Why the indirection/dereferencing operator (*) is used with a pointer of any data type except a string? 'char'
In the following program the letter 'A' and the string "Computer" are stored using the character pointers but only for the pointer 'a' the dereferencing operator (*) is used. Why is that?
#include <stdio.h>
int main()
{
char *a, *b;
*a = 'A';
b = "Computer";
printf("%c %s", *a, b);
return 0;
}
Why is the * operator not needed while assigning a string to a pointer?
To answer your question, we need to get rid of some syntactic sugar first and know exactly what b is:
char *b;
b = "Computer";
is (almost) equivalent to
const char *b;
const char str[] = {'C', 'o', 'm', 'p', 'u', 't', 'e', 'r', '\0'};
b = &str[0];
Meaning b is a pointer to the first element in an array of char elements. Or simpler, b just points to a char
printf("%c", *b) expects a char (%c), and by using *b we are giving printf the first char from the string (or char array) "Computer"
printf("%s", b) expects a pointer (%s). We are thus providing b which points to "Computer". Why? Because under the hood, printf starts at a location, reads a character and goes to the next location (which is b + 1). It does that until printf reads the value 0 somewhere along its path.
So the core idea is that you're indeed dealing with a pointer, but printf needs itself a pointer to go through an array of char
Note that the charatcer '0' is not the number 0, but the number 0 is equal to the character '\0' which is what you sometimes see in char arrays like in my example.
As an extra on why the above snippets are not exactly the same: A string in code is stored in a read-only location, whereas the assignment as an array of characters is stored in modifiable memory. The const keyword ensures immutability, but both of these strings are still stored in entirely different locations and thus behaviour might not be the same.