Why won't this while loop loop?

[JonnyFrond]

---

Hi guys,

Having real problem here, I don't understand why this won't loop?

What am I doing wrong?


#include <stdio.h>

int main()

{

int num = 14; /* initialise number to be guessed and make number 14 */
int guess = 0; /* initialise variable for guessing number */
int i = 0;

printf("Try to guess the number i am thinking of");

while (i == 0)
{
printf("\nEnter a number between 1 and 100");
scanf("%d", &guess); /* prompt user to enter a number */

if (guess > 0 & guess < 101 & guess != num) /* Analyse user input and check results */
printf("\nGuess you didnt get it try again");

else
printf("\nGuess you must have guessed right!");
i = 1;
}
return 0;

}


I'm sure this is how I have done this before in C++ and it worked fine. I might be mistaken though.

Any help would be most welcomed.

Kind regards

Jonny

[mcross1882]

Alright well I just compiled your code and it looped fine for me. (Until I enter 14 of course) I made no alterations to your code, so maybe something wrong with the compiler?

Matt

[zombykillah]

Cmon it can't loop just read the code ...

while (i == 0) {
 ...
 i = 1;
}

[mcross1882]

Close, but that's just part of the error. I forgot to add that I added brackets to the condition statement. Without properly adding brackets to the If/Then the conditions wont execute properly. This code works...

#include <cstdio>
int main()
{

	int num = 14; /* initialise number to be guessed and make number 14 */
	int guess = 0; /* initialise variable for guessing number */
	int i = 0;

	printf("Try to guess the number i am thinking of");

	while (i == 0) {
		printf("\nEnter a number between 1 and 100");
		scanf("%d", &guess); /* prompt user to enter a number */

		if (guess > 0 & guess < 101 & guess != num) { /* Analyse user input and check results */
			printf("\nGuess you didnt get it try again");
		}else {
			printf("\nGuess you must have guessed right!");
			i = 1;
		}
	}
	return 0;

}

[zombykillah]

Your code still ain't working...
Try 101 or any bigger digit ... you will always get a you guessed it.

Well but the while will loop now.

[mcross1882]

Well of course the input expected is between 0 and 100 therefore any number less or greater than that range will give an unexpected response. (In this case false to the condition) If you want to do it safely then make the code like this which will force the user to stay in range...

#include <cstdio>
int main()
{

	int num = 14; /* initialise number to be guessed and make number 14 */
	int guess = 0; /* initialise variable for guessing number */
	int i = 0;

	printf("Try to guess the number i am thinking of");

	while (i == 0) {
		printf("\nEnter a number between 1 and 100\n");
		scanf("%d", &guess); /* prompt user to enter a number */

		if (guess >= 0 && guess <= 100 && guess != num) { /* Analyse user input and check results */
			printf("\nGuess you didnt get it try again");
		}else if(guess < 0 || guess > 100) {
			printf("\nDigit out of range! Try again...");
		}else {
			printf("\nGuess you must have guessed right!");
			i = 1;
		}
	}
	return 0;

}

[elija]

Also, you only have single & here instead of two.

if (guess > 0 & guess < 101 & guess != num)

& = bitwise operator <-- Sets a bit where the corresponding bit in both operands = 1
&& = logical operator <-- This is the one to use if you mean if this is true and this is true

As an learning exercise, try looking at the difference between 1 & 2 and 1 && 2

In binary 1 = 0001 and 2 = 0010 so using this for the following examples and remembering that the & will set the result bit to 1 where both are 1 and the && will test the truth (ie where != 0)

0001 &
0010 
====
0000

0001 &&
0010
====
0011

The first evaluates to false and the second to true


I'm not sure I've explained that well