First time with pointers in C (problem in output)

Question

0.00/5 (No votes)

See more:

Hello everyone!!!
Today, I started using pointers in C language and I have a problem in the output.

I have this
[^]error[^]

I tried to change %d to %ls as it says ,but in the output it doesn't show me the addresses .
Where the problem is?
Thanks in advance !!!

What I have tried:

Code:

C++

#include <stdio.h>

int main()
{
    int x;
    int *p;
    p = &x;

    *p = 75;
    printf("price of x = %d ,\t price of pointer p = %d ,\t address of x => %d ,\t address of pointer p = %d \n" , x , *p , &x , p);

    x = 999;
    printf("price of x = %d ,\t price of pointer p = %d ,\t address of x => %d ,\t address of pointer p = %d \n" , x , *p , &x , p);

    *p++;
    printf("price of x = %d ,\t price of pointer p = %d ,\t address of x => %d  ,\t address of pointer p = %d \n" , x , *p , &x , p);

    return 0;
}

Posted 20-Jul-20 8:19am

Nick_is_asking

Updated 20-Jul-20 20:45pm

KarstenK

v2

Add a Solution

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

OriginalGriff · Accepted Answer · 2020-07-20T08:37:00

Solution 1

Try this:

C++

printf("Price of x = %d\nPrice via pointer p = %d\nAddress of x = %p\nAddress pointed to by p = %p\n", x, *p, &x, p);

Posted 20-Jul-20 8:37am

OriginalGriff

Comments

Nick_is_asking 20-Jul-20 14:45pm

It works. Thank you!!!

Nick_is_asking 20-Jul-20 14:52pm

One more question.If x = 8 and I write this: *p++; then I recieve this:
price of x = 9 , price of pointer p = -971355120 , address of x => 0x7ffec61a4c0c , address of pointer p = 0x7ffec61a4c10
=============================
.but I expected this:
price of x = 9 , price of pointer p = 9 , address of x => 0x7ffec61a4c0c , address of pointer p = 0x7ffec61a4c10

or am I wrong?

OriginalGriff 20-Jul-20 15:00pm

You are wrong: it's down to operator precedence. The "postincrement operator" (p++) is higher than the "dereference operator" (*p) so *p++ increments the pointer, not the content of the pointer, then returns the value that now points to. If you want to increment x via p, then you need brackets to force the order you want: (*p)++

Rick York 20-Jul-20 16:34pm

you should avoid writing expressions like that because they can be prone to error. It is better to first dereference the pointer (*p) and use that value. Then increment the pointer when you are done with its value.

Nick_is_asking 20-Jul-20 15:19pm

ok I understand, thanks!!

OriginalGriff 20-Jul-20 15:52pm

You're welcome!

CPallini 20-Jul-20 16:03pm

5.