Thread: Switch and case problem.

Code:

#include <stdio.h>

	int fx(int a);

	int main(void)
{
	int x;

		printf("Enter an integer: ");
		scanf("%d", &x); /* assume user enters 5 */

	switch( x )
	{
		case 8: printf("%d\n", fx(4) );
		case 6: printf("%d\n", fx(6) ); 
		case 5: printf("%d\n", fx(8) ); 
		case 3: printf("%d\n", fx(10) );
		default: printf("%d\n", fx(12) );
	}
}
	int fx(int a)
{
	return (a - 4);
}

Output

Code:

4
6
8

Ok, I have no clue how the program gets the output. First I thought that for each of the cases, it gives a number to use as 'a'. But then I cant see how the number 5 activates 4, 6, and 8. Can someone explain it to me?

try putting break; after each case.

Well, the output is correct according to the example in the book, my problem is that I dont understand how the program came to that conclusion.

Well, assume x is 5 as the comment says and see where you end up from there.

hint: switch(5) does what?

Look at the values passed into the fx function, there's your answer.

So basically, in the printf cases, I plug that number in as 'a' down in the definition of the function? If so, then the first three 4, 6, and 8 values are all below 5...which I think would make sense why they appear in the output. To test this, I tried entering '1' as 'x' and the output gives me 12, and I assume that came from the 'default printf'. God I feel stupid.