Thread: Pointer question

Code:

#include<stdio.h>


int main (void)
{
	int number = 10;
	int *pointer;
    
	pointer = &number;
	
    printf("The value of var is: %d \n", number);
 
    printf("The address of var : %p \n", &pointer);
	
	pointer++;
	
	
	printf("The address of var : %p \n", &pointer);


 
    return 0;
}

The value of var is: 10
The address of var : 6422312
The address of var : 6422312

so address of variable is 6422312 and int is 4 bytes
if i increment pointer it should be 6422316
Why it show the same address after incrementing pointer ?

First off do you realize that you're not printing the address of *pointer but you're trying to print the address of **pointer?

Perhaps you should see about increasing your compiler warning levels and fixing any warnings, such as:

||=== Build: Debug in chomework (compiler: GCC 8-1) ===|
main.c||In function ‘main’:|
main.c|13|warning: format ‘%p’ expects argument of type ‘void *’, but argument 2 has type ‘int **’ [-Wformat=]|
main.c|18|warning: format ‘%p’ expects argument of type ‘void *’, but argument 2 has type ‘int **’ [-Wformat=]|
||=== Build finished: 0 error(s), 2 warning(s) (0 minute(s), 0 second(s)) ===|

Two things, first do you see that "but argument 2 has type ‘int **’" part of the warning. It is telling you that you are passing a double pointer, not a single pointer (caused by that misplaced ampersand). Second you are using an incorrect type for the format specifier, invoking undefined behaviour, you need to cast that variable to a void* when using the "%p" format specifier.

Originally Posted by jimblumberg

First off do you realize that you're not printing the address of *pointer but you're trying to print the address of **pointer?

Perhaps you should see about increasing your compiler warning levels and fixing any warnings, such as:

@jimblumberg my compiler doesn't show any warnings

Code:

#include<stdio.h> 
 
int main (void)
{
    int number = 10;
    int *pointer;
     
    pointer = &number;
     
    printf("The value of var is: %d \n", number);
  
    printf("The address of var : %p \n", &pointer);
     
    pointer++;
     
     
    printf("The address of var : %p \n", pointer);
 
 
  
    return 0;
}

Below is the output of code

C:\Users\Vajra\Desktop>gcc -o test test.c


C:\Users\Vajra\Desktop>test
The value of var is: 10
The address of var : 0061FF28
The address of var : 0061FF30

Originally Posted by Salem

Try it with
gcc -Wall -Wextra -o test test.c

It tried this and it doesn't give any warning

back to the question

The value of var is: 10
The address of var : 0061FF28
The address of var : 0061FF30

if i increment pointer it should be 0061FF32 int data type is 4 bytes why it is become 0061FF30

Code:

#include<stdio.h> 
  
int main (void)
{
    int number = 10;
    int *pointer;
      
    pointer = &number;

    printf("The size of an int is %zu\n", sizeof(int));  /* added this line just to make sure int is 4 "bytes" */
      
    printf("The value of var is: %d \n", number);
   
    printf("The address of var : %p \n", pointer);  /* Fixed this line here */
      
    pointer++;
      
      
    printf("The address of var : %p \n", pointer);
  
  
   
    return 0;
}

Output:

Code:

The size of an int is 4
The value of var is: 10 
The address of var : 0xffba888c 
The address of var : 0xffba8890

I don't see the problem now: 0xffba888c + 4 = 0xffba8890

Originally Posted by Hodor

Thanks hodor
Output:

Code:

The size of an int is 4
The value of var is: 10 
The address of var : 0xffba888c 
The address of var : 0xffba8890

I don't see the problem now: 0xffba888c + 4 = 0xffba8890

understood

The address of var : 0xffba888c = 4290414732 decimal
The address of var : 0xffba8890 = 4290414736 decimal