Output does not print the correct variable addresses

Question

Despite using double as datatype and correct format specifiers the output does not print the correct variable addresses output consists of just zeros

#include <stdio.h>

void main() {
    double a[5] = { 6.0, 7.0, 8.0, 9.0, 10.0 };
    double *p;
    p = a;
    printf("%lf \n", p);
    for (int i = 0; i < 5; i++) {
         printf("%lf \n", p++);
    }
}

Output:

You can't use `%lf` to print pointers. Are you trying to print the pointers themselves, or the pointed-to values? — Steve Summit, Feb 06 '23 at 17:02
@SteveSummit thank you so much I understood my mistake I was trying to print the address not value , hence using %d fixed it! — Sharanya Thambi, Feb 06 '23 at 17:26
@SharanyaThambi Okay, but in that case, you want `%p`. You should not use `%d` to print addresses, and on an increasing number of machines, it won't work at all. — Steve Summit, Feb 06 '23 at 17:31

score 0 · Answer 1 · answered Feb 06 '23 at 17:10

I fixed it with 2 simple changes, described in the comments.

#include<stdio.h>
void main()
{
    double a[5]={6.0,7.0,8.0,9.0,10.0};
    double *p;
    p=a;
    printf("%p \n",p);       // Changed formatter to %p  to print addresses
    for(int i=0;i<5;i++)
    {
         printf("%p \n",p++); // Changed formatter to %p to print addresses
    }
}

Output

0x7fffdd887250 
0x7fffdd887250 
0x7fffdd887258 
0x7fffdd887260 
0x7fffdd887268 
0x7fffdd887270

`printf("%p \n",p);` --> `printf("%p \n",(void*)p);` – Support Ukraine Feb 06 '23 at 18:04 — Support Ukraine, Feb 06 '23 at 18:04

Vlad from Moscow · Answer 2 · 2023-02-06T17:46:33.103

These calls of printf

printf("%lf \n",p);

and

printf("%lf \n",p++);

invokes undefined behavior because there is used the wrong conversion specifier "%lf" (where by the way the length modifier has no effect even when used with object of the type double) with pointers.

Either you should use the conversion specifier %p like

printf("%p \n", ( void * )p);

and

printf("%p \n", ( void * )p++);

Or if you want to output addresses as integers you can include headers <stdint.h> and <inttypes.h> (the last header already includes the first header and write

#include <stdint.h> #include <inttypes.h>

//...

printf("%" PRIuPTR "\n", ( uintptr_t )p);

and

printf("%" PRIuPTR "\n", ( uintptr_t )p++);

Pay attention to that according to the C Standard the function main without parameters shall be declared like

int main( void )

Here is a demonstration program.

#include <stdio.h>
#include <stdint.h>
#include <inttypes.h>

int main( void )
{
    double a[] = { 6.0, 7.0, 8.0, 9.0, 10.0 };
    const size_t N = sizeof( a ) / sizeof( *a );

    double *p = a;


    printf( "%p", ( void * )p );
    printf( " <-> %" PRIuPTR "\n", ( uintptr_t )p );

    for ( size_t i = 0; i < N; i++ )
    {
        printf( "%p", ( void * )p );
        printf( " <-> %" PRIuPTR "\n", ( uintptr_t )p++ );
    }
}

The program output is

00AFFE40 <-> 11533888
00AFFE40 <-> 11533888
00AFFE48 <-> 11533896
00AFFE50 <-> 11533904
00AFFE58 <-> 11533912
00AFFE60 <-> 11533920

score 0 · Answer 3 · edited Feb 09 '23 at 12:19

0

maybe make the printf of for loop like this:

for(int i=0;i<5;i++)
{
printf("%lf \n",p[i]++);
}

OUTPUT:

edited Feb 09 '23 at 12:19

buddemat

4,552
14
29
49

answered Feb 06 '23 at 21:52

Ecug Uehx

1

Output does not print the correct variable addresses

3 Answers3

Output