Extremly simple C program,why I am getting output like this?

Question

Code:

#include <stdio.h>
int main()
{
  long cn=1;
  char ch;
  while((ch=getchar())!=EOF)
  {
    printf("%ld\t%c\n",cn++,ch);
  } 
}

When I input word "secret" and hit enter it shows count up to 7 and not 6,why?

`EOF` must differ from any character! Thus `getchar` returns `int`, not `char`. Enable compiler warnings and take them seriously! — too honest for this site, Oct 12 '15 at 13:30
[Why must the variable used to hold getchar's return value be declared as int?](http://stackoverflow.com/q/18013167/995714) — phuclv, Oct 12 '15 at 13:51

score 11 · Accepted Answer · answered Oct 12 '15 at 13:01

11

Because the "enter" character is read as well. This is in fact a "newline", ASCII code 10 (or hex 0A).

answered Oct 12 '15 at 13:01

Bart Friederichs

Thank you,I was making reading Ritchie C book and making a quiz for www.embeddedfunda.com – piyush ashtikar Oct 12 '15 at 13:10

1 Answers1