QUESTION 15 · Arrays and Functions

Why pass the length separately with an array parameter?

2 marks

Answer

A parameter written as int a[] is adjusted to an int * parameter and does not carry the array’s element count.

Pass n separately and ensure it does not exceed the actual array capacity.
sizeof on that parameter gives the pointer size, not the original array size.

Explain passing a one-dimensional array to a function with its length.

5 marks

Answer in brief

A function can process an array when it receives access to its elements and enough information to keep its accesses within bounds.

Function interface

long long total(const int a[], int n);

Meaning

In a parameter declaration, const int a[] is adjusted to const int *a. The function receives a pointer value identifying the first element. The array contents are not automatically copied. n states how many elements the function is allowed to read. const prevents modification through this parameter’s element access.

Example trace

main holds 4, 5 and 6 and calls total(a, 3). The function visits a[0], a[1] and a[2], adding them to its local accumulator. It returns 15. main prints the result. The array remains unchanged because this function only reads it.

Important points

Pass a valid length that fits the actual object.
Do not use sizeof a / sizeof a[0] inside this adjusted parameter to discover the caller’s array length.
A non-const int parameter written as int a[] can be used to modify the caller’s elements.
C still passes a value: the pointer is copied, while the elements remain in the original array.

Complexity

The function reads n elements once, giving O(n) time. It does not build a second array; only a sum and counter are added, giving O(1) auxiliary space. Copying a pointer argument does not mean copying n elements.

C program and output

Download .c Open compiler

/* CodeBhavya | PPS AR25 | Learning example */
#include <stdio.h>
long long total(const int a[], int n)
{
    long long sum = 0;
    for (int i = 0; i < n; ++i) { sum += a[i]; }
    return sum;
}

int main(void)
{
    int a[3] = {4, 5, 6};
    printf("Total = %lld\n", total(a, 3));
    return 0;
}

Sample input and output

This example uses values already written in the program.

Output

Total = 15