Can a pointer to a string be used in a printf?

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

you should do "printf("S: %s\nPTR: %s\n", s, ptr); " instead of printf("S: %s\nPTR: %s\n", s, *ptr);

difference between ptr and *ptr is: ptr gives you the address in the memory of the variable you are pointing to and *ptr gives rather the value of the pointed variable In this case is *ptr = ptr[0]

this code will show what i mean:

printf("\tS: %s\n\tPTR: %s\n\tAddress of the pointed Value: %x\n\tValue of the whole String: %s\n\tValue of the first character of the String: %c\n", s, ptr,ptr,ptr,*ptr);

Answer 2

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.

Answer 3

printf("%s\n", ptr);

Is this what you want?

By the way, from printf(3), here's the documentation for the s conversion specifier (i.e %s):

If no l modifier is present: The const char * argument is expected to be a pointer to an array of character type (pointer to a string). Characters from the array are written up to (but not including) a terminating null byte ('\0'); if a precision is specified, no more than the number specified are written. If a precision is given, no null byte need be present; if the precision is not specified, or is greater than the size of the array, the array must contain a terminating null byte.

Answer 4

In my experience you should get segmentation fault when you try to use %s directive with *p.