Regarding switch

[myk45]

Hello,

#include <stdio.h>

int main(void)
{
        const int j = 2;
        int i;
 
        switch (i) {
                case 1:break;
                case j:break;
                default:break;
        }
 
        return 0;

}

i tried executing the above code, and i got the following error:
error: case label does not reduce to an integer constant

Well, since i have declared j as const , should it not take it as a constant?
Can anyone please explain?

Thanks.

[gerard4143]

Try this

#include <stdio.h>

#define j 2

int main(void)
{

        int i;
 
        switch (i) {
                case 1:break;
                case j:break;
                default:break;
        }
 
        return 0;

}

or

#include <stdio.h>

enum choices {a, b, c};

int main(void)
{

	int i = 0;

	fputs("enter a value->", stdout);
	fscanf(stdin, "%d", &i);

	switch (i) 
	{
	case a:
	{
		fputs("a\n", stdout);
		break;
	}
	case b:
	{
		fputs("b\n", stdout);
		break;
	}
	default:break;
	}

	return 0;

}

[myk45]

Thanks for the reply.

Yes. This works fine. But im wondering why the former did not work inspite of the const qualifier.

[gerard4143]

Thanks for the reply.

Yes. This works fine. But im wondering why the former did not work inspite of the const qualifier.

Try running this code with a 'const int'

#include <stdio.h>

int main(void)
{
	int *iptr = (int*)NULL;
	const int i = 1234;

	fprintf(stdout, "i->%d\n", i);

	iptr = (int*)&i;

	*iptr = 898989;

	fprintf(stdout, "i->%d\n", i);

	return 0;

}

[creeps]

Because in C, const != "constant", but "read-only". That is, it is legal to initialize a const variable using

void f(int n) {
    
    const int m = n / 2;
}

but its value is dependent of n. So if we call f passing first 2 as argument, and then 4, the values of m will differ. But a switch statement requires integer constants which are known before the program executes. In the above example, this is clearly not the case.

[myk45]

@creeps
Thanks a lot! i got it. i had misunderstood the const qualified to mean "constant".
Thanks for the clarification.

@gerard4143
Thanks.

[gerard4143]

Because in C, const != "constant", but "read-only". That is, it is legal to initialize a const variable using

void f(int n) {
    
    const int m = n / 2;
}

but its value is dependent of n. So if we call f passing first 2 as argument, and then 4, the values of m will differ. But a switch statement requires integer constants which are known before the program executes. In the above example, this is clearly not the case.

I don't like the term "read-only" because in this case it was read/write if you knew enough to step around the compiler protection.

[nezachem]

> step around the compiler protection...
... straight into the undefined behaviour.

[gerard4143]

> step around the compiler protection...
... straight into the undefined behaviour.

Yes true enough..But I think we moved away from the original problem of why

#define j 2

or

enum choice {a, b, c};

works but

const int j = 2;

does not...IMHO I really feel its because the enums and the define are literals which are written into the instructions while const int j is a variable which has to be fetched from memory...