Why Does `sizeof` Give Different Array Sizes Inside and Outside a C Function?

Why Array Size Calculation Differs Inside a Function in C

When passing an array to a function, C automatically decays the array into a pointer to its first element. This pointer carries the address of the first array element but not its size.

Consider this example:

#include <stdio.h>

void test(int arr[]) {
    int arrSize = (int)(sizeof(arr) / sizeof(arr[0])); // Incorrect
    printf("%d\n", arrSize);
}

int main() {
    int point[3] = {50, 30, 12};

    int arrSize = (int)(sizeof(point) / sizeof(point[0])); // Correct
    printf("%d\n", arrSize); // Prints 3

    test(point);

    return 0;
}

In main, sizeof(point) correctly gives the size of the entire array (12 bytes). However, within test, sizeof(arr) yields 4 bytes (the size of an integer pointer), resulting in an incorrect array size calculation (missing one element).

To resolve this, the array size must be explicitly passed as a separate parameter:

void test(int arr[], size_t elems) {
    int arrSize = elems;
    /* ... */
}

int main() {
    int point[3] = {50, 30, 12};
    /* ... */
    test(point, sizeof(point) / sizeof(point[0]));
    /* ... */
}

Note that sizeof(point)/sizeof(point[0]) works for stack-allocated arrays but not for dynamically allocated arrays since it relies on the underlying decay mechanism.

The above is the detailed content of Why Does `sizeof` Give Different Array Sizes Inside and Outside a C Function?. For more information, please follow other related articles on the PHP Chinese website!