CODEWITHSUNDEEP
c
Cror2014
Cror2014

Reputation: 447

getchar and EOF

I was working on the following example of C code from Deitel & Deitel. It seems that the code is supposed to print the characters entered before EOF, in the reverse order. But I have to press EOF (ctrl+z in windows) several times and Enter key to get it done. Could you please let me know why it does not respond at the first EOF?

 #include <stdio.h>

 int main( void )
 {
     int c;
     if ( ( c = getchar() ) != EOF ) {
         main();
         printf( "%c", c );
     } /* end if */

     return 0; 
 }

Upvotes: 2

Views: 182

Answers (2)

Luis Colorado
Luis Colorado

Reputation: 12708

Well getchar(3) is a function that operates in buffer mode, so you have to input some characters, and press the character used to signal end of data (^D in unix, ^Z in windows)

The problem here is that windows console driver is not specified the same way as the unix tty driver, so the behaviour will, in general, not be the same... Try to test the program in a real unix environment (or linux) and see if the input, at least is reversed, as the example said.

In unix, the behaviour of terminal input is that ^D is interpreted as soon as it is pressed, but if some input is in the driver buffer before it, it will make those input available to the program (so you'll have to press it a second time to signal EOF condition, which consists in a read(2) that results in 0 characters actually read). In case you have pressed <Return> before ^D (return makes all data available to the application, with the difference that the \n char is also appended to the data read), the input buffer is empty, so the EOF condition comes inmediately, after the <return> char.

In windows, you need to press <return> for anything to be read (the ^Z to be interpreted), and things complicate.

By the way, I have executed your program on a BSD unix system, with the following result:

$ a.out
apsiodjfpaosijdfa
^D
afdjisoapfjdoispa$ _

Explanation: first line is the input line "apsiodjfpaosijdfa", followed by a \n, and ^D signalling end of input. All this data goes to the application at once, and getchar() then processes it character by character. It prints the \n first (making the line to appear below ^D) and then the input chars, reversed. As there was no \r at the beginning of data, no return is issued at end, and the prompts appears next to the output. The final _ signals the position of the cursor at the end.

If you don't want to deal with end of data characters (or don't have any unix at hand to make the test) you can use a text file to test your program (no eof char, only the actual end of the file), by redirecting program input from a file, like in this example that uses the original source code as input:

$ a.out <pru.c

}
;0 nruter     

/* fi dne */ }     
;) c ,"c%" (ftnirp         
;)(niam         
{ ) FOE =! ) )(rahcteg = c ( ( fi     
;c tni     
{
) diov (niam tni

>h.oidts< edulcni#$ _

Upvotes: 1

Malcolm McLean
Malcolm McLean

Reputation: 6404

It's to do with they way the windows console is passing the CTRL+Z to the program. It's probably waiting for you to compose a line and not recognising a line until it has a non- CTRL-Z character in it. So it waits until you accidentally press space and also enter.

Just echo everything in a scratch program to see exactly what is going on.

Upvotes: 0

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