How is memory allocation being done in this C program?

Question

#include<stdio.h>
#include<stdlib.h>
int main()
{
    int *a[10];
    a[2] = (int*)malloc(sizeof(int));
    a[2][3]=4;
    printf("%d", a[2][3]);
    return 0;

}

I have given only the memory equivalent of a single int to the pointer variable. How is that I am able to access an element at index 3 for that pointer of a single int?

[Please don't cast the return value of `malloc()` in C](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc). — unwind, Feb 05 '16 at 10:23
Use a memory debugger to catch such issues, such as BoundsChecker or Valgrind. — jxh, Feb 05 '16 at 10:24
I am not trying to be a klutz here. I just saw this C code, and didn't know if array bounds checking was even optional for languages, as I was naturally expecting an error. — goelakash, Feb 05 '16 at 10:38
@goelakash Everything is optional for languages, basically. :) It's up to the language designers and implementors to shape the semantics of the language (within the bounds of what the computer can do, of course). — unwind, Feb 05 '16 at 10:40
Accessing unallocated memory is a subtle because it will end up in **undefined** behavior. Your program can: 1. **Crash Immediately** (*in case you request memory that does not belong to your process*) 2. **Temporarily works** 3. **Works perfectly for the next 100 years.** The final behavior depends on many factors, such as the **combination of compilers/hardware/system**. You example is very simple and works because you are accessing few bytes out of bounds and this is very likely to be a portion of memory that is reserved (but not allocated) to your process — Davide Spataro, Feb 05 '16 at 10:41

score 1 · Answer 1 · answered Feb 05 '16 at 10:22

1

You can try but the behaviour of your program will be undefined.

C does not perform that kind of checking at runtime.

answered Feb 05 '16 at 10:22

Bathsheba

231,907
34
361
483

score 1 · Accepted Answer · answered Feb 05 '16 at 10:22

1

That's because nothing is preventing you accessing arrays out of bound and this invokes undefined behavior. Any expected or unexpected behavior of program can be seen.

answered Feb 05 '16 at 10:22

haccks

104,019
25
176
264

score 1 · Answer 3 · answered Feb 05 '16 at 10:23

You're accessing some randomly set RAM element. This is a common C programming mistake and is one of the reason for many security flaws.

The Heartbleed bug would be a nice example, where people were able to read a good portion of the server RAM, by accessing elements outside of the array structure.

How is memory allocation being done in this C program?

3 Answers3