Thread: Passing an Array to a function

Well strings are arrays of characters.

So when you say
char foo[ ] = "hello";
strlen( foo );

If you look up strlen, you'll see it takes a char* pointer.

An identical call would be
strlen( &foo[0] );

Also, since arrays are always passed as a pointer, whatever you do to the array inside the function (via the pointer), the effect will be stored in the array declared by the caller.

You don't pass the whole array to a function. You pass a pointer to the first element in the array to the function. Try something like this:

Code:

void func(int nums[10])
{
  int i;

  for(i = 0;i < 10;++i)
    nums[i] = i;
}

int main(void)
{
  int i;
  int nums[10];

  func(nums);  // An equivalent call would be func(&nums[0]);
  for(i = 0;i < 10;++i)
    printf("&#37;d\n", nums[i]);

  return 0;
}

The parameter in func() can actually be expressed in different ways, and the size of the last dimension can actually be omitted. These definitions would all work:

Code:

void func(int nums[10])
void func(int nums[])
void func(int *nums)

Since you're passing a pointer to the first element in the array, your function doesn't have its own local copy of the array. Any changes you make to the array in your function are reflected in the calling function.