What is ‘array decaying’ in C?

It is sometimes said that arrays in C are basically pointers. This is not true. What is true is that a value of an array type can decay to a value of a pointer type.

Take this:

#include <stdio.h>

void takes_arr_pointer_1(int* arr) {
  printf("in takes_arr_pointer_1, sizeof(arr) = %d\n", sizeof(arr));
}

void takes_arr_pointer_2(int arr[]) {
  printf("in takes_arr_pointer_2, sizeof(arr) = %d\n", sizeof(arr));
}

int main() {
  int arr[100];
  printf("in main, sizeof(arr) = %d\n", sizeof(arr));
  takes_arr_pointer_1(arr);
  takes_arr_pointer_2(arr);
  return 0;
}

This prints:

% ./a.out
in main, sizeof(arr) = 400
in takes_arr_pointer_1, sizeof(arr) = 8
in takes_arr_pointer_2, sizeof(arr) = 8

In the context of main, the arr variable is an array of 100 ints. On my machine, sizeof(int) = 4, so sizeof(arr) = 400.

In the context of both other functions, the arr variable is a pointer to an int. Thus, sizeof(arr) is the size of a pointer, which on my machine is 8 bytes.

We were able to pass the int arr[100] to both functions, even though they accept pointers. The conversion of arr from an array type to a pointer type is known as array decaying, i.e. the array has “decayed” to a pointer.

The two parameter definitions int* arr and int arr[] are actually the same. Perhaps this is where the confusion comes from.

This page copyright James Fisher 2016. Content is not associated with my employer. Found an error? Edit this page.

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