Question

I am thinking of something like:

#include <stdio.h>
#include <conio.h>
#include <stdlib.h>

int main(void) {
    //test pointer to string
    char s[50];
    char *ptr=s;
    printf("\nEnter string (s): ");
    fgets(s, 50, stdin);
    printf("S: %s\nPTR: %s\n", s, *ptr);

    system("PAUSE");
    return 0;
}

Or should I use a for loop with *(s+i) and the format specifier %c?
Is that the only possible way to print a string through a pointer and a simple printf?

Update: The printf operates with the adress of the first element of the array so when I use *ptr I actually operate with the first element and not it’s adress. Thanks.

Answer 1

The "%s" format specifier for printf always expects a char* argument.

Given:

char s[] = "hello";
char *p = "world";
printf("%s, %s\n", s, p);

it looks like you’re passing an array for the first %s and a pointer for the second, but in fact you’re (correctly) passing pointers for both.

In C, any expression of array type is implicitly converted to a pointer to the array’s first element unless it’s in one of the following three contexts:

• It’s an argument to the unary “&” (address-of) operator
• It’s an argument to the unary “sizeof” operator
• It’s a string literal in an initializer used to initialize an array object.

(I think C++ has one or two other exceptions.)

The implementation of printf() sees the "%s", assumes that the corresponding argument is a pointer to char, and uses that pointer to traverse the string and print it.

Section 6 of the comp.lang.c FAQ has an excellent discussion of this.