Getting unexpected output in C while accessing array. Getting mostly 0 as output

Question

This is my code

#include<stdio.h>


int main()
{
  int n,a[n],op;
  printf("Enter size of the array\n");
  scanf("%d",&n);
  printf("Enter the array elements\n");
  for(int i=0;i<n;i++)
  {
    scanf("%d",&a[n]);
  }
  printf("Select array element to display from 1 to %d\n",n);
  scanf("%d",&op);
  if(op-1>=0&&op-1<n)
  {
    printf("Element is %d\n",a[op-1]);
  }
  else
  {
    printf("Invalid Entry\n");

  } 
  return 0;
}

Output

Enter size of the array 
5
Enter the array elements
2
5
6
1
5
Select array element to display from 1 to 5
2
Element is 0

Why is this happening? I couldn't figure this out

paxdiablo · Answer 1 · 2021-08-17T22:57:01.303

Consider your code:

int n,a[n],op;

Given that the value of n is undefined at this point, the variable length array (VLA) created by a[n] will be of some arbitrary size. C doesn't magically reach back in time from the point where you request a value of n and adjust the array size :-)

You need to create the VLA after you've set the value of n, reorganising your code at the start of main with something like this:

int n,op;
printf("Enter size of the array\n");
scanf("%d",&n);
int a[n];

That way, n will be a known value when you use it to create the VLA.

Additionally, your scanf("%d",&a[n]); is undefined behaviour because n will always be beyond the end of the array (which has indexes 0 through n-1 inclusive). You should be using i as the index there.

There are other logic issues such as checking that the scanf succeeded and that you didn't enter a non-positive number but that's usually okay for educational code (assuming that's what this is). As you develop as a coder, you will naturally begin to think of these possibilities and harden your code accordingly.

By way of example, this is a better way to enter n, although you could recover from non-numeric data as an optional extra (see here for a nifty line input solution in C):

int n = -1;
while (n < 1) {
    printf("Enter array size (positive integer): ");
    if (scanf("%d", &n) != 1) {
        puts("\nERROR: non-numeric");
        exit(1);
    }
}

@RyanKafoor, in your example you have static memory allocation, because you allocate your array on stack. And dynamic memory allocation could be a solution: you can declare 'int *a' and after you got n, dynamicly allocate array with malloc. — Igor_M, Aug 17 '21 at 14:14
@RyanKafoor: dynamic allocation *is* possible, you just need to ensure the size is correct. That's the case whether it's dynamic allocation from `malloc` or via VLAs (which are arguably not dynamic but we'll not get into that discussion here). — paxdiablo, Aug 17 '21 at 22:58

score 2 · Answer 2 · answered Aug 17 '21 at 13:38

2

scanf("%d",&a[n]);

should be scanf("%d",&a[i]);

answered Aug 17 '21 at 13:38

heysujal

118
7

True, however, friends don't let friends use scanf(). – Jim D Aug 17 '21 at 13:44
@Jim, there are certain *uses* of `scanf` that are problematic, such as with `%s`, but the function itself is not unsafe. – paxdiablo Aug 17 '21 at 23:08
@paxdiablo One of the key issues when using scanf() is that conversion errors can be challenging to detect. Quoting from the man page "return the number of input items...this can be fewer than provided for, or even zero, in the event of an early matching failure." The second issue is that scanf() is a bloated function (which can be important on memory constrained platforms). Finally, scanf() is fundamentally an interpreter and presents a sizable attack surface. There are much more reliable alternatives for this type of input. – Jim D Aug 18 '21 at 01:22

programmerc3981143 · Answer 3 · 2021-08-17T14:17:40.197

The error is in this part of your code:

  int n,a[n],op; 
  
  // int n is not defined as anything so it is a garbage value
  // you specify the size of the array to be n
  // which allocates an undefined size for the array

Also in your first loop you reassign &a[n], which is undefined since it goes beyond the size of the array and even if it were valid it would just reassign a single element, it should be &a[i].

to fix it read the value of n before defining array 'a'. The changes i made are down below:

#include<stdio.h>


int main()
{
  int n; // change
  printf("Enter size of the array\n");
  scanf("%d",&n);

  int a[n],op; // change

  printf("Enter the array elements\n");
  for(int i=0;i<n;i++)
  {
    scanf("%d",&a[i]); // change
  }
  printf("Select array element to display from 1 to %d\n",n);
  scanf("%d",&op);
  if(op-1>=0&&op-1<n)
  {
    printf("Element is %d\n",a[op-1]);
  }
  else
  {
    printf("Invalid Entry\n");

  } 
  return 0;
}

Getting unexpected output in C while accessing array. Getting mostly 0 as output

3 Answers3