How can a character pointer without any address specification hold data?

The following C program is not supposed to work by my understanding of pointers but it does.

#include<stdio.h>
main() {
    char *p;
    p = "abcdefghijk";
    printf("%s", p);
}

Outputs:

abcdefghijk

The char pointer variable p is pointing to something random as I have not assigned any address to it like p = &i; where i is some char array.

That means if I try to write anything to the memory address held by the pointer p it should give me segmentation fault since it is some random address not assigned to my program by the OS.

But the program compiles and runs successfully. What is happening?

Solution

In C a string literal like "abcdefghijk" is actually stored as an (read-only) array of characters. The assignment makes p point to the first character of that array.

I note that you mention p = &i where i would be an array. That is in most cases wrong. Arrays naturally decays to pointers to their first element. I.e. doing p = i would be equal to p = &i[0].

While both &i and &i[0] would result in the same address, it is semantically very different. Lets take an example:

char array[10];

With the above definition doing &array[0] (or just plain array as explained just above) you get a pointer to char, i.e. char *. When doing &array you get a pointer to an array of ten characters, i.e. char (*)[10]. The two types are very different.