Output does not print the correct variable addresses - arrays

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:
0.000000
0.000000
0.000000
0.000000
0.000000
0.000000

Answer 1

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

Answer 2

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

Answer 3

maybe make the printf of for loop like this:
for(int i=0;i<5;i++)
{
printf("%lf \n",p[i]++);
}
OUTPUT:
0.000000
6.000000
7.000000
8.000000
9.000000
10.000000