Why does this C program print weird characters in output?

Question

Dear all,

I've the following program:

#include <stdio.h>

int main()
{
        int ch;
        while( ch = getchar() != '\n') {
                printf("Read %c\n",ch);
        }  
        return 0;
}

No matter what I enter I get:

Read  

Why is this happening and what is that weird char that I see?

Stackoverflow is not printing the weird char. You can see it here: http://ideone.com/EfZHr

Answer 1

You need to place parenthesis as:

while( (ch = getchar()) != '\n')

Precedence of != is greater than that of =

while( ch = getchar() != '\n')

is same as:

while( ch = (getchar() != '\n') )

which reads a char compares it with newline and then assigns the result of comparison to ch. Now the result of comparison is 0 (when newline is entered) or 1 (when anything else is entered)

The weird char you're seeing is the control char with value 1, there is no printable symbol for ASCII 1, so I guess its the shell that prints the weird char with value 0001 in it.

You can confirm it by piping your program output to octal dump (od) :

$ echo 'a' | ./a.out | od -bc         # user entered 'a'
0000000 122 145 141 144 040 001 012
          R   e   a   d     001  \n
here you go  ----------------^


$ echo '\n' | ./a.out | od -bc        # user entered '\n'
0000000

GCC when used with -Wall warns you as:

warning: suggest parentheses around assignment used as truth value

Answer 2

ch = getchar() != '\n'

Writing this will cause unexpected behavior depending on the languages operator precedence. In C = is evaluated after != so ch will be true or false. Try:

(ch = getchar()) != '\n'

Answer 3

C (and C++) interpret the while loop as:

while( ch = (getchar() != '\n')) {

So ch gets the value 1 (for true), which is an unprintable character. You should use explicit parenthesis to fix the precedence:

 while( (ch = getchar()) != '\n') {