Can't pass a char array to a function

Question

I am having difficulty passing a char array to a function:

This is code is found inside a function, which calls another function createbitshape:

char ans[8];
char bitshp[8];
...
fgets(ans, 10, stdin);
createbitshape(random_num, bitshp);
printf("bitshp outside: %p\n", &bitshp);

Here is createbitshape:

void createbitshape(int n, char bitshp[]){
    int i, count;
    count = 0;
    i = 1<<(sizeof(n) * 2 - 1);
    for ( ; i > 0; i >>=1 )
    {
        if (n & i)  /* check if any of the bits of n is not 0 .*/
            bitshp[count] = '1';
        else
            bitshp[count] = '0';
        count++;
    }
    bitshp[8] = '\0';
    printf("bitshp inside: %p\n", &bitshp);

The prototype is:

 void createbitshape(int, char[]);

When I run the code, I see two different addresses of bitshp:

bitshp inside: 0x7fff854d8b80
bitshp outside: 0x7fff854d8b70

How come ? Does createbitshape allocates another memory room ? How do I change this code such that createbitshape writes the content to bitshp defined in the calling function?

(p.s. I know similar questions have been already asked, but I simply don't get how to translate the answers there to my case...)

Answer 1

Your confusion is because the assumption is incorrect, there is no pass-by-reference in C, only pass-by-value.

When arrays are passed as function argument, they are automatically converted to a pointer that points to their first element. It looks like pass-by-reference as you can modify what the pointer points, but it's not, it's still pass-by-value, the value of the pointer itself.

That means the function argument bitshp is a new pointer, it's supposed to have a different address than the bitshp outside the function. But they do share the same value.

Answer 2

Yu Hao answer is great, I'm glad you accepted it.

I'd like to complement it with a brief explanation on why the addresses differ inside and outside. Its a combination of two things:

• Array variables decay to pointers when they are passed as function arguments:

  static size_t my_sizeof(char array[32]) {
      return sizeof(array);
      /* Always returns either 4 or 8 depending on your architecture */
  }
  

• When you use the address-of operator on an array, it will return the same address:

  char array[8];
  printf("%p %p\n", array, &array);
  /* Will output the same value twice */