C - function parameters and pointers

Question

Consider two cases of calling a function:

void Convert(int number, int **staticPointer)
{
    int * dynamicPointer = malloc(sizeof(int));
    *dynamicPointer = number;
    *staticPointer = dynamicPointer;
}

int main()
{
    int *p;

    Convert(5, &p);

    printf("The number is: %d", *p);

    free(p);
}

The above works beautifully with no problem.

Now the case below:

void Convert(int number, int *staticPointer)
{
    int * dynamicPointer = malloc(sizeof(int));
    *dynamicPointer = number;
    staticPointer = dynamicPointer;
}

int main()
{
    int *p;

    Convert(5, p);

    printf("The number is: %d", *p);

    free(p);
}

This causes SEGFAULT. Notice that I do call the function in different ways with different parameters, and I do understand that the difference is in the calling of the function.

Can somebody please explain to me why I have to call the function as in the first example, and why the other example doesn't work (it crashes in main)?

The purpose of my program is to reference a static pointer to a specific memory address through a function.

The first one changes the pointer in the `main()` function; the second changes the copy of the pointer in the `Convert()` function, but doesn't change the pointer in the `main()` function, so the pointer in `main()` is uninitialized and you are lucky that you got a crash to tell you are doing it wrong (that wasn't guaranteed). — Jonathan Leffler, May 11 '16 at 14:27

Sourav Ghosh · Answer 1 · 2016-05-11T14:36:09.843

C uses pass by value for function parameter passing.

In the first case, by passing the address of the pointer p, you were able to change the pointer itself. So, im main(), your changes are reflected.
In he second case, the pointer itself cannot be changed, as it itself is passed-by-value. Once you return from the Convert() function, any changes made to p will be lost. So, in main(), you'll be accessing invalid memory (*p) which nvokes undefined behavior. Hence the crash.

score 0 · Answer 2 · answered May 11 '16 at 14:29

0

There are two things you're changing:

Function type
Function invocation

Since you want to modify the pointers value, you have to provide a pointer to the pointer to int.

See resources on pointers to pointers.

answered May 11 '16 at 14:29

Aif

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C - function parameters and pointers

2 Answers2