Problem printing ASCII control characters

Hello, I am trying to finish an assignment that reads input as a stream of characters and prints it as a decimal number followed by the ASCII code. If the code comes prior to the "space" it needs to print the control character notation.

I am having problems printing some ot the codes for characters 8-20. Mainly 8 (BS), and 13 (CR). On 13 it prints 10 (LF).

Any ideas where I'm going wrong ("pulling my few remaining hairs out!")?

Here is my code:

Code:

#include <stdio.h>
#include <ctype.h>

int main(void)
{
	int ch;
	int chcount = 0;  //character count

	printf("Please enter your characters:  \n");
	
	while ((ch = getchar()) != EOF)
	{
		if (chcount >= 8){
			printf("\n");
			chcount = 0;
		}	
		
		if (isprint(ch)){
			printf("%d = %c, " , ch, ch);
			chcount++;
		}					
		
		else if(ch == 9){
			chcount++;
			printf("%d = Horizontal Tab, ", ch);
		}
		
		else if(ch == 10){
			chcount++;
			printf("%d = Newline, ", ch);
		}
		
		else if(ch == 13){
			chcount++;
			printf("%d = Carriage Return, ", ch);
		}
				
		else if(iscntrl(ch)){
			chcount++;
			printf("%d = ^%c, ", ch, ch+64);
		}

		else if(isspace(ch)){
			chcount++;
			printf("%d = ^%c, ", ch, ch+64);
		}
	
	}

	printf("program has ended\n");
	getchar();

	return 0;
}

I suspect you're on Windows because you have a getchar() at the end of the program. For some reason, Windows treats text files and binary files differently, which means that it does newline translation for text streams: in C, '\n' is newline, but CRLF is newline on Windows. Thus your getchar() might not actually read what's in a redirected file/typed at the keyboard, because it will have been translated. I suspect that's your ^H problem as well, although I'm not familiar enough with Windows to say for sure.

C99 allows you to try to reopen standard input/output as binary with freopen(), but it's not guaranteed to work. You could also switch to using fopen() and force the user to give a filename, since you can tell fopen() to operate in binary mode. Or, perhaps, see if there is a non-standard function on your platform to force standard input not to do any sort of translation.

Thank you for the reply, yes I am using windows and microsoft visual studio. I am a newbie at this and just following instructions for the assignment. we have not worked with any C99 options.

I did get this bit of information but I cannot figure out how to keep this from happening:

"characters such as backspace which cannot be entered from the keyboard because the terminal program interprets them before handing them to your program as input"

It seems like there should be a simpler soulution than what I've come up with ( which doesnt completly work anyway).

TIA

So regarding backspace, you mean that you try to hit backspace and nothing shows up? Yes, that is expected and unavoidable, at least with standard C. You're generally going to get your input line-at-a-time when the user is typing, and any backspaces are handled by the terminal, not your program.

It seems you want to respond immediately when a key is hit. This is not possible in standard C. There are system-specific functions that do this, but you'll have to a) consult your system's documentation (or check the FAQ) and b) find out what your instructor allows.

Your instructor might expect you to use Windows-only functions, wittingly or not (unfortunately, many instructors don't have a good grasp of the language). You need to figure out what's allowed for your class.

I have modified my code using getch() to give me unechoed unbuffered input but I have now lost the ^z EOF function. I only find one paragraph in my book (C Primer Plus) on getch(). Can anyone tell me how to stop the program? Thanks!

Code:

printf("Please enter your characters:  \n");
	
	while ((ch = getch()) != EOF)
		
	{
		if (chcount >= 8){
			printf("\n");			
			chcount = 0;
		}	
		
		if (isprint(ch)){
			printf("%d = %c, " , ch, ch);
			chcount++;
		}					
		
		else if(ch == '\t'){
			printf("%d = \\t, ", ch);
			chcount++;
		}
		
		else if(ch == '\n'){
			chcount++;
			printf("%d = \\n, ", ch);
		}

		else if(ch == '\r'){
			chcount++;
			printf("%d = \\r, ", ch);
		}
				
		else if(iscntrl(ch)){
			chcount++;
			printf("%d = ^%c, ", ch, ch+64);
		}

		else if(isspace(ch)){
			chcount++;
			printf("%d = ^%c, ", ch, ch+64);
		}
	
	}

	printf("program has ended\n");
	getchar();

	return 0;
}

You could try....

Code:

while ((ch = getch()) != '\z')

Originally Posted by CommonTater

You could try....

Code:

while ((ch = getch()) != '\z')

Thanks Mr Tater. That will end the program on "z". I actually figured that out but I was trying to get it to work using the standard windows "^Z". When I use ^Z if prints out "26 = ^Z"

Stumped newbie

Did you try it? '\z' is ^z .... or it should be.

If that's not working just substitute the explicit number 26.

You might also want to try using do {... } while() moving the exit condition to the end of the loop.

\z is not valid C. \ does not introduce a control character, but instead starts an escape sequence. Only a select number of letters may follow the backslash.

yes the '\z', '^Z', or '26' do not work. Using 'z' or something like '#' will end the program just fine. Also if i wanted to use a switch statement how would I code that?
TIA

Originally Posted by d2rover

yes the '\z', '^Z', or '26' do not work. Using 'z' or something like '#' will end the program just fine. Also if i wanted to use a switch statement how would I code that?
TIA

Look at the DO... WHILE structure.... it would go in your while() statement at the bottom.

And it's not '26' it's just 26. (i.e. a number not a character)

Originally Posted by cas

\z is not valid C. \ does not introduce a control character, but instead starts an escape sequence. Only a select number of letters may follow the backslash.

Thanks... I found that out after posting. It's not standard C.

Odly I seem to remember using it.... Oh well.

Originally Posted by CommonTater

Look at the DO... WHILE structure.... it would go in your while() statement at the bottom.

And it's not '26' it's just 26. (i.e. a number not a character)

Thanks Tater, I thought I had tried every combination 26 worked fine!