Arrays in C program

Question

Here is my C Program Code :

#include<stdio.h>

unsigned int count = 1; 

int main(void)
{
  int b = 10;
  int a[3];
  a[0] = 1;
  a[1] = 2;
  a[2] = 3; 

  printf("\n b = %d \n",b);
  a[3] = 12; 
  printf("\n b = %d \n",b); 

  return 0;
}

And the output as

b=10 
b=12

Can anyone explain why it is and the reason behind the error.

Answer 1

You are writing beyond the bounds of the array. This would invoke undefined behavior .

Answer 2

You are writing at a[3] = 12; to an out of bound index, which is UB and since its UB it means anything can happen so chances are you have just over-written the value at a[3] which is the location of b .

Answer 3

Your program has undefined behavior . Depending on your result, your memory looked like this:

  a[0]   a[1]   a[2]    b   
+------+------+------+----+
|  1   |  2   |  3   | 12 |
+------+------+------+----+

b is the same as a[3] since it "sits" next to it in the memory.

This is only one possible scenario, your program could has a "segmentation fault" or anything else as well.

Answer 4

It's because you're accessing a[3] , which is the 4th element, but you only defined an array of size 3. As a result, you're getting a pointer to garbage data, which happens to point to the memory location of b . So when you assign a[3] , you're assigning the memory location of b :

#include<stdio.h>

unsigned int count = 1; 

int main(void)
{
  int b = 10;
  int a[3];
  a[0] = 1;
  a[1] = 2;
  a[2] = 3; 

  printf("\n b = %d \n",b);
  a[3] = 12; 

  printf("\n b = %d \n",b); 

 return 0;
}

The pointer memory locations point to the same thing for a[3] and b :

b = 10 
a[3]:0xa
b:0xa

b = 12 

Now try actually creating an array with four elements. Change a[3] to a[4] , so that a[3] isn't a garbage location. You'll see it works.

Answer 5

The so-called "Undefined Behaviour" can include many kinds of behaviour. Amongst them can be:

• nothing bad happens
• you get a segmentation fault (or any other access violation)
• you destroy/tamper other data structures.

You have the latter here: the variable b happens to be in memory directly after the array, so it is modified due to this wrong access.

Answer 6

Maybe you can define a pointer to the address of a[3], and output it. Then you will know the answer.

Answer 7

because if you prints address of both variable b and a[3] like,

printf("%u",&b);  
printf("\n %u",&a[3]); 

then it gives same output for both variable address

Answer 8

Your program results in undefined behavior, answers here explain this "undefined behavior" trying to give some meaning to your output,which is not always true. In-fact i try to run your program and I can see b is printed 10 in both printf and finally i get run time error (I used Visual Studio 2010)