Question

Forgive me if this is a simple error, but I am just learning C at the moment and I just made a trivial program to get a grasp of pointers.

I have a simple piece of code that yields an output I expect

int x = 4;
int *p;
p = &x;
printf("%d\n\n",*p);

//output is 4 as expected

But when I try to do the same with a char array even though I follow the same logic…

char x[] = "Hello, Stack Overflow!";
char *p[];
p = &x;
printf("%s\n\n",*p);

//this gives me an error when compiling as follows
//
// run.c:15: error: incompatible types when assigning to type ‘char *[1]’ from type ‘char (*)[23]’

Answer 1

Multiple errors. First, char *p[] doesn’t declare a pointer-to-an-array (as you think it does) – it rather declares an array of char pointers.

Second, since x is an array, &x will evaluate to the same numerical value as x (since arrays cannot be passed by value, only by pointer, they decay into pointers when passed to a function). What you need to do is rather

const char *x = "Hello SO!";
const char **p = &x;
printf("%s\n", *p);

This is the easy solution (making the string literal, that is, an array of chars, decay into a pointer). There’s another solution that requires a bit more thinking. From the compiler error message, you can see that the type of &x is char (*)[]. So what you have to declare here is a pointer-to-array:

char x[] = "Hello SO!"; // x is a char array
char (*p)[] = &x; // p is a pointer to a char array
printf("%s\n", *p); // printf accesses *p, so it prints the underlying char array - correct!