What does this number stand for

Question

I was simply trying out this code which is saw in a tutorial

#include <stdio.h>

int main() {
    int a = 34;
    int *ptra = &a;
    printf("%d\n", ptra);

    return 0;
}

Here when I run this code I get a number as an output. The number was 6422216 .

What does this number stand for?

(I was trying to learn pointer arithmetics)

Answer 1

It is the address of your variable a

let's understand using the above diagram where int a=10 is referred using pointer ptr

int a=10;
int *ptr=&a;
printf("%d",*ptr); //10 which is value of variable a
printf("%p",(void *)ptr);  //0x7fffa0757dd4 which is the address of variable a

Answer 2

The behavior of the code sample is actually undefined because you pass a value of type int * where printf expects an int as the argument for conversion %d .

The code may be corrected as printf("%p\\n", (void *)ptra); or possibly printf("%llu\\n", (long long)ptra);

The value printed is the address of the variable a , which is system specific and may change from one run of the program to another, as is the case on OS/X that implements address space randomisation to increase the difficulty for hackers to exploit some program flaws.

To cut a long story short, the answer to your question What does this number stand for? is the number is the numeric value of the address of variable a which is stored in the pointer ptr . Since addresses can be wider than the int type, you should use a longer type such as long long , preferably unsigned:

printf("%llu\n", (long long)ptra);

Answer 3

Your program prints out the memory address of where the integer a is stored and I'll explain why but firstly you need to know what the symbols stand for.

* - is used to declare a pointer and when placed before a address it will give you what is stored at that address

& - will give you the address of the variable written after it

So, first you assign the integer a value of 34

int a = 34;

Then you create a pointer of a integer called ptra and assign it the memory address of where the integer a is stored

int *ptra = &a;

And finally you print the value of ptra which is the memory address of where the integer a is stored

printf("%d\n", ptra);

If you would wan to print out the value of a through the integer, then you should replace ptra with *ptra in the printf function.