Can't use && when comparing char?

ricks's Avatar, Join Date: May 2007
Newbie Member
Hi everyone,
I'm a beginner working through Oualline's Practical C Programming, and one of the exercises (6-3) is to wite a program that will report a letter grade, including + or -, given a numeric grade.

The problem I encountered is that the char comparisons are not working as I expected. Of course, I'm a newbie, so I probably shouldn't expect anything! But in any case, it's not like the regular int comparisons. On the int, I could do this:
Code:
		if ((0 <= percent) && (60 >= percent))
but it doesn't work on the char. on the char i can only do one:
Code:
		if (digit2 > '5')
Can't use the && for some reason. It doesn't create any compilation problem, but it doesn't work. I was wondering if someone can offer some insight on this.

The reason I'm using a char, is because I want the second digit of the inputted numeric grade. Depending on what it is, the program will append, a +, a - or nothing.

I've included the whole program below. The problematic section is the one marked with the comment "not working right now."

Any help would be appreciated. I look forward to learning from you.

Rick

Code:
/*****************************************
 * Exercise 6-3 p.93
 * Given a numeric grade, print the letter
 * 05/06/2007 16:00:47
 * *****************************************/
#include <stdio.h>

char	line[4];	/* standard input -- enough for 3 numbers */
int	percent;	/* inputted numeric grade */
int	digit2;		/* second digit of input */
char	plus;		/* holds +, - or empty space char */


int main() {
	/* print prompt and get the input */
	printf("Welcome to the grade calculator\n\n");
	printf("Input a numeric grade: ");
	fgets(line, sizeof(line), stdin);
	sscanf(line, "%d", &percent);
	sscanf(line, "%c%c", &digit2, &digit2); /* second assignment overwrites the first; char for int seems okay */
	
	/* set variable plus to +, - or neither */
	/* not working right now 05/06/2007 17:07:44 */
	if ((1 <= digit2) && (3 >= digit2))
		plus = '-';
	else if ((4 <= digit2) && (7 >= digit2))
		plus = ' ';
	else if ((8 <= digit2) && (0 >= digit2))
		plus = '+';
	else
		plus = ' ';
	
	/* start debugging */
	if (digit2 > '5') {					
		plus = '+';					
		printf("%c Greater than 5 %c ", plus, plus);	
	}
	printf("%c", digit2);	
	printf("%c", plus);	
	/* end debugging */

	/* see where the input falls */
	if ((0 <= percent) && (60 >= percent))
		printf("F%c", plus);
	else if ((61 <= percent) && (70 >= percent))
		printf("D%c", plus);
	else if ((71 <= percent) && (80 >= percent))
		printf("C%c", plus);
	else if ((81 <= percent) && (90 >= percent))
		printf("B%c", plus);
	else if ((91 <= percent) && (100 >= percent))
		printf("A%c", plus);
	else
		printf("You entered an invalid percentage value.\n");


	return(0);
}
DaWei's Avatar, Join Date: Dec 2006
Team Leader
1, 3, 4, 7, 8, and 0 are not char. They are int. The appropriate chars would be '1', '3', etc.
aVague's Avatar, Join Date: May 2007
Go4Expert Member
so that s why you should write '1', '3' , ...'0' instead of 1,3...,0

operation like '1'<'3' is equal to same inequality for codes of '1' and '3' ( code of '1' is lower than code of '3')
but there can be some other problems...
aVague's Avatar, Join Date: May 2007
Go4Expert Member
btw
code is integer, of course
aVague's Avatar, Join Date: May 2007
Go4Expert Member
and 1 more
operation '3'>4 is aviable
it means code of char '3' ( 51) compares with 4 , result is true 51>4
ricks's Avatar, Join Date: May 2007
Newbie Member
aVague and DaWei, Thanks so much!
Putting single quotes around the numbers took care of the problem. Definitely, a newbie mistake