I have written this code and I am not getting any error, so I came to know that it's entirely programmer's botheration to keep a check bounds check. but I am confused by the fact that in static memory allocation the int a[10] has allocated i.e. 40 bytes during compile time so how come it can allocate more memory to it during runtime, because it's also printing the same.
#include<stdio.h>
main()
{
int a[10];
for(i=0;i<=100;i++)
{
a[i]=i;
}
}
Nope, this is undefined behavior. And well that's why it may work and it may not. (Depends on the moon). You should avoid it. Don't think that this is making your program work by allocating more memory - it doesn't.
If you write this, you are writing in some part of your program's memory which will in turn result in errorneous result or if it is trying to write to some memory which is non-modifiable, you will get error. In either way you are invoking undefined behavior - anything may happen from crashing your code to a correct execution. Don't rely on that.
Well it is writing to the memory out of the bound of the array where it is not supposed to write. That is what is happening. That is why this is not what you should be doing. You are basically changing part of memory that program is being assigned and moreover which you are not supposed to change.
