Can't use && when comparing char?

Discussion in 'C' started by ricks, May 8, 2007.

1. ricksNew Member

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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 */
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);
}

```

2. DaWeiNew Member

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1, 3, 4, 7, 8, and 0 are not char. They are int. The appropriate chars would be '1', '3', etc.

3. aVagueNew Member

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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...

4. aVagueNew Member

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btw
code is integer, of course

5. aVagueNew Member

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and 1 more
operation '3'>4 is aviable
it means code of char '3' ( 51) compares with 4 , result is true 51>4

6. ricksNew Member

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aVague and DaWei, Thanks so much!
Putting single quotes around the numbers took care of the problem. Definitely, a newbie mistake