difference between sizeof(array) before and after passing to function in C

Question

y is an array of integers with 10 members (size=40 Bytes) but when I pass it to my_function() as x , x has 10 members but if I try to get it's size (sizeof(x)) I will get 8 Bytes as result.

Could you please explain why this happens ?

void my_function(int x[]) {
  printf("%d\n", sizeof(x)/sizeof(x[0]));
}

int main() {

  int y[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
  printf("%d\n", sizeof(y)/sizeof(y[0]));  // Output : 10
  my_function(y);  // Output : 2

  return 0;
}

my_function() just gets a pointer to an integer which is 8 bytes on your system. It has no way of knowing the size of the structure it is pointing to. — jmq, Oct 09 '19 at 19:40
`sizeof` shows the size of the *type*, not the data sent in. The type doesn’t have size of the array so it is just a pointer and there’s no size information available. You always need to give the size to the function separately — Sami Kuhmonen, Oct 09 '19 at 19:45

score 0 · Accepted Answer · answered Oct 09 '19 at 19:44

Arrays decay to pointers when passed as function parameters, so in my_function the statement printf("%d\n", sizeof(x)/sizeof(x[0])); is equivalent to printf("%d\n", sizeof(int*)/sizeof(int));. On your machine and with your compiler, the size of the pointer is equal to twice the size of an int.

difference between sizeof(array) before and after passing to function in C

1 Answers1