Question

I have came across this problem when using pointer to pointer to a char:

void setmemory(char** p, int num)
{
    *p=(char*)malloc(num);
}

    void test(void)
    {
        char* str=NULL;
        setmemory(&str,100);
        strcpy(str,"hello");
        printf(str);
    }

int main()
{
    test();
    return 0;
}

The code above is correct,but I can’t figure it out why using a pointer to a pointer char** p here? Why just using a pointer to a char instead? so I change this snippet into below and found it not working ,can anyone tell me why? thanks!

void setmemory(char* p, int num)
{
    p=(char*)malloc(num);
}

void test(void)
{
    char* str=NULL;
    setmemory(str,100);
    strcpy(str,"hello");
    printf(str);
}

int main()
{
    test();
    return 0;
}

Answer 1

One thing to note here is – When we say pointers, we generally tend to think in terms of pass by reference but not necessarily. Even pointers can be passed by value

char* str is local to test and char* p is local to setmemory . So the changes you do in setmemory will not be visible in test if you dont send a pointer to a pointer.

You can make it work with a single pointer like this

 char * setmemory(char* p, int num) // p is a new pointer but points at the same
                                    // location as str
{
    p=(char*)malloc(num); // Now, 'p' starts pointing at a different location than 'str'
    strcpy(p ,"hello");  // Copy some data to the locn 'p' is pointing to
    return p; // Oops. The poor `str` is still pointing at NULL :( 
              // Send him the address of the newly allocated area
}

void test(void)
{
    char* str=NULL;
    str=setmemory(str,100); // We are passing a pointer which is pointing to NULL

    printf(str); //Now str points to the alloced memory and is happy :)
}

int main()
{
    test();
    return 0;
}

Note that in setmemory we are returning a local pointer, but it is not a problem ( no dangling pointer problems ) as this pointer points to a location on heap and not on stack