Question

I have a very simple program to print the chars in a string but for some reason it is not working:

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

void * print_chars(char *process_string) {
    int i;
    int string_len;

    string_len = strlen(process_string);

    printf("String is %s, and its length is %d", process_string, string_len);

    for(i = 0; i < string_len; i++) {
        printf(process_string[i]);
    }

    printf("\n");
}

int main(void) {
    char *process_string;

    process_string = "This is the parent process.";

    print_chars(process_string);

    return 0;
}

When I run it in Netbeans, I get the following:

RUN FAILED (exit value 1, total time: 98ms)

If I remove the line

printf(process_string[i]);

the program runs but nothing prints out to the console (obviously).

Any ideas what I’m missing here?

Answer 1

There are a couple of problems.

One is that you’re not seeing any output from the printf("String is %s, and its length is %d", ...). This is because standard output is line buffered by default, and you are not including a newline, so it never actually decides that there’s a line ready to print. If you change the format string to add a \n, you will see the output from this command.

The second is that you are passing a char into the first argument of printf(), where it expects a char *. This causes it to crash, as it tries to interpret that character as a pointer. You want to pass something like printf(process_string) instead. However, it’s generally a bad idea to pass a variable string directly into the first argument of printf(); instead, you should pass a format string that includes %s, and pass the string in as the corresponding argument: printf("%s\n", process_string). Or, if you want to print it character by character, printf("%c", process_string[i]), followed by a printf("\n") to flush the buffer and actually see the output. Or if you’re doing it character by character, putchar(process_string[i]) will be simpler than printf().