How to print variable addresses in C?

c pointers memory-address

164,962

Solution 1

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);

Solution 2

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!

Solution 3

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);
}

164,962

Author by

nambvarun

Updated on July 09, 2022

Comments

nambvarun almost 2 years

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

nambvarun over 13 years

that's what i originally thought but the tutorial on this website (cis.temple.edu/~ingargio/cis71/code/addresses.c) is telling me to use %d... is it wrong.
Carl Norum over 13 years

@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.
pmg over 13 years

+1: ... but for portability (and to follow the SHALL from the Standard) don't forget to cast the address. printf("%p\n",(void*)&a);
Carl Norum over 13 years

@pmg, is the cast necessary? I thought conversions to and from void * are safe and automatic in C (6.3.2.2 paragraph 1).
nambvarun over 13 years

@pmg what do u mean by portability?
hlovdal over 13 years

By portability is meant the possibility to compile/run the program with different compilers/operating systems/computers. en.wikipedia.org/wiki/Portability_%28software%29
pmg over 13 years

Thanks @hlovdal; that is indeed what I mean.
pmg over 13 years

@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.
Carl Norum over 13 years

@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?
hlovdal over 13 years

@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);.
pmg over 13 years

@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*)
hlovdal over 13 years

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."
nambvarun over 13 years

@pmg where do u learn these fascinating concepts? Could u please reference me to some books. Thanks for the fascinating discussion.
Keith Thompson about 11 years

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.)
Alexander over 6 years

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.
Cristian Gutu over 5 years

Why do we have to cast to (void*)?