How to print variable addresses in C?

Question

When i run this code.

#include 

void moo(int a, int *b);

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

    x = 1;
    y = &x;

    printf(\"Addr         


        

Answer 1

Looks like you use %p: Print Pointers

Answer 2

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);

Answer 3

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!

Answer 4

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;
}

Answer 5

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);
}