How to print variable addresses in C?

Question

When i run this code.

#include <stdio.h>

void moo(int a, int *b);

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

    x = 1;
    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);
    moo(9, y);
}

void moo(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);
}

I keep getting this error in my compiler.

/Volumes/MY USB/C Programming/Practice/addresses.c:16: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
/Volumes/MY USB/C Programming/Practice/addresses.c:17: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int **’
/Volumes/MY USB/C Programming/Practice/addresses.c:17: warning: format ‘%d’ expects type ‘int’, but argument 3 has type ‘int *’
/Volumes/MY USB/C Programming/Practice/addresses.c: In function ‘moo’:
/Volumes/MY USB/C Programming/Practice/addresses.c:23: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int *’
/Volumes/MY USB/C Programming/Practice/addresses.c:24: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘int **’
/Volumes/MY USB/C Programming/Practice/addresses.c:24: warning: format ‘%d’ expects type ‘int’, but argument 3 has type ‘int *’

Could you help me?

Thanks

blargman

Those say "warning", not "error". That means that your program will still run. But do fix the warnings per the answer below. — Jim Balter, Mar 13 '11 at 00:13

score 100 · Accepted Answer · edited Aug 31 '15 at 13:28

100

You want to use %p to print a pointer. From the spec:

p The argument shall be a pointer to void. The value of the pointer is converted to a sequence of printing characters, in an implementation-defined manner.

And don't forget the cast, e.g.

printf("%p\n",(void*)&a);

edited Aug 31 '15 at 13:28

Giorgi Moniava

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answered Mar 12 '11 at 23:57

Carl Norum

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that's what i originally thought but the tutorial on this website (http://www.cis.temple.edu/~ingargio/cis71/code/addresses.c) is telling me to use %d... is it wrong. – nambvarun Mar 12 '11 at 23:58
3

@blargman, yes, it's wrong. You might be able to coerce things into working by typecasting, but since `%d` is for printing signed integers, it's probably not a good choice. – Carl Norum Mar 13 '11 at 00:00
8

+1: ... but for portability (and to follow the **SHALL** from the Standard) don't forget to cast the address. `printf("%p\n",(void*)&a);` – pmg Mar 13 '11 at 00:07
3

@pmg, is the cast necessary? I thought conversions to and from `void *` are safe and automatic in C (6.3.2.2 paragraph 1). – Carl Norum Mar 13 '11 at 00:12
@pmg what do u mean by portability? – nambvarun Mar 13 '11 at 00:12
By portability is meant the possibility to compile/run the program with different compilers/operating systems/computers. http://en.wikipedia.org/wiki/Portability_%28software%29 – hlovdal Mar 13 '11 at 00:15
Thanks @hlovdal; that is indeed what I mean. – pmg Mar 13 '11 at 00:19
6

@Carl: in variadic functions the compiler cannot check the types of arguments against the expected types. The cast to `void*` is not automatic: what is automatic in variadic functions is a few low-range integers to `int`, and `float` to `double`. – pmg Mar 13 '11 at 00:24
@pmg - that makes sense. I found more about it in 6.5.2.2 after asking here. Have you ever worked on a machine where `void *` was incompatible with any other pointer types? – Carl Norum Mar 13 '11 at 00:28
@Carl: And similar to variadic functions, if you happen to call a function which the compiler have not seen a prototype for, it will assume int for all parameters, and sizeo(int) might be different from sizeof(void *), so it is important to be aware of these two cases. This specifically also applies to `NULL` which might be defined as just `0` which is of type int. E.g. `ret = execlp("ls", "ls", "-l", (void *)NULL);`. – hlovdal Mar 13 '11 at 00:36
@Carl: no, I have only worked with well-behaved machines. In ancient days, with TurboC (or whatever) I had a few programs where function pointers and data pointers had different size: `sizeof (void(*)()) != sizeof (void*)` – pmg Mar 13 '11 at 00:39
Actually the man page for execlp also says this: "The list of arguments must be terminated by a NULL pointer, and, since these are variadic functions, this pointer must be cast (char *) NULL." – hlovdal Mar 13 '11 at 00:41
@pmg where do u learn these fascinating concepts? Could u please reference me to some books. Thanks for the fascinating discussion. – nambvarun Mar 13 '11 at 00:44

score 9 · Answer 2 · answered Jun 13 '13 at 06:20

9

When you intend to print the memory address of any variable or a pointer, using %d won't do the job and will cause some compilation errors, because you're trying to print out a number instead of an address, and even if it does work, you'd have an intent error, because a memory address is not a number. the value 0xbfc0d878 is surely not a number, but an address.

What you should use is %p. e.g.,

#include<stdio.h>

int main(void) {

    int a;
    a = 5;
    printf("The memory address of a is: %p\n", (void*) &a);
    return 0;
}

Good luck!

answered Jun 13 '13 at 06:20

Ron

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The value `0xbfc0d878 ` *is* a number. `(void*)0xbfc0d878` is not. And `%p` is likely to use a human-readable representation that looks like a number (typically hex) but that doesn't mean pointers are numbers. (BTW, the question was answered more than 2 years ago.) – Keith Thompson Jun 13 '13 at 06:38
Why do we have to cast to `(void*)`? – Cristian Gutu Feb 01 '19 at 16:23

score 1 · Answer 3 · edited Mar 20 '18 at 06:44

1

To print the address of a variable, you need to use the %p format. %d is for signed integers. For example:

#include<stdio.h>

void main(void)
{
  int a;

  printf("Address is %p:",&a);
}

edited Mar 20 '18 at 06:44

Vinz

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answered Mar 20 '18 at 01:48

Amarendra Deo

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While this code may answer the question, providing additional context regarding **how** and **why** it solves the problem would improve the answer's long-term value. – Alexander Mar 20 '18 at 03:15

score 0 · Answer 4 · answered Mar 12 '11 at 23:59

0

Looks like you use %p: Print Pointers

answered Mar 12 '11 at 23:59

skaz

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score 0 · Answer 5 · answered Apr 22 '20 at 14:02

I tried in online compiler https://www.onlinegdb.com/online_c++_compiler

int main()
{
    cout<<"Hello World";
    int x = 10;
    int *p = &x;
    printf("\nAddress of x is %p\n", &x); // 0x7ffc7df0ea54
    printf("Address of p is %p\n", p);    // 0x7ffc7df0ea54

    return 0;
}

How to print variable addresses in C?

5 Answers5

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