How come we have to dereference the pointer returned from strbrk(), but don't have to derefence the pointer returned from strrchr()?

So I have the code below:

#include <stdio.h>
#include <string.h>


int main(void){
    const char *string1 = "This is a test";
    const char *string2 = "beware";
    printf( "%s\"%s\"\n'%c'%s\n\"%s\"\n", "Of the characters in ", string2,
            *strpbrk( string1, string2 ),
            " appears earliest in ", string1 );
    /// How come for the above example we had to dereference the pointer returned from strbrk()
    ///Below we didn't have to derefence the pointer returned from strrchr() and so we didn't have to
    /// put * in front of the function to dereference it.

    puts("");
    const char *stringg1 = "A zoo has many animals including zebras";
    int c = 'z';
    printf( "%s\n%s'%c'%s\"%s\"\n",
            "The remainder of string1 beginning with the",
            "last occurrence of character ", c,
            " is: ", strrchr(stringg1, c));

}

The output is:

Of the characters in "beware"
'a' appears earliest in 
"This is a test"

The remainder of string1 beginning with the
last occurrence of character 'z' is: "zebras"

My question is why do we have to dereference the pointer in a function like strpbrk() and use *, for it to work properly, but for the bottom bit of code that gets the last occurrence of a string, we don't have to dereference it and use *. In fact, if we dereference strrchr() it causes an error and for the program to crash. From my understanding, strrchr() returns a pointer, so to access the value store in the pointer location, we have to dereference it no?

I read another post and think it may have something to do with the fact that in the top portion, where I use strpbrk() I am only printing a character, whereas in the bottom portion I'm printing a string. If this is the cause, can someone explain why?

I am learning C so please answer using beginner words.

Solution

It has nothing to do with strbrk and strchr since both return a string. It has to do with %s vs %c.

You wanted to print the entire string returned by strbrk, so you used %s and provided (a pointer to) the string.

You wanted to print the first character of the string returned by strbrk, so you used %c and provided the first character of the string returned by strbrk (*p ≡ *( p + 0 ) ≡ p[0]).

#include <stdio.h>
#include <string.h>

int main( void ) {
   const char *s = "The brown fox jumped";

   printf( "%c\n", *strpbrk( s, "jf" ) );   // f
   printf( "%s\n", strpbrk( s, "jf" ) );    // fox jumped

   printf( "%c\n", *strchr( s, 'e' ) );     // e
   printf( "%s\n", strchr( s, 'e' ) );      // e brown fox jumped

   printf( "%c\n", *strrchr( s, 'e' ) );    // e
   printf( "%s\n", strrchr( s, 'e' ) );     // ed
}

Note the blindly dereferencing the result of these functions is unsafe since they can return NULL. It's also unsafe to blindly provide the value returned by these to %s for the same reason.