Thread: No error, but the result is unexpected

Hi,

I really don't understand whats wrong with this condition:

Code:

	// validate input
	if	(isdigit(iResponse) && (iResponse >= 1 || iResponse < 10)) {
		printf("\nThank you\n"); 
	} else {
		printf("\nEnter a number between 1 and 10!\n"); 
	}

111 and 5 both display the "Enter a number between.." message.

So I want the user to input a NUMBER only and BETWEEN 1 and 10. What am I doing wrong here?

Thanks in advance.

So I want the user to input a NUMBER only and BETWEEN 1 and 10. What am I doing wrong here?

Code:

(iResponse >= 1 || iResponse < 10)

|| is either
&& is only

Think if it this way, if response is >= 1 or < 10, then it will pretty much always be true right ?
example -

- 1 true( < 10)
3 true( > 1 && < 10)
55 true( > 1)

basically || evaluates to true if either of the two operands is true and && evaluates to true only when both of them are.

If I change the above to:

Code:

if	(isdigit(iResponse) && iResponse >= 1 || iResponse < 10) {
		printf("\nThank you\n"); 
	} else {
		printf("\nEnter a number between 1 and 10!\n"); 
	}

..then it seems to work, but still, it takes 0 as valid whereas I want numbers >1 ? But I wonder why won't the first (posted above) condition work?

Many Thanks

like I said, You need to do this -

Code:

(isdigit(iResponse) && (iResponse >= 1 && iResponse < 10))

Originally Posted by Spidey

Code:

(iResponse >= 1 || iResponse < 10)

|| is either
&& is only

Think if it this way, if response is >= 1 or < 10, then it will pretty much always be true right ?
example -

- 1 true( < 10)
3 true( > 1 && < 10)
55 true( > 1)

basically || evaluates to true if either of the two operands is true and && evaluates to true only when both of them are.

Ohk. But I use &&, it then returns false for every number:

Code:

	if	(isdigit(iResponse) && (iResponse>1 && iResponse<10)) {
		printf("\nThank you.\n"); 
	} else {
		printf("\nEnter a number between 1 and 10!\n"); 
	}

what would be the correct condition?

Ok, this is weird, I'm doing you said and its still not working:

Code:

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

int main() {

	int iResponse; // initialize
	
	// get input
	printf("\nPlease enter a number: ");
	scanf("%d", &iResponse);
	
	// validate input
	if	(isdigit(iResponse) && (iResponse >= 1 && iResponse < 10)) {
		printf("\nThank you.\n"); 
	} else {
		printf("\nEnter a number between 1 and 10!\n"); 
	}
	
	// generate random number
	// srand(time());
	
return 0;
}

Still shows "Enter a number .." still for ANY number. Something to do with my compiler/enviornment? I'm using Xcode on Leopard.

Thanks

Originally Posted by Salem

Beginner C Question Help
Another recent poster who got completely the wrong idea as to what isdigit() actually did.

If you've used scanf("%d"), then you have ALREADY used all the digits you're going to have anyway.

Ah ok, got it now. :-)

The book "C programming for the absolute beginner" didn't explain this very well. I guess I'll have to dump and read "C for Dummies" instead.

Thanks anyway Salem and Spidey!

Why won't you try C Primer Plus Fifth Edition by Stephen Prata? I think it's a wonderful book to understand C. It will help you a lot, and then you can start reading The C Programming Language Second Edition by Dennis Ritchie and Brian Kernighan. I wish you the best, and think very well about the two books mentioned above. If you can't afford to buy them new, try to look for used versions of them. Amazon, eBay, and other virtual stores have them available. ;-) Take care!

You don't need to start on a whole new book on C. Just ask about the difference between the integer iResponse and the char buffer you're trying to apply isdigit() to. Unfortunately your use of scanf() makes that char buffer hard to see and understand.

Also a review of the difference between boolean AND and OR is a good idea.

Then continue on with whatever book you have.