Sizeof array through function in C

Question

I'm not sure why I cannot use sizeof(array) when passing the array through my function only outputs a value of 1, instead of 1000000. Before passing the array to the function, I printed out the sizeof(array) to get 4000000 and when I try to printout the sizeof(array) in the function, I only get 4. I can iterate through the array in both the function and the main to display all values, I just cannot display sizeof within the function. Did I pass the array through incorrectly?

#include <stdio.h>
#include <stdlib.h>

int
searchMe(int numbers[], int target)
{
    printf("%d\n", sizeof(numbers));
    int position = -1;

    int i;
    int k = sizeof(numbers) / sizeof(numbers[0]);   

    for (i = 0; i < k && position == -1; i++)
    {
        if (numbers[i] == target)
        {
            position = i;
        }
    }
    return position;
}

int
main(void)
{
    int numbers[1000000];
    int i;
    int search;
    for (i = 0; i < 1000000; i++) {
        numbers[i] = i;
    }

    printf("Enter a number: ");
    scanf("%d", &search);

    printf("%d\n", sizeof(numbers));
    int position = searchMe(numbers, search);

    printf("Position of %d is at %d\n", search, position);
    return 0;
}

Answer 1

int searchMe(int numbers[], int target)

is equivalent to

int searchMe(int *numbers, int target)

There is a special C rules for function parameters that says parameters of array type are adjusted to a pointer type.

It means in your program that sizeof numbers actually yields the size of the int * pointer type and not of the array.

To get the size you have to add a third parameter to your function and explicitly pass the size of the array when you call the function.