urgent! variable address

my teacher gave out an assignment pertaining to variable address. she wants us to add something to the program to display the address of the variables a, b, and c. and i can't seem to figure out how to do it and i'm on a tight schedule and i need it before May 11,2009. here is the program...

Code:

#include<stdio.h>

void main(){
int a,b,c,i;
a=1;b=2;c=3;

trace(&a,&b,c);printf("%d %d %d\n",a,b,c);
trace(&a,&b,c);printf("%d %d %d\n",a,b,c);
trace(&c,&b,a);printf("%d %d %d\n",a,b,c);

for(i=1;i<=3,i++)
    trace(&a,&b,c);
printf("%d %d %d\n",a,b,c);
getch();}


void trace(int *x, int *y, int z){
int t;

z+=*x;
*x=*y+z;
t=z;
z=*y;
*y=*x;
*x=t;
}

pls help me...

You know how to print an integer variable:

Code:

printf("%d", yourInteger);

What does this do?

Code:

printf("%p, &yourInteger);

A few things to note:

• Your schedule is your problem. Avoiding telling the world that you urgently need an answer by a given date, especially since the more malicious among us would only give you an answer after the stated date, whereas if they did not know that it was urgent, they would urgently reply.
• The main function should be declared with a return type of int, not void. Correspondingly, you should return 0 at the end of the main function, but this is optional with respect to the 1999 edition of the C standard.
• You should indent your code properly and consistently. Along the same lines, it is usually bad practice to place multiple statements on the same line.
• You should declare (i.e., provide a function prototype for) the trace function before calling it, as you do in the main function.
• You should not use getch since it is non-standard and unnecessary.

Now, on to your problem: you are looking to use the %p format specifier for printf. This format specifier is for printing a pointer to void, so all you need to do is to use it as you use the %d format specifier, except that now you would cast the address of the desired variable to void*.

EDIT:

Originally Posted by Adak

What does this do?

It results in undefined behaviour.

Originally Posted by Adak

What does this do?

Code:

printf("%p, &yourInteger);

Originally Posted by laserlight

It results in undefined behaviour.

As a matter of fact, it will simply fail to compile.

Originally Posted by grumpy

As a matter of fact, it will simply fail to compile.

As a matter of fact it compiles without errors or warnings, and runs just fine!

Code:

#include <stdio.h>

int main(void) {
  int i = 0;

  printf("%p", &i);

  i = getchar();
  return 0;
}

Originally Posted by Adak

As a matter of fact it compiles without errors or warnings, and runs just fine!

Code:

#include <stdio.h>

int main(void) {
  int i = 0;

  printf("%p", &i);

  i = getchar();
  return 0;
}

I think grumpy was talking about

Code:

printf("%p, &yourInteger);//without ending quotes

It sounds to me as a typo but actually you wrote it in your previous post.
And yes the code you wrote above works without any error and warning. Although casting is not used.

Originally Posted by Adak

As a matter of fact it compiles without errors or warnings, and runs just fine!

Originally Posted by BEN10

And yes the code you wrote above works without any error and warning. Although casting is not used.

According to the MinGW port of gcc 3.4.5:

Code:

test.c: In function `main':
test.c:6: warning: void format, different type arg (arg 2)

You should increase your warning levels and/or test with a range of compilers. Yet, a clean compile at the highest warning level does not indicate that undefined behaviour is not involved.

Ah! The missing double quotation char, strikes again! << Just seeing if Grumpy was awake >> j/k

Why would I want or need to cast the address of the variable, before printing it?

Laserlight, what do you need to code up on your compiler, to print out the address of a variable, which has no warnings or errors?

My compiler's warnings are turned all the way up, already. I never turn them down.

Originally Posted by Adak

Why would I want or need to cast the address of the variable, before printing it?

You need to cast because %p is used to print a pointer to void, but you want to print a pointer to int. Okay, but you might not need to cast, but then you would be relying on undefined behaviour.

Originally Posted by Adak

Laserlight, what you you need to code up on your compiler, to print out the address of a variable, which has no warnings or errors.

What I described in post #3: a cast to void*.

Code:

#include <stdio.h>

int main(void) {
  int i = 0;

  printf("%p", (void*)&i);

  i = getchar();
  return 0;
}