Issues on using pointers

Question

I just got started learning about pointers and was executing some code:

#include <stdio.h> int main(void){ int num = 12; // initialize num to 12 int *pnum = NULL; // initialize the pointer pnum = &num; // stores the address of num in pnum printf("the address of the number is %p\n", &num); // display the address of the number printf("the address of the pointer is %p\n", &pnum); // display the address of the pointer printf("the value of the pointer is %p\n", pnum); // display the value of the pointer printf("the value the pointer is pointing to is %d\n", *pnum); // display the value the pointer is pointing to return 0; }

In this code above, It prints out 0xffffcbec for the address of the number and value of the pointer, and 0xffffcbe0 for the address of the pointer. I want to know the reason. I feel like this is related to some incorrect inputs of data types.

I use VScode btw.

I want to know the reason. The reason for what? Its quite unclear what you're asking. — tkausl
– tkausl, Commented Aug 3, 2022 at 15:11
The reason it prints 0xffffcbe0 for the address of pnum is because the address of pnum is 0xffffcbe0. — William Pursell
– William Pursell, Commented Aug 3, 2022 at 15:22
pointer is a var himself and so on have a dedicated address where it store the pointed address. so, 0xffffcbe0 hold your pointer data and his content store 0xffffcbec. it seem that the linker has stored pointer var before your int in your program memory, all is nice. — r043v
– r043v, Commented Aug 3, 2022 at 15:23

Some programmer dude · Accepted Answer · 2022-08-04 12:37:35Z

The value of pnum is the value of &num, i.e. the location of the variable num. Therefore &num and pnum will print the same.

To make it easier to understand and visualize, I recommend you draw it all out:

 +-------+ +------+ +-----+ | &pnum | ---> | pnum | ---> | num | +-------+ +------+ +-----+

That is:

&pnum is pointing to pnum
pnum (which is the same as &num) is pointing to num

Also, the pointer-to operator & and the dereference operator * are each others opposites.

So pnum is the value of &num. And *pnum is the value of num.

And *&num is plain num, as the operators cancel out each other.

I am sorry everyone who confused on my question. I was gonna clarify my question but stackoverflow was crashing at the time. SO, I was wondering, is there nothing wrong with my code? Because I thought it should print 8-bits but 0xffffcbec looks like more than 8-bits( 8-bits means basically 8 characters right?) and I assumed that there was an error. I want to know if the code is how it is supposed to happen.
@TrevorSong A bit is a single 0 or 1. A byte is a collection of (usually) 8 bits. An address on a 32-bit system is 4 bytes, and 8 bytes on a modern 64-bit system. A single hexadecimal digit represents 4 bits.

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