在C中，如果将双指针分配给值，为什么指针变量为0？

Question

int main() 
{
    int var = 10;
    int **ptr;

    **ptr = var;

    printf("%x\n",ptr);
    printf("%d\n",**ptr);

    return 0;
} 

The above code prints: 0 and 10

Why is ptr 0 ?

How is the code showing **ptr = 10, if ptr is 0 ?

I tried printing *ptr.

This gave segmentation fault (since ptr is 0).

So, again why not throw segmentation fault at **ptr ?

Could this be related to compiler ?

PS: I am using https://www.onlinegdb.com/online_c_compiler to run this code.

Answer 1

Pointer ptr is not initialized. Your code invokes undefined behaviour because you are dereferencing an uninitialized pointer.

Answer 2

You assign the pointer to pointer itself, not the the object referenced.

int v = 10;

int *ptr = &v;
int **ptrptr = &ptr;

printf("%d\n", **ptrptr);

So, again why not throw segmentation fault at **ptr ?

By incident. This is UB and your pointer has a random value. This value points to another location with the random value. Then you dereference that second one. So if those random values point to the place in the memory allocated to your program it will not segfault. If not it will segfault.

Why it is printing 0 and 10 ? Because in this case compiler has optimized all the pointer operations out and assumed that not initialized pointers are NULL, just printing the constant values:

main:
        push    {r4, lr}
        mov     r1, #0
        ldr     r0, .L3
        bl      printf
        mov     r1, #10
        ldr     r0, .L3+4
        bl      printf
        mov     r0, #0
        pop     {r4, pc}
.L3:
        .word   .LC0
        .word   .LC1
.LC0:
        .ascii  "%x\012\000"
.LC1:
        .ascii  "%d\012\000"