What's happening here in this for loop [closed

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Closed 7 years ago.

I was trying something else but suddenly stuck with this infinite loop . Please suggest an answer with explanation that what's going on here in the below for loop

#include<stdio.h>

int main()
{
 int x=0;
 int i;
 int array[5];
 for(i=0;i<=5;i++)
 {
  array[i]=x;
  printf("#%d value set in index %d\n",x,i);
 }

 return 0;
}

When I remove = sign in the condition of for loop It works fine.

But when I put this it goes to infinite loop , Why? Accessing extra element (more than its limit) in array is undefined behaviour or what ? Any help will appreciated. Thanks in advance.

~

Answer 1

Your for loop should go from 0 to 4 as your array has 5 elements

#include<stdio.h>

int main()
{
 int x=0;
 int i;
 int array[5];
 for(i=0;i<5;i++)
 {
  array[i]=x;
  printf("#%d value set in index %d\n",x,i);
 }

 return 0;
}

Answer 2

You are trying to store 6 int values in an array of size 5, which is illegal.

So when you try to write at the 6th position of the array, you are writing into the variable i with the value of x (which is 0). Now in your next iteration since i is 0, it is less than the condition specified in the for loop. This causes your loop to run all over again and again!

Answer 3

Why you're going into an infinite loop, in this specific case, is actually pretty easy to understand, take a look at the addresses on your stack:

int main( ) 
{
    int x = 0;
    int i;
    int array[5];
    printf("&x = %#x, &i = %#x, array = %#x, array+4 = %#x\n", &x, &i, array, array+4);

The result of this printf() will show you the addresses of your variables and the start and end of the array:

&x = 0xbfac9cec, &i = 0xbfac9ce8, array = 0xbfac9cd4, array+4 = 0xbfac9ce4

So in order, your stack looks like:

 var        address
***********************
array[0]    0xbfac9cd4
array[1]    0xbfac9cd8
array[2]    0xbfac9cdc
array[3]    0xbfac9ce0
array[4]    0xbfac9ce4
  i         0xbfac9ce8
  x         0xbfac9cec

Now your loop is writing 0-5 (6 elements), there are only 5 elements in your array so writing to the 6th actually overwrites the next thing on the stack which is i in this case. That makes this line:

array[i]=x;

The same as writing this:

i = x;

That will store 0 (in your case) to i, and restart the loop, so you're going to see it loop forever and print "0 stored to index 0", then "index 1", 2, 3, 4 then restart again when you set i=x;

Answer 4

Undefined behavior happens when i==5. array has valid indexes 0..4 - arrays in C are 0-based.

If you replace <= with <, you iterate through valid indexes.

Accessing extra element (more than its limit) in array is undefined behaviour or what ?

Yes.

Answer 5

You're writing 6 ints to an array with space for 5. The 6th write is outside the bounds of the array so its effect is unpredictable. In your case, its writing to the next sizeof(int) bytes of stack. This is the memory used for i, the loop counter, which gets reset to 0.

As you say in your question, the fix for this is to replace the <= exit condition of your for loop with <.

Answer 6

To avoid causing errors like this as easily, here's two good rules for writing for loops over actual (local) arrays:

1. Iteration starts at index 0, since C's arrays are 0-based.
2. Use <, always. Never <=.
3. Don't repeat the size, let the compiler compute it using sizeof array / sizeof *array. Note the asterisk in the second term.

So, this loop should have been written:

for(i = 0; i < sizeof array / sizeof *array; i++)

and then you would have been safe.

Note that this only works for "real" arrays that have a size visible to sizeof, if you've let the array "collapse" into a pointer it won't work.

Also note that sizeof is not a function, so no ()s are necessary around its argument in cases like these.