Error evaluating 'char'

This is a discussion on Error evaluating 'char' within the C Programming forums, part of the General Programming Boards category; I wrote a program that is supposed to evaluate a character that the user types in and tell the user ...

  1. #1
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    Error evaluating 'char'

    I wrote a program that is supposed to evaluate a character that the user types in and tell the user if it's a number, a letter, etc.
    The problem is that the program always goes to 'else' and returns the wrong answer!

    Code:
    #include <stdio.h>
    
    int main()
    {
        int cod;
        char c;
    
        cod= (int) c;
    
        printf("Type one character");
        scanf("%c", &c);
        if (cod<=57 && cod>=48){
            printf("You typed a number\n");
        }else if (cod>=97 && cod<=122) {
            printf("You typed a letter\n");
        }else if (cod>=33 && cod<=46){
            printf("You typed a punctuation sign\n");
            } else {
            printf("What you typed was not a letter, a punctuation sign or a number\n");
            }
    return 0;
    }
    Thanks
    Alastor

  2. #2
    Registered User Tonto's Avatar
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    Code:
    	printf("Type one character");
    	scanf("%c", &c);
    
    	if (cod <= '9' && cod >= '0')
    	{
    		printf("You typed a number\n");
    	}
    	// etc etc...
    No need to convert the char to an integer value or use cryptic "magic-number" ASCII codes.

  3. #3
    ATH0 quzah's Avatar
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    Code:
    cod= (int) c;
    This does nothing at all. Perhaps if you actually did it after you read some value into c... Even if you do though, the cast is pointless.

    Next, there's nothing at all that says the numbers 48 - 57 have to represent numbers, or that 33 through 46 have to be letters, or that anything that doesn't fit into those ranges is punctuation. You should instead be using the functions included in <ctype.h> to check the value.


    Quzah.
    Hope is the first step on the road to disappointment.

  4. #4
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    I was following the ASCII values Quzah, I tried another program before...

    Code:
    #include <stdio.h>
    
    int main()
    {
        int cod;
        char c;
    
        c='C';
        cod = (int) c;
       printf("C is %c and cod is %d", c,cod);
       }
    Since cod returned a numerical value (In ASCII of course) I thought that the program I posted before would work
    Thanks for the help!

    EDIT: Tonto, what if I want to check for letters?

    cod>='a' && cod<='z' ?

  5. #5
    ATH0 quzah's Avatar
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    There's no guarintee you're using the ASCII character set. The C standard doesn't care anything about what character set you use. Thus, if you hard-code some numbers in, it may not work the way you expect if someone else (or you) compiles it elsewhere. While it may be sufficient for your simple program, you should try and genralize (IE: standardize) your code as much as possible.


    Quzah.
    Hope is the first step on the road to disappointment.

  6. #6
    cwr
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    Prefer:
    Code:
    isdigit(cod)
    isalpha(cod)
    To:
    Code:
    (cod >= '0' && cod <= '9')
    (cod >= 'a' && cod <= 'z')

  7. #7
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    so that's what isalpha is for...

  8. #8
    cwr
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    Of course, isalpha will include uppercase letters as well, so you'd actually want islower() if you just wanted lowercase.

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