help with switch case

[cprog12]

Hi, Im fairly new to programming with C and was wondering if anyone could tell me whats wrong with this 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:

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.

[tabstop]

Because "h" is not an integer constant. 'h' is an integer constant.

[cprog12]

Thanks for that.

[cprog12]

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

[CommonTater]

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.

[cprog12]

ok thanks Tater.

[cprog12]

it wont compile:

:~$ 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’

[cprog12]

If it helps, heres the new 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;
	}
}

[CommonTater]

If it helps, heres the new 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.

[cprog12]

kk, it works now thanks for your help.

[Jesius]

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

[quzah]

Yep, my bad... don't need the & for the input variable.

You do if you are using %c and a single char:

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.

[CommonTater]

You do if you are using %c and a single char:

char c;
scanf( "%c", &c );

Would you believe too many eggnogs?

[iMalc]

Would you believe too many eggnogs?

Frankly I can hardly believe anyone drinks that stuff...

[CommonTater]

Frankly I can hardly believe anyone drinks that stuff...

Well, I put up with it for the rum hidden within...