Algorithm for insertion in one-dimensional array, what is happening here?

Question

#include <stdio.h>

int main()
{
int arr[5], size, i, pos, elem;

printf("Enter the size of array(less than 100)\n");
scanf("%d",&size);

printf("\nEnter the elements of array one by one\n");

for ( i = 0; i < size; i++)
scanf("%d",&arr[i]);

printf("\nEnter the position of insertion\n");
scanf("%d",&pos);

printf("\nEnter the element\n");
scanf("%d",&elem);

for ( i = size - 1 ; i >= pos - 1; i--)
    arr[i + 1] = arr[i];

arr[pos - 1] = elem;

printf("\nInserted array is\n");

for ( i = 0; i <= size; i++)
    printf("\t%d",arr[i]);
    printf("\n");



return 0;
}

Since the array is int arr[5] it shouldn't be possible to insert an element at any index not in range [0,1,2,3,4]. Why am I able to insert at arr[6] and get correct answer still?.

It is undefined behaviour. You can't expect any particular outcome from running this program. — juanchopanza, Feb 14 '15 at 17:47
There's nothing to stop you from writing beyond the limits of an array, but it will lead to [*undefined behavior*](http://en.wikipedia.org/wiki/Undefined_behavior). — Some programmer dude, Feb 14 '15 at 17:47

score 1 · Accepted Answer · answered Feb 14 '15 at 17:47

Accessing array out of bound leads to undefined behvaior.

You might see it working sometimes and might not work sometimes.(You might end up with a crash even)

So just go by the standard and stop accessing array out of bound. Get rid of the undefined behavior

score 0 · Answer 2 · answered Feb 14 '15 at 17:47

You are writing to memory that you did not allocate. Eventually, when you hit memory that you do not have write permissions for, you will get a segmentation fault.

C/C++ doesn't do bounds checking for you like other languages such as Java. It leaves it to the operating system to figure out when you're accessing memory you're not supposed to.

Algorithm for insertion in one-dimensional array, what is happening here?

2 Answers2