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In this array insertion code, the value of usedsize does not get updated when I increment it inside the void insert function. I want to know why that keeps happening and why the value of arr[i] gets updated when this doesn't?

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

void insert(int arr[], int element, int index, int usedsize){
    for(int i=usedsize;i>=index;i--){
        arr[i+1]=arr[i];
    }
    arr[index]=element;
    usedsize++;

}

void traverse(int arr[], int size){
    for(int i=0;i<size;i++){
        printf("%d",arr[i]);
        printf("\t");
    }

}
int main()
{
    int narr[25]={5,7,12,34,45};
    int index = 3;
    int usize = 5;
    int element = 21;
    insert(narr,element,index,usize);
    traverse(narr,usize);

    return 0;
}
EfCross
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    It's a copy of the passed variable, if you want to update the value the function argument must be a pointer. – anastaciu May 04 '22 at 19:26
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    C literally passes arguments only by value. You need to pass a reference explicitly. – the busybee May 04 '22 at 19:42
  • @EfCross "In this array insertion code, the value of usedsize does not get updated when I increment it inside the void insert function." You are mistaken. The variable usedsize is updated within the function.:) – Vlad from Moscow May 04 '22 at 19:49
  • Arrays are converted to a pointer to the first (zeroth) element of the array — and the key term is 'pointer'. You have to explicitly pass a pointer to an integer such as `usize` in `main()` — `insert(narr, element, index, &usize);` and `void insert(int arr[], int element, int index, int *usedsize)` (and then use `*usedsize` in the function). – Jonathan Leffler May 04 '22 at 20:03
  • Does anyone have a good canonical duplicate for the problem of (not) passing pointers to functions in order to get the value updated in the calling function? – Jonathan Leffler May 04 '22 at 20:04

1 Answers1

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Like the commenters noted, the variable usedsize in the insert function is a copy.

why does the value of arr[i] get updated?

In C, the array name indicates the address of first element and arrays are always passed as pointers, even if you use the square bracket notation 0.

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

void insert(int arr[], int element, int index, int *usedsize) {
    printf("Address arr[%d] is %p\n", 0, &arr[0]);
    printf("Address usedsize is %p\n", usedsize);
    for(int i=*usedsize; i>=index; i--) {
        arr[i+1]=arr[i];
    }
    arr[index]=element;
    (*usedsize)++;
}

int main() 
{
    int narr[25]= {5,7,12,34,45};
    int index = 3;
    int usize = 5;
    int element = 21;
    printf("Address arr[%d] is %p\n", 0, &narr[0]);
    printf("Address usize is %p\n", &usize);
    insert(narr,element,index,&usize);
    traverse(narr,usize);
    return 0;
}

Original code:
- Address arr[0] is 0x7ffdf2f8b510
- Address usize is 0x7ffdf2f8b504
- Address arr[0] is 0x7ffdf2f8b510
- Address usedsize is 0x7ffdf2f8b4cc
Now:
- Address arr[0] is 0x7fff0c03d740
- Address usize is 0x7fff0c03d734
- Address arr[0] is 0x7fff0c03d740
- Address usedsize is 0x7fff0c03d734
aloha_erich
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