What is the difference between EOF and ordinary integer?

below is the code

Code :

#include <stdio.h>



int main(void)
{
    int ch开发者_运维知识库;

    while((ch = getchar()) != 'h')
        putchar(ch);


    return 0;
}

Question :

1.)So as usual i just run this code , due to curiosity when the program prompt for input , i insert ^z(CTRL + Z) which is EOF (Windows 7 Command Prompt) , but all i get is infinite looping of character printing.

2.)From the code , my logic is that since i input ^z to the program , it'll just evaluate the logic (ch = getchar()) != 'h' and value true or 1 will be returned and character ^z will be printed out.But instead different result is yield.

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When you press ^Z, the program notices the input stream was closed, but you keep on using getchar(), so you get EOF. This loops infinitely, since you can't input 'h' anymore. Note that only 'h' (not 'A', not ^M, and also not ^Z) can stop the program, since you loop if you don't get a 'h'.

In other words, if you want to stop if anything else but 'h' is entered, then do

do
{
    ch = getchar();
    putchar(ch);
} while (ch == 'h');

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EOF is typically defined as -1 (but it's implementation specific). When you call putchar(-1), it will be converted to unsigned char, and become the value 255, which is then output (ÿ unless I'm mistaken).

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EOF is an integer with an implementation-defined negative value. Your code loops indefinitely because only entering 'h' would make it finish. But once it reads EOF, which is != 'h', it enters an infinite loop because the standard input is closed and there is nothing that can make it return anything else.