C Simple Code Involving getchar() and putchar() Unexpected Output

Question

As I was following an example from a book,

#include <stdio.h>

main()
{
        int c;

        c = getchar();
        while (c != EOF) {   
                putchar(c)
                c = getchar();
        }
}

I thought it would make more sense to read character first, then print it so switched putchar and getchar around

c = getchar();
putchar(c);

now when I run it, what happens is the first output of putchar is missing the first character of c? This is the output:

kingvon@KingVon:~/Desktop/C$ ./a.out
first letter is missing?
irst letter is missing?
but now it is not 
but now it is not

This is interesting, why does this happen?

The standard idiom make it a little clearer: `while( (c = getchar()) != EOF ){ putchar(c); }` If you prefer, you could write it `while( c = getchar(), c != EOF ){ putchar(c); }` — William Pursell, Dec 31 '20 at 00:06
Another problem with your suggested change is that you will call `putchar(EOF)` eventually — M.M, Dec 31 '20 at 00:08

score 2 · Accepted Answer · answered Dec 31 '20 at 00:04

2

Because you're getting a character before the loop. That means c is equal to that first character, but in the loop it's getting every character after that. So,

Get: f
Start the loop
Get: i
Print: i
And so on

answered Dec 31 '20 at 00:04

Dock

444
5
13

score 0 · Answer 2 · answered Dec 31 '20 at 00:03

The problem is that now you're not printing the character that you read with getchar() before the loop, so you don't print the first character.

If you want to do the getchar() first, put it into the while() condition.

#include <stdio.h>

main()
{
    int c;

    while ((c = getchar()) != EOF) {   
        putchar(c)
    }
}

C Simple Code Involving getchar() and putchar() Unexpected Output

2 Answers2