getc and fgetc confusion

Hi all,

a small confusion is there for me regarding the getc and fgetc functions. i have written this small program

Code:

int main(int argc, char *argv[])
{
    int c;
    while((c = fgetc(stdin)) != EOF)
    {
        fputc(c,stdout);
    }

return 0;
}

Let us say i have input "qwmb" followed with [enter]. Then the output of the program is "qwmb" and the control would go to the new line. now my doubt is once you entered the input as q then what is the value the variable c is holding and if i enter w is the variable c not over written? if i press [enter] how does the variable c hold the complete value "qwmb"?
could somebody please explain me this concept?

thanks in advance,
regards,
satya

The variable c doesn't hold anything but one character at a time. The reason why your program does this is because of input buffering: fgetc() just returns the next character in the stream. When you enter "qwmb" and press return, each subsequent call to fgetc() will get the next character ('q', 'w', ...), and each subsequent call to fputc() will then output this character.

So, to answer your question, the variable c is actually used to hold all the letters, but one at a time.

Your c variable is overwritten in each iteration, but stdin is line buffered. Which means what you type is entered into a buffer, and when you hit enter, fgetc() reads from that buffer one character at a time.

thank you all for the replies. Now it opened up lot of questions for me why line buffering is required and etc, etc, i will try to google and try to get answers.

thanks once again,
regards,
satya