help me

[indocoding.com]

i am computer newbie, i use linux for my work, i wanna learn c programming language, but i have warning with compailing on linux, please tell me about pointer variable on linux,
example code

#include <stdio.h>

void rischan(int a, int * b);

int main(void) {
  int x;
  int *y;

  x=2;
  y=&x;
  printf("Address of x = %d, value of x = %d\n", &x, x);
  printf("Address of y = %d, value of y = %d, value of *y = %d\n", &y, y, *y);
  rischan(6,y);
}

void rischan(int a, int *b){
  printf("Address of a = %d, value of a = %d\n", &a, a);
  printf("Address of b = %d, value of b = %d, value of *b = %d\n", &b, b, *b);
}

the warning like this

c.c: In function ‘main’:
c.c:11: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
c.c:12: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int **’
c.c:12: warning: format ‘%d’ expects type ‘int’, but argument 3 has type ‘int *’
c.c: In function ‘rischan’:
c.c:17: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
c.c:18: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int **’
c.c:18: warning: format ‘%d’ expects type ‘int’, but argument 3 has type ‘int *’

please tell me...why...?..thanks, and iam sorry my english is poor...

[quzah]

Try:

printf("Address of x = %p, value of x = %p\n", (void *)&x, (void *)x);

Quzah.

[Adak]

Your program is telling printf() to print the value of the pointer (which is an address), as an int. Of course, addresses use the %p format, not %d, because addresses are not int's.

Welcome to the forum, indocoding.com!

Your English is perfectly understandable.