help with switch case

This is a discussion on help with switch case within the C Programming forums, part of the General Programming Boards category; Hi, Im fairly new to programming with C and was wondering if anyone could tell me whats wrong with this ...

  1. #1
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    help with switch case

    Hi, Im fairly new to programming with C and was wondering if anyone could tell me whats wrong with this code:
    Code:
    #include <stdio.h>
    
    main() {
    	char input;
    	char h[200];
    	char b[200];
    	
    	printf("enter input: ");
    	scanf("s", input);
    	
    	switch (input) {
    		case "h":
    			printf("hi!\n");
    			break;
    		case "b":
    			printf("bye\n");
    			break;
    		default:
    			printf("incorrect input\n");
    			break;
    	}
    }
    and output:
    Code:
    switch.c:12: error: case label does not reduce to an integer constant
    switch.c:15: error: case label does not reduce to an integer constant
    Thanks for reading.

  2. #2
    and the Hat of Guessing tabstop's Avatar
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    Because "h" is not an integer constant. 'h' is an integer constant.

  3. #3
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    Smile

    Thanks for that.

  4. #4
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    i compiled and ran it and when i type h or b it prints incorrect input all the time, any idea why this is??

  5. #5
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    Quote Originally Posted by cprog12 View Post
    i compiled and ran it and when i type h or b it prints incorrect input all the time, any idea why this is??
    Try scanf( "%c" , &input);

    Your formatting string is probably what's causing the trouble.

    Also you have declared h and b arrays but are not using them.

  6. #6
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    ok thanks Tater.

  7. #7
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    it wont compile:
    Code:
    :~$ gcc '/home/bren/Program Files/switch.c'
    /home/bren/Program Files/switch.c: In function ‘main’:
    /home/bren/Program Files/switch.c:9: warning: format ‘%c’ expects type ‘char *’, but argument 2 has type ‘int’

  8. #8
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    If it helps, heres the new code:
    Code:
    #include <stdio.h>
    
    main() {
    	char input;
    	char h;
    	char b;
    	
    	printf("enter input: ");
    	scanf("%c", input);
    	
    	switch (input) {
    		case 'h':
    			printf("hi!\n");
    			break;
    		case 'b':
    			printf("bye\n");
    			break;
    		default:
    			printf("incorrect input\n");
    			break;
    	}
    }

  9. #9
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    Quote Originally Posted by cprog12 View Post
    If it helps, heres the new code:
    Code:
    #include <stdio.h>
    
    main() {
    	char input;
    	char h;
    	char b;
    	
    	printf("enter input: ");
    	scanf("%c", input);
    	
    	switch (input) {
    		case 'h':
    			printf("hi!\n");
    			break;
    		case 'b':
    			printf("bye\n");
    			break;
    		default:
    			printf("incorrect input\n");
    			break;
    	}
    }
    Yep, my bad... don't need the & for the input variable.

    You still don't need to define variables char b; and char h; ... your program, as shown actually makes no use of them.

  10. #10
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    kk, it works now thanks for your help.

  11. #11
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    Code:
    	char input;
    	char h;
    	char b;
    	
    	printf("enter input: ");
    	scanf("%c", &input);

  12. #12
    ATH0 quzah's Avatar
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    Quote Originally Posted by CommonTater View Post
    Yep, my bad... don't need the & for the input variable.
    You do if you are using %c and a single char:
    Code:
    char c;
    scanf( "%c", &c );
    The reason you don't need it in the first example is because he's reading into an array, and scanning for %s; which, by using the name of the array, gives you a pointer to the first element. All of scanf's arguments are to be pointers (addresses).

    Edit: Think of what the function actually does: It takes arguments, and puts values in them. The only way you can do that, is if you are passing the address of the variable you want to fill.

    Quzah.
    Last edited by quzah; 12-25-2010 at 06:32 PM.
    Hope is the first step on the road to disappointment.

  13. #13
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    Quote Originally Posted by quzah View Post
    You do if you are using %c and a single char:
    Code:
    char c;
    scanf( "%c", &c );
    Would you believe too many eggnogs?

  14. #14
    Algorithm Dissector iMalc's Avatar
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    Quote Originally Posted by CommonTater View Post
    Would you believe too many eggnogs?
    Frankly I can hardly believe anyone drinks that stuff...
    My homepage
    Advice: Take only as directed - If symptoms persist, please see your debugger

    Linus Torvalds: "But it clearly is the only right way. The fact that everybody else does it some other way only means that they are wrong"

  15. #15
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    Quote Originally Posted by iMalc View Post
    Frankly I can hardly believe anyone drinks that stuff...
    Well, I put up with it for the rum hidden within...

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