Unexpected behaviour of a c program?

Question

This code is not giving the output as expected and clearly i cant understand the program behaviour.Plz help me in understanding this.Any Help will be appreciated. You can see the output of the program https://code.hackerearth.com/31d954z?key=fce7bc8e07fc10a3d7a693e45a5af94a here.

1.In the last comment, I can't find why the elements of array are not updated.

2.In the body of func on printing 'a' it gives some unexpected output.

For e.g., if i pass j = 2 and a[0] = 1

       After j = j+1 , which results in j = 3;

       a=a+j should result in a = 4

       but instead it result in a = 13.

#include <stdio.h>

void func(int j,int *a)
{
 j=j+1;
 printf("%d\t%d\n",a,j);     //prints j updated value and a
 a=a+j;
 printf("a = %d\n ",a);      //prints value of a
}


void main()
{
 int a[5] ={1,2,3,4,5},i,j=2;
 for (i =0;i<5;i++ )
 func(j, a[i]);
 for (i =0;i<5;i++ )   
 printf("%d\t",a[i]);    //Prints the array as 1 2 3 4 5
}

On running this code,Output is :

1 3 // Here a = 1 and j = 3 a = 13 //but after addition a = 13 2 3 a = 14 3 3 a = 15 4 3 a = 16 5 3 a = 17 1 2 3 4 5 //array elements not updated

Answer 1

I want to understand the code behavior.

Your code produces Undefined Behavior, thus you should stop what you are doing and debug it.

---

When you want to index arrays you do it like this:

a[i]

where i is the index and a your array. So if you want to access the first element, you need to do a[0], when you want to index the 3rd element, you do a[2] and so on.

---

However, what you may want to do is something like passing the i-th element, add it and print it, only.

So, you should enable your compiler warnings:

prog.c: In function 'func':
prog.c:6:11: warning: format '%d' expects argument of type 'int', but argument 2 has type 'int *' [-Wformat=]
  printf("%d\t%d\n",a,j);     //prints j updated value and a
          ~^
          %ls
prog.c:8:15: warning: format '%d' expects argument of type 'int', but argument 2 has type 'int *' [-Wformat=]
  printf("a = %d\n ",a);      //prints value of a
              ~^
              %ls
prog.c: At top level:
prog.c:12:6: warning: return type of 'main' is not 'int' [-Wmain]
 void main()
      ^~~~
prog.c: In function 'main':
prog.c:16:10: warning: passing argument 2 of 'func' makes pointer from integer without a cast [-Wint-conversion]
  func(j, a[i]);
          ^
prog.c:3:6: note: expected 'int *' but argument is of type 'int'
 void func(int j,int *a)
      ^~~~

and then modify your code accordingly, for example like this:

#include <stdio.h>

void func(int j,int a)
{
 j=j+1;
 printf("%d\t%d\n",a,j);   
 a=a+j;
 printf("a = %d\n ",a);
}


int main(void)
{
 int a[5] ={1,2,3,4,5},i,j=2;
 for (i =0;i<5;i++ )
 func(j, a[i]);
 for (i =0;i<5;i++ )   
 printf("%d\t",a[i]);
}

which outputs:

1   3
a = 4
 2  3
a = 5
 3  3
a = 6
 4  3
a = 7
 5  3
a = 8
 1  2   3   4   5