Random elements showing up on the array in c

Question

So just like in python list I wanted to try and implement using c (I am new to c programming). This is my code

// File name:- list.c

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

int main() 
{
    int myNumbers[] = {25, 50, 75, 100};
    // int length = len(myNumbers);
    push(myNumbers, 200, len(myNumbers));
    for(int i = 0; myNumbers[i]; i++){
        printf("array[%d] = %li\n", i, myNumbers[i]);
    };
    len(myNumbers);

    return 0;
}


// File name:- l.c

#include "l.h"
#include <stdio.h>

int len(int arr[])
{
    int i;
    for(i = 0; arr[i]!='\0'; i++){
        continue;
    };
    printf("Length is: %d\n", i);
    return i;
}

void push(int list[], int value, int length)
{
    // int length = len(list);
    list[length] = value;
}

The above code does give me the result I expect, i.e

Length is: 4
array[0] = 25
array[1] = 50
array[2] = 75
array[3] = 100
array[4] = 200
Length is: 5

Whereas when int myNumbers[] = {25, 50, 75, 100, 125}; or anything more than 4 values in the array...

The Result is given unexpected random values like:-

Length is: 5
array[0] = 25
array[1] = 50
array[2] = 75
array[3] = 100
array[4] = 125
array[5] = 200
array[6] = 2782528512
array[7] = 1952226512
array[8] = 1
Length is: 9

How to fix this issue? I had even tried by directly passing the length and even calling the function without passing the length, but none of them worked... I went through the code for any logic error, I wasn't able to find any..

I expect this as my result...

Length is: 5
array[0] = 25
array[1] = 50
array[2] = 75
array[3] = 100
array[4] = 125
array[5] = 200
Length is: 6

Answer 1

C does not automatically terminate arrays with zero (except for string literals). If you want a zero to mark the end of your array, you must put it there.

C does not automatically grow arrays. Your push routine will not make space in the array for a new element. It will corrupt your program.

To tinker with pushing numbers onto a list like that, you can declare myNumbers to have a lot of space, as with int myNumbers[100] = {25, 50, 75, 100, 0}; to reserve space for 100 elements and to mark the end of the first few with a zero. As you learn more about C, you will learn about other ways to manage memory.

Answer 2

The array defined by int myNumbers[] = { 25, 50, 75, 100 }; has exactly 4 elements and does not have a null terminator. Hence calling len() or trying to push an element both have undefined behavior.

You could define myNumbers as int myNumbers[100] = { 25, 50, 75, 100 }; and your code would work up to 99 elements because the remaining elements will have been initialized to 0.

Note also that the printf format specifier %li expects a long int argument value, which myNumbers[i] is not. You should use %i or %d to output this int value as a decimal number.

Answer 3

The issue in your code is related to array bounds and memory access. When you declare an array in C like this: int myNumbers[] = {25, 50, 75, 100};, it has a fixed size of 4 elements. You cannot directly add elements to it beyond its initial size without causing undefined behavior, which is why you are getting unexpected random values.

To fix this issue, you need to handle dynamic memory allocation to increase the size of the array when you want to add more elements. Here's a modified version of your code to achieve this using dynamic memory allocation:

// File name: list.c

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

int main() 
{
    int* myNumbers = (int*)malloc(4 * sizeof(int)); // Allocate memory for 4 elements
    int length = 4;
    myNumbers[0] = 25;
    myNumbers[1] = 50;
    myNumbers[2] = 75;
    myNumbers[3] = 100;
    
    push(myNumbers, 125, &length);
    push(myNumbers, 200, &length);
    
    for (int i = 0; i < length; i++) {
        printf("array[%d] = %d\n", i, myNumbers[i]);
    }
    
    free(myNumbers); // Free the dynamically allocated memory
    
    return 0;
}

// File name: l.c

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

void push(int* list, int value, int* length)
{
    list[*length] = value;
    (*length)++;
}

// File name: l.h

#ifndef L_H
#define L_H

void push(int* list, int value, int* length);

#endif

In this version, the myNumbers array is allocated dynamically using malloc, and its size is initially set to 4. The length variable keeps track of the number of elements in the array. When you call the push function, it adds the new element to the end of the array and increments the length variable. This way, you can add elements beyond the initial size without any issue.

Remember to free the dynamically allocated memory using free(myNumbers) when you're done with the array to avoid memory leaks.

Answer 4

you can declare int myNumbers[] with more space like -

int myNumbers[100] = {25, 50, 75, 100, 125};