Unexpected input when using getchar(), and unexpected output using putchar()

Question

I am new to c and understand very little of it however as far as i understand it this code should either print the character i enter (it should print it twice), or print the number representation of the bits for my character then my character (before asking for input again for 100 loops) however it seems to do neither.

instead it prints some random numbers (i assume the representation as a number) then the letter, then 10.

i am using gcc to compile it in on ubuntu 18.04 running on wsl if that makes any difference at all. once again im a total newb so i don't know if that is even a possible point of error.

Here is my code:

#include <stdio.h>
int c;


 int main() {
    
 
    for(int x =0; x < 100; x++){
        c = getchar();
        printf("%d", c);
        putchar( c );

    }
}

example:

input: f

output: 102f10

or

input: r

output: 114r10

The number which is printed before the character (since you used `%d`) is its [ASCII value](https://en.wikipedia.org/wiki/ASCII). `getchar()` function returns a character (`char`). So it should be stored in a char type variable, not an int. — svtag, Jul 21 '21 at 09:47

score 2 · Accepted Answer · answered Jul 21 '21 at 09:53

After you entered a character that is read by this call

c = getchar();

then the input buffer stored also the new line character '\n' that appeared due to pressing the Enter key.

Thus in this loop

for(int x =0; x < 100; x++){
    c = getchar();
    printf("%d", c);
    putchar( c );

}

if you entered for example the character 'f' then this call

    printf("%d", c);

outputted its internal code

after that the next call

    putchar( c );

outputted the character itself.

Now the input buffer contains the new line character '\n'. And in the next iteration of the loop its internal representation

is outputted by the call

    printf("%d", c);

Instead of the call

c = getchar();

use

scanf( " %c", &c );
        ^

pay attention to the blank in the format string. In this case white space characters as for example the new line character '\n' will be skipped.

@Newbie You marked it as a valid answer. That's good, but maybe also "up" the answer saying it is a "good one". To others, sorry if you consider this comment being SPAM, but I think the answer deserves a good score. — U. Windl, Jul 21 '21 at 14:13

score 1 · Answer 2 · answered Jul 21 '21 at 09:45

1

That is because you're printing the ASCII representation as integer of the character you're inputting by doing printf("%d",c). If you want the character you inputted to be printed on the console as a character only, you should just remove printf and use putchar.

answered Jul 21 '21 at 09:45

Nastor

638
4
15

oh okay that explains the first part of random numbers and is as i expected and along with the comment about 10 being the enter key that explains all the numbers – Newbie Jul 21 '21 at 09:52

score 1 · Answer 3 · answered Jul 21 '21 at 12:17

1

When your program's output is confusing, with stuff jammed together, it helps to break things up instead. Try this program instead:

int main() {
    for(int x =0; x < 100; x++){
        c = getchar();
        printf("got character %c = %d\n", c, c);
    }
}

answered Jul 21 '21 at 12:17

Steve Summit

45,437
7
70
103

...or use a debugger to single-step your program. – U. Windl Jul 21 '21 at 14:14

Unexpected input when using getchar(), and unexpected output using putchar()

3 Answers3