Why am i getting weird output from my pointer?

Question

So i am going to do some string manipulation, but when i try to print out what my pointer is pointing to at time time, i get really weird output.

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

int main (){

    size_t n = 10;
    char *mystring = malloc(10);
    int line = 0;
    int tokens = 0;
    char *ptr;

    if(mystring==NULL){
            fprintf(stderr, "No memory\n");
            exit(1);
    }

    while(getline(&mystring, &n, stdin)>0){
             printf("len = %lu, mystring = %s\n", strlen(mystring), mystring);
             printf("let's tokenize this string\n line = %d tokens = %d\n", line, tokens);

            ptr = mystring;

            printf("ptr = %ch\n", ptr[0]);
    }

return 0;
}

This is what my output looks like when i enter hello

len = 6, mystring = hello

let's tokenize this string line = 0 tokens = 0

ptr = hh

what's with the double hh?

I believe you are looking for `%c`. The first H is from the first character in `hello` while the second is from the `h` in your `printf`. — duncan, Oct 24 '16 at 17:06
i see, i feel like an idiot lol. Thank you for pointing that out to me — Christian Soto, Oct 24 '16 at 17:08
it would be a good idea for you to NOT malloc() but rather: `char *mystring = NULL; size_t n = 0;` and call free() and reset the values at the end of each pass through the while loop, — user3629249, Oct 25 '16 at 07:46
because the function: `getline()` can read right past a NUL char, the code should use the `n` variable for the length, not the results of a call to `strlen()` — user3629249, Oct 25 '16 at 07:49
the posted code never sets the variables `line` and `tokens` after their initial initialization to 0 — user3629249, Oct 25 '16 at 07:52

score 3 · Accepted Answer · answered Oct 24 '16 at 17:06

3

That is what you're printing. You are printing ptr = <ptr[0]>h. There is an extra h after %c in your printf() statement. You probably want to use:

        printf("ptr = %c\n", ptr[0]);

answered Oct 24 '16 at 17:06

kichik

33,220
7
94
114

score 0 · Answer 2 · answered Oct 24 '16 at 17:07

0

The format specifier for printing a character is %c. You have %ch, which is saying "print the character given by the next parameter, then print an 'h'".

Get rid of the extra h in your format string:

printf("ptr = %c\n", ptr[0]);

answered Oct 24 '16 at 17:07

dbush

205,898
23
218
273

score 0 · Answer 3 · edited May 23 '17 at 10:27

printf("ptr = %ch\n", ptr[0]);
ptr = hh

what's with the double hh?

The first h comes from the first character of the hello string, corresponds to ptr[0], and is there since you used %c in the printf format string.

The second h is there since you put it there in the printf format string, after %c.

The correct printf call would be something like:

printf("ptr = %c\n", ptr[0]);

If you want to print an address (pointer), you may want to consider this SO question (you can use the %p format specifier).

Why am i getting weird output from my pointer?

3 Answers3