C Passing Arrays to Functions

You can pass a whole array to a function. In the function parameter we write the array name with empty brackets. We should also send the size, because the function does not know it.

Example: sum of an array

#include <stdio.h>

int sum(int arr[], int size) {
    int total = 0;
    for (int i = 0; i < size; i++) {
        total += arr[i];
    }
    return total;
}

int main(void) {
    int marks[] = {70, 85, 60, 95};

    printf("Total = %d\n", sum(marks, 4));
    return 0;
}

Output:

The function changes the original array

An array is not copied. The function gets the address of the first element. So if the function changes a value, the original array also changes.

#include <stdio.h>

void addBonus(int arr[], int size) {
    for (int i = 0; i < size; i++) {
        arr[i] += 5;
    }
}

int main(void) {
    int marks[] = {70, 85, 60};

    addBonus(marks, 3);
    for (int i = 0; i < 3; i++) {
        printf("%d ", marks[i]);
    }
    printf("\n");
    return 0;
}

Output:

Why sizeof does not work inside the function

Inside the function arr is only a pointer. So sizeof(arr) gives the size of a pointer, not of the array. Calculate the size in the place where the array is declared, and pass it as an argument.

Passing a 2D array

For a 2D array, the column size must be written in the parameter.

#include <stdio.h>

void show(int m[][3], int rows) {
    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < 3; j++) {
            printf("%d ", m[i][j]);
        }
        printf("\n");
    }
}

int main(void) {
    int m[2][3] = {{1, 2, 3}, {4, 5, 6}};
    show(m, 2);
    return 0;
}

Output:

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