How to print variable addresses in C?

问题

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

回答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);

回答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!

回答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);
}

回答4:

Looks like you use %p: Print Pointers

来源：https://stackoverflow.com/questions/5286451/how-to-print-variable-addresses-in-c

标签

pointers

memory-address