Perché questo codice C ha un comportamento inaspettato

Question

Ho scritto il seguente codice C semplice, che compila e funziona bene.Comunque si comporta in un modo che non capisco.Digioso un personaggio, stampa lo sullo schermo.Ma Quando premo il tasto di ritorno stampa l'intera riga .Quindi, se digito le lettere A, B e C, ABC è stampata sulla riga di comando due volte.Perché lo fa?

#include <stdio.h>


int main(){


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

return 0;

}

.

Answer 1

It's your terminal, not the program.

When you press a key, your terminal prints it, but doesn't pass it to the program.

When you press enter, the terminal pass the whole line to the program, and the program prints the line.

EDIT: if you use Unix/Linux/etc, you can write stty -icanon -echo to disable that terminal behavior. The -echo turns off the printing, and the -icanon turns off the buffering.

Answer 2

Because your terminal is line buffered.

It doesn't send data to your program until it encounters a newline, though it will echo the character to the screen so you can see they key you hit.

Answer 3

What you are seeing is a combination of a few things.

1. When you type a character, unless you expressedly stop it from happening, it will print to the screen.
2. Most print statements won't actually print anything until a new line is printed.

So, the text you are seeing is coming from the type commands, but the putchar() string printed is happening all at once.

Answer 4

You're reading from the standard input stream which is line buffered.

Try this alternate code and you'll understand better what is going on:

#include <stdio.h>
int main(){
    int c;
    while((c=getchar())!=EOF){
        printf("got char %c\n", c);
    }
    return 0;
}

You need non-line buffered input. This is going to depend on your platform, but here is an answer on Linux: non-buffering stdin reading