Initializing arrays in C

Question

The following C program creates an array which has fixed number of elements. Even when we assign three more value which exceeding the number of maximum array elements works fine.

#include <stdio.h>

int main(){

//declares a fixed sized array
int balance[5];

//initializing
balance[0] = 1;
balance[1] = 2;
balance[2] = 3;
balance[3] = 4;
balance[4] = 5;
//three more
balance[5] = 6;
balance[6] = 7;
balance[7] = 8;


printf("%d \n", balance[0]);
printf("%d \n", balance[1]);
printf("%d \n", balance[2]);
printf("%d \n", balance[3]);
printf("%d \n", balance[4]);
printf("%d \n", balance[5]);
printf("%d \n", balance[7]);

//and it works fine!

return 0;
}

But when we try to assign more than three element it throws an exception

#include <stdio.h>

int main(){

//declares a fixed sized array
int balance[5];

//initializing
balance[0] = 1;
balance[1] = 2;
balance[2] = 3;
balance[3] = 4;
balance[4] = 5;
//three more
balance[5] = 6;
balance[6] = 7;
balance[7] = 8;
//one more again
balance[8] = 9;


printf("%d \n", balance[0]);
printf("%d \n", balance[1]);
printf("%d \n", balance[2]);
printf("%d \n", balance[3]);
printf("%d \n", balance[4]);
printf("%d \n", balance[5]);
printf("%d \n", balance[7]);
printf("%d \n", balance[8]);

//throws an exception!

return 0;
}

But after declaring and initializing an array by using integer literals we can insert more elements without getting an error.

#include <stdio.h>

int main(){

//declares and initialize a fixed sized array
int balance[5] = {1,2,3,4,5};

//let's add few more
balance[5] = 6;
balance[6] = 7;

//and even more
balance[7] = 8;
balance[8] = 9;
balance[9] = 10;

printf("%d \n", balance[0]);
printf("%d \n", balance[1]);
printf("%d \n", balance[2]);
printf("%d \n", balance[3]);
printf("%d \n", balance[4]);
printf("%d \n", balance[5]);
printf("%d \n", balance[6]);
printf("%d \n", balance[7]);
printf("%d \n", balance[8]);
printf("%d \n", balance[9]);

//and it works fine!

return 0;
}

What is the reason for this behavior of C language?

Answer 1

Your code invokes Undefined Behavior as you try to access array out of bounds, which is unsafe memory, and hence the outcome is undefined.

Answer 2

You are accessing unallocated memory which will invoke undefined behavior. Anything could happen. These are all out of bound access

balance[5] = 6;
balance[6] = 7;
balance[7] = 8;
balance[8] = 9;

Answer 3

Array out of bound access leads to undefined behavior. Your array can hold 5 integer elements so the valid index for your array is 0 to 4 anything other than this is array out of bound access.

Answer 4

I'd like to give a more generic answer. Everyone here got' the right answer, but i want to explain why & be a little more lengthy about the explanation. Let's say you have a char array of 3 cases.

  char tab[3];

In terms of memory, what we got is a 3 cases of data allocated. Since it's an array, it's alignated in your memory. Let's initialise that array :

tab[0] = 'A';
tab[1] = 'B';
tab[2] = 'C';

In your memory, it looks something like that :

Now, what happens if you write tab[6] ? Since it's C, you can clearly do that, no problem whatsoever. (There are no electric-fence ;) )

In that case, tab[6] has not been allocated to your program. You can have any scenario from now on :

• Maybe this data is critic to the system, and your program wont have access to that memory (ie Segmentation Fault)
• Maybe your programm has another array somewhere, and you could get the luck of the draw and actually get something out of there.
• (others thing worst than that)
• ???

Basically when you provoke an Out of Bounds error, you cause unexpected behaviour. What you should do is always look after thoses mistakes are they can be deadly killers of your beloved application.