Simple c program to accept and print the character.
int c;
while((c=getchar())!=EOF)
{
putchar(c);
}
I am not getting why it accept input when I press Ctrl+Z at the end of line
ex Hello(press Ctrl+Z)
hello (some symbol)
but it work properly after leaving a line then pressing Ctrl+Z.
And I am using Window 7
When you call getchar() it in turn ends up making a read() system call. The read() will block until it has some characters available. The terminal driver only makes characters available when you press the return key or the key signifying end of input. At least that is what it looks like. In reality, it's more involved.
On the assumption that by ctrl-Z you mean whatever keystroke combination means "end of input", the reason is the way that the read() system call works. The ctrl-D (it's ctrl-D on most Unixes) character doesn't mean "end of input", it means "send the current pending input to the application".
When you've typed something in before pressing ctrl-D, that input gets sent to the application which is probably blocked on the read() system call. read() fills a buffer with the input and returns the number of bytes it put in the buffer.
When you press ctrl-D without any input pending (i.e. the last thing you didwas hit return or ctrl-D, the same thing happens but there are no characters, so read() returns 0. read() returning 0 is the convention for end of input. When getchar() sees this, it returns EOF to the calling program.
This answer in Stack Exchange puts it a bit more clearly
https://unix.stackexchange.com/a/177662/6555
You have not said what system you are working on, [U|Li]nix or Windows. This answer is Windows specific. For [Li|U]nix, replace references to ctrl-z with ctrl-d.
While using a terminal, Ctrl-z will not produce an EOF (-1) (see good answers from Haccks & JeremyP for detailed whys), so the loop will not exit the way you have it written. However, you can put a test for ctrl-z in your while loop condition to exit...
int main ()
{
int c=0;
puts ("Enter text. ctrl-z to exit:");
while(c != 26) //(26 is the ASCII value for ctrl-z)
{
putchar(c);
c = getchar();
}
return 0;
}
By the way, here is a table showing the values for ASCII control characters.
I found the answer on wiki:
In Microsoft's DOS and Windows (and in CP/M and many DEC operating systems), reading from the terminal will never produce an EOF. Instead, programs recognize that the source is a terminal (or other "character device") and interpret a given reserved character or sequence as an end-of-file indicator; most commonly this is an ASCII Control-Z, code 26.
#include <stdio.h>
int main()
{
int c;
while((c=getchar())!=26)
{
putchar(c);
}
}
You can use ASCII value of CTRL-Z.Now it won't take input after pressing CTRL-Z.
getchar() fution read single character by Pressing Ctrl+Z sends the TSTP signal to your process, means terminate the process (unix/linux)
Related
Following "The C Programming Language" by Kernighan and Ritchie, I am trying to enter the program described on page 18 (see below).
The only changes I made were to add "int" before "main" and "return 0;" before closing the brackets.
When I run the program in Terminal (Mac OS 10.15) I am prompted to enter an input. After I enter the input I am prompted to enter an input again - the "printf" line is apparently never reached and so the number of characters is never displayed.
Can anyone help me with the reason why EOF is never reached letting the while loop exit? I read some other answers suggesting CTRL + D or CTRL + Z, but I thought this shouldn't require extra input. (I was able to get the loop to exit with CTRL + D).
I have also pasted my code and the terminal window below.
#include <stdio.h>
int main(){
long nc;
nc = 0;
while( getchar() != EOF )
++nc;
printf("%ld\n", nc);
return 0;
}
From pg. 18 of "The C Programming Language
My screenshot
You already have the correct answer: when entering data at the terminal, Ctrl-D is the proper way to indicate "I'm done" to the terminal driver so that it sends an EOF condition to your program (Ctrl-Z on Windows). Ctrl-C breaks out of your program early.
If you ran this program with a redirect from an actual file, it would properly count the characters in the file.
EOF means end of file; newlines are not ends of files. You need to press CTRL+D to give the terminal an EOF signal, that's why you're never exiting your while loop.
If you were to give a file as input instead of through the command line, then you would not need to press CTRL+D
Adding to the two good answers I would stress that EOF does not naturally occur in stdin like in other files, a signal from the user must be sent, as you already stated in your question.
Think about it for a second, your input is a number of characters and in the end you press Enter, so the last character present in stdin is a newline character not EOF. For it to work EOF would have to be inputed, and that is precisely what Ctrl+D for Linux/Mac or Ctrl+Z for Windows, do.
As #DavidC.Rankin correctly pointed out EOF can also occur on stdin through bash piping e.g. echo "count this" | ./count or redirecting e.g. ./count < somefile, where somefile would be a text file with the contents you want to pass to stdin.
By the way Ctrl+C just ends the program, whereas Ctrl+D ends the loop and continues the program execution.
For a single line input from the command line you can use something like:
int c = 0;
while((c = getchar()) != EOF && c != '\n'){
++nc;
}
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;
}
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#$ _
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.
I'm making a simple program in C that reads an input. It then displays the number of characters used.
What I tried first:
#include <stdio.h>
int main(int argc, char** argv) {
int currentChar;
int charCount = 0;
while((currentChar = getchar()) != EOF) {
charCount++;
}
printf("Display char count? [y/n]");
int response = getchar();
if(response == 'y' || response == 'Y')
printf("Count: %d\n",charCount);
}
What happened:
I would enter some lines and end it with ^D (I'm on Mac). The program would not wait at int response = getchar();. I found online that this is because there is still content left in the input stream.
My first question is what content would that be? I don't enter anything after pressing ^D to input EOF and when I tried to print anything left in the stream, it would print a ?.
What I tried next:
Assuming there were characters left in the input stream, I made a function to clear the input buffer:
void clearInputBuffer() {
while(getchar() != '\n') {};
}
I called the function right after the while loop:
while((currentChar = getchar()) != EOF) {
charCount++;
}
clearInputBuffer();
Now I would assume if there is anything left after pressing ^D, it would be cleared up to the next \n.
But instead, I can't stop the input request. When I press ^D, rather than sending EOF to currentChar, a ^D is shown on the terminal.
I know there is a probably a solution to this online, but since I'm not sure what exactly my problem is, I don't really know what to look for.
Why is this happening? Can someone also explain exactly what is going on behind the scenes of this program and the Terminal?
man 3 termios - search for VEOF. That will tell you what it actually does.
If you need more explanation, I'll start by saying the ISO C stdin stream has a default buffer, so any bytes read are stored into that buffer unless this behavior is somehow overridden (e.g. setvbuf).
The getchar function will read from this default buffer unless the buffer has no characters in it left to read. In that case, it will call the read function to actually store new data into that buffer and return the number of bytes read.
However, your terminal has its own input buffer. It will wait for an input sequence recognized as an end-of-line (EOL) delimiter. This is where things get interesting. If ICANON is enabled, and you use Ctrl+D with bytes in the terminal's input buffer already, then you effectively will send all of that pending bytes to the program, as if you had entered an end-of-line delimiter. The read function will receive those bytes and store them in the input buffer used for stdin, resulting in getchar returning an appropriate value.
If Ctrl+D is pressed with no pending bytes in the terminal's input buffer, no data will be sent, read will return 0, and EOF gets returned by getchar after getchar sets the end-of-file indicator for the stdin stream.
Given the two behaviors of Ctrl+D, it follows that pressing it twice will send all pending bytes on the first key press, effectively emptying the terminal's input buffer, followed by the second key press sending 0 bytes to read, which means getchar returns EOF and the end-of-file indicator for stdin is set.
If an error occurs (e.g. stdin was closed), read itself will return -1, and getchar will return EOF after setting the error indicator for the stdin stream. The following may help to illustrate the idea of how it works, though there's likely a lot more going on behind the scenes with the TTY itself than just waiting for an EOL or VEOF and sending data after either one is detected:
Of course, if ICANON isn't set on the controlling terminal, then you will never receive EOF unless your input is not from a terminal because suddenly certain special key sequences like Ctrl+D won't be recognized as special key sequences since the feature is turned off.
For a bit more completeness, please note that the ICANON bit and termios stuff in general do not necessarily apply much on Windows. The Windows Command Prompt uses Ctrl+Z for one thing, and the Windows operating system has no concept of terminals other than things like the _isatty C runtime function that is used to detect whether a file descriptor points to a file description that involves a console handle.
Pressing Ctrl+Z with data pending will effectively cancel any remaining input that follows it, though an end-of-line character (Ctrl+M or Enter) still needs to be pressed for the data to be sent unless processed input was disabled by using the SetConsoleMode Windows API function.
If pressed with no input data pending and sent by entering an end-of-line character, it acts as EOF. For example, hello^Z1234^M results in hello^Z being read, and everything including the ^M end-of-line character is ignored. ^Z1234^M or just ^Z^M will trigger EOF.
Operating systems are weird.
Ctrl+D is a bit weird on Unix -- it's not actually an EOF character. Rather, it's a signal to the shell that stdin should be closed. As a result, the behavior can be somewhat unintuitive. Two Ctrl+Ds in a row, or a Return followed by a Ctrl+D, will give you the behavior you're looking for. I tested it with this code:
#include <stdio.h>
int main(void) {
size_t charcount = 0;
while (getchar() != EOF)
charcount++;
printf("Characters: %zu\n", charcount);
return 0;
}
Edited to include chux's format character suggestion.
You can do it (also) this way:
fseek(stdin,0,SEEK_END);
This works fine for me.
I'm working out of the K&R book, and I'm on the code example of how to count characters from a stream of text. I copied their code and tried running it, but when the command line prompts you for characters, the loop doesn't exit and thus will never print out the character count. Is there an error here I'm not catching?
#include <stdio.h>
main()
{
long nc;
nc = 0;
while(getchar() != EOF) {
++nc;
}
printf("%1d\n", nc);
}
Whenever you want to stop it just send the EOF signal to the shell.
Ctrl+d in Linux or Ctrl+z on Windows.
By the way (as additional info) Ctrl+c send SIGINT to a process in Linux and on Windows it does something similar.
Have you tried to press Ctrl+D (on Linux) or Ctrl+Z (on Windows)? If yes then It will come out of loop for sure. On pressing these keys, it will return EOF and loop will terminate.
Try ending your character stream with CNTL-Z (end of file character). Just hitting Enter results in a CR which is just another character to count
#include <stdio.h>
int main() {
char read = ' ';
while ((read = getchar()) != '\n') {
putchar(read);
}
return 0;
}
My input is f (followed by an enter, of course). I expect getchar() to ask for input again, but instead the program is terminated. How come? How can I fix this?
The Terminal can sometimes be a little bit confusing. You should change your program to:
#include <stdio.h>
int main() {
int read;
while ((read = getchar()) != EOF) {
putchar(read);
}
return 0;
}
This will read until getchar reads EOF (most of the time this macro expands to -1) from the terminal. getchar returns an int so you should make your variable 'read' into an integer, so you can check for EOF. You can send an EOF from your terminal on Linux with ^D and I think on windows with ^Z (?).
To explain a little bit what happens. In your program the expression
(read = getchar()) !='\n'
will be true as long as no '\n' is read from the buffer. The problem is, to get the buffer to your program, you have to hit enter which corresponds to '\n'.
The following steps happen when your program is invoked in the terminal:
~$\a.out
this starts your program
(empty line)
getchar() made a system call to get an input from the terminal and the terminal takes over
f
you made an input in the terminal. The 'f' is written into the buffer and echoed back on the terminal, your program has no idea about the character yet.
f
f~$
You hit enter. Your buffer contains now 'f\n'. The 'enter' also signals to the terminal, that it should return to your program. Your progam
reads the buffer and will find the f and put it onto the screen and then find an '\n' and immediatley stop the loop and end your program.
This would be standard behaviour of most terminals. You can change this behaviour, but that would depend on your OS.
getchar() returns the next character from the input stream. This includes of course also newlines etc. The fact that you don't see progress in your loop unless you press 'Enter' is caused by the fact that your file I/O (working on stdin) doesn't hand over the input buffer to getchar() unless it detects the '\n' at the end of the buffer. Your routine first blocks then handles the two keystrokes in one rush, terminating, like you specified it, with the appearance of '\n' in the input stream. Facit: getchar() will not remove the '\n' from the input stream (why should it?).
after f you are putting "enter" which is '/n'.
so the loop ends there.
if you want to take another character just keep on putting them one after the other as soon as enter is pressed the loop exits.
You've programmed it so the loop ends when you read a \n (enter), and you then return 0; from main which exits the program.
Perhaps you want something like
while ((read = getchar()) != EOF) {
putchar(read);
}
On nx terminals you can press Control-D which will tell the tty driver to return the input buffer to the app reading it. That's why ^D on a new line ends input - it causes the tty to return zero bytes, which the app interprets as end-of-file. But it also works anywhere on a line.