what is teh output of the snippet code qnd how??

Question

#include <stdio.h>

int main() {

static int a[]={10,20,30,40,50};
static int *p[]={a,a+3,a+4,a+1,a+2};
int **ptr=p;
ptr++;
printf("%d%d",ptr-p,**ptr);
return 0;

}

Answer 1

#include <stdio.h>

int main() {
static int a[]={10,20,30,40,50};
printf("%d %d %d\n", *a, a[0], a[3]); // 10 10 40
printf("%p %p %p\n", a, &a[0], &a[3]); // 0x404040 0x404040 0x40404c
printf("%ld\n", sizeof(int)); // 4
static int *p[]={a,a+3,a+4,a+1,a+2};
printf("%p %p %p\n", *p, p[0], p[1]); // 0x404040 0x404040 0x40404c
printf("%p %p %p\n", p, &p[0], &p[1]); // 0x404060 0x404060 0x404068
printf("%ld\n", sizeof(int*)); // 8
int **ptr=p;
printf("%p %p\n", ptr, p); // 0x404060 0x404060
ptr++; // 0x404060 add 1 * sizeof(int*)
printf("%p %p\n", ptr, p); // 0x404068 0x404060
printf("%d %ld\n", 0x404068 - 0x404060, ptr-p); //8 1, I don't know the reason, subtract two array pointer will get the number of elements between them.(address1 - address2)/sizeof(element)
printf("%p %p %d\n", ptr, *ptr, **ptr); // 0x404068 0x40404c 40
printf("%ld %d",ptr-p,**ptr); //1 40
return 0;
}

Answer 2

Martin Taylor i always run the codes before asking but the procedure of pointers here was vague forme, i know the output but not the procedure and debugging.

Answer 3

The tip is very explicit.

printf("%ld %d", ptr-p, **p);

./Playground/file0.c: In function 'main':
./Playground/file0.c:15:10: warning: format '%d' expects argument of type 'int', but argument 2 has type 'long int' [-Wformat=]
15 | printf("%d %d", ptr-p, **p);
| ~^ ~~~~~
| | |
| int long int
| %ld

Answer 4

output is my question not the sololearn's compiler error.
i donno why the output is 140 𝓕𝓛𝓨

Answer 5

is it the Clion debugging 𝓕𝓛𝓨 ?
its nice but no comprehensible for mi sir

Answer 6

Nariman Tajari
I don't understand the word 'Clion'.

we know
& means get the address
* means get the value of the address

0x40404c is the address of 40
0x404068 is the address of 0x404068 (40's address)