Thread: why function ignores the parameter

I don't know why function ignores the parameter in my code and loop run 10 times. I expect it should be run 5 times only

Code:

 #include<stdio.h>

void foo ( int i)
{
 for (i = 0; i < 10; i++)
 {
     printf("%d\n", i);
 }
}


int main(void)
{   
    foo(5);
    return 0;
}

Code:

0
1
2
3
4
5
6
7
8
9

5 is passed but you are re-assigning the value with 0 in the for loop.

Try this version:

Code:

#include<stdio.h>

void foo ( int i)
{
 for (; i < 10; i++)
 {
     printf("%d\n", i);
 }
}


int main(void)
{
    foo(5);
    return 0;
}

Originally Posted by rstanley

5 is passed but you are re-assigning the value with 0 in the for loop.

Thanks for clearing doubts

This is my code to print array

Code:

#include <stdio.h>

int main (void)
{
 int i, array[] = {1,2,3,4,5};
 
 for (i = 0; i < 5; i++)
 {
  printf("array element : %d  \n", array[i]);
 }
 return 0;
}

just now changed line inside the loop. I wonder why the line inside the loop printing the values form 0 to 4

Code:

 #include <stdio.h>

int main (void)
{
 int i, array[5] = {1,2,3,4,5};
 
 for (i = 0; i < 5; i++)
 {
  printf("array element : %d  \n", array[5]);
 }
 return 0;
}

Code:

array element : 0
array element : 1
array element : 2
array element : 3
array element : 4

The subscripts of a 5 element array are 0 - 4, not 1 - 5!

It is illegal to access any "element" outside of the array, so:

Code:

printf("array element : %d  \n", array[5]);

Is illegal!!! You are accessing data outside of the array, and you are lucky in this case by actually accessing the loop variable 'i'!!!

When I compile your second example on Linux with gcc version 12.2.0, it prints 5 zeros!

Doctor, Doctor, it hurts when I access data outside of the array!

Don't do it! ;^)