that it is pass by reference and I changed myArray's values with pointer. I also used new-operator to initialize that pointer.

After that, I changed code, I initialized variable myArray in main method with new-operator. (before it is in exampleMethod().)

void exampleMethod(char *array) {
    array[0] = 'h';
    array[1] = 'e';
    array[2] = 'l';
    array[3] = 'l';
    array[4] = 'o';
    array[5] = 0; // for terminating
    fprintf(stdout, "exampleMethod(): myArray is:%s.\n", array);
}
void main() {
    char* myArray = new char[6];;
    exampleMethod(myArray);
    fprintf(stdout,"main(): myArray is:%s.\n", myArray);
}

Surprisingly, code is running properly. It gives this ouput:

exampleMethod(): myArray is:hello.
main(): myArray is:hello.

Why did not previous code run in such a way that I expected? I compiled and run it with Visual Studio 2010 that is Visual C++ project. I also tried it with Visual Studio 2015.

You're passing a copy of the pointer: void exampleMethod(char *array) — iBug
– iBug, Commented Jan 29, 2018 at 9:22
I used traditional C way traditional C way would have been void exampleMethod(char **array) and exampleMethod(&myArray); — user2672107
– user2672107, Commented Jan 29, 2018 at 10:05

iBug · Accepted Answer · 2018-01-29 09:24:13Z

2

You're passing a copy of the pointer by void exampleMethod(char *array), so any change to the pointer in exampleMethod() will not affect the pointer in main().

You may want to pass it by reference (add an ampersand & before the identifier to make it a reference):

void exampleMethod(char * &array)

So in this way any modification to the pointer in exampleMethod will apply to the pointer in main(), too, as they are the same object now.

And a side note: Don't forget to delete[] the array whenever you get it from dynamic allocation.

answered Jan 29, 2018 at 9:24

iBug

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