Array index out of bound in C

Question

Why does C differentiates in case of array index out of bound

#include 
int main()
{
    int a[10];
    a[3]=4;
    a[11]=3;//doe         


        

Answer 1

The segfault is not an intended action of your C program that would tell you that an index is out of bounds. Rather, it is an unintended consequence of undefined behavior.

In C and C++, if you declare an array like

type name[size];

You are only allowed to access elements with indexes from 0 up to size-1. Anything outside of that range causes undefined behavior. If the index was near the range, most probably you read your own program's memory. If the index was largely out of range, most probably your program will be killed by the operating system. But you can't know, anything can happen.

Why does C allow that? Well, the basic gist of C and C++ is to not provide features if they cost performance. C and C++ has been used for ages for highly performance critical systems. C has been used as a implementation language for kernels and programs where access out of array bounds can be useful to get fast access to objects that lie adjacent in memory. Having the compiler forbid this would be for naught.

Why doesn't it warn about that? Well, you can put warning levels high and hope for the compiler's mercy. This is called quality of implementation (QoI). If some compiler uses open behavior (like, undefined behavior) to do something good, it has a good quality of implementation in that regard.

[js@HOST2 cpp]$ gcc -Wall -O2 main.c
main.c: In function 'main':
main.c:3: warning: array subscript is above array bounds
[js@HOST2 cpp]$

If it instead would format your hard disk upon seeing the array accessed out of bounds - which would be legal for it - the quality of implementation would be rather bad. I enjoyed to read about that stuff in the ANSI C Rationale document.

Answer 2

Just to add what other people are saying, you cannot rely on the program simply crashing in these cases, there is no gurantee of what will happen if you attempt to access a memory location beyond the "bounds of the array." It's just the same as if you did something like:

int *p;
p = 135;

*p = 14;

That is just random; this might work. It might not. Don't do it. Code to prevent these sorts of problems.

Answer 3

That's not a C issue its an operating system issue. You're program has been granted a certain memory space and anything you do inside of that is fine. The segmentation fault only happens when you access memory outside of your process space.

Not all operating systems have seperate address spaces for each proces, in which case you can corrupt the state of another process or of the operating system with no warning.

Answer 4

As JaredPar said, C/C++ doesn't always perform range checking. If your program accesses a memory location outside your allocated array, your program may crash, or it may not because it is accessing some other variable on the stack.

To answer your question about sizeof operator in C: You can reliably use sizeof(array)/size(array[0]) to determine array size, but using it doesn't mean the compiler will perform any range checking.

My research showed that C/C++ developers believe that you shouldn't pay for something you don't use, and they trust the programmers to know what they are doing. (see accepted answer to this: Accessing an array out of bounds gives no error, why?)

If you can use C++ instead of C, maybe use vector? You can use vector[] when you need the performance (but no range checking) or, more preferably, use vector.at() (which has range checking at the cost of performance). Note that vector doesn't automatically increase capacity if it is full: to be safe, use push_back(), which automatically increases capacity if necessary.

More information on vector: http://www.cplusplus.com/reference/vector/vector/