Using int for character types when comparing with EOF
c
Solution
Your program doesn't work fine; it won't be able to distinguish between `EOF` and `255`.
The reason it appears to work correctly is because `char` is probably `signed` on your platform, so it's still capable of representing `-1`.
Problem
Quoting from Kernighan and Ritchie's 'The C Programming Language' Page 16 - ``` #include<stdio.h> main() { int c; c = getchar(); while(c!=EOF) { putchar(c); c = getchar(); } getchar(); return 0; } ``` "The type `char` is specifically meant for storing such character data, but any integer type can be used. We used `int` for a subtle but important reason. The problem is distinguishing the end of the input from valid data. The solution is that `getchar` returns a distinctive value when there is no more input, a value that cannot be confused with any real character. This value is called `EOF`, for "end of file". We must declare `c` to be a type big enough to hold any value that `getchar` returns. We can't use `char` since `c` must be big enough to hold `EOF` in addition to any possible `char`. Therefore we use `int`.". I looked up in stdio.h, it says `#define EOF (-1)` The book conclusively states that `char` cannot be used whereas this program "works just fine" (See EDIT) with `c` as `char` data type as well. What is going on? Can anyone explain in terms of bits and signed values? EDIT: As Oli mentioned in the answer, the program cannot distinguish between `EOF` and `255`. So it will not work fine. I want to know what's happening - Are you saying that when we do the comparison c!=EOF, the EOF value gets cast to a char value = 255 (11111111 in binary; i.e. the bits 0 through 7 of EOF when written in 2's complement notation)?