Thread: What's the best way to handle many program options?

Here's some sample code I just wrote. Take a look at it and you'll see what I'm trying to explain. Does it do what you want it to do?

Code:

#include <stdio.h>

struct data_t {
    int n;
    int (*func)(int x, int n);
    const char *name;
};

int func_add(int x, int n) { return x + n; }
int func_multiply(int x, int n) { return x * n; }
int func_subtract(int x, int n) { return x - n; }

void permutations(int x, struct data_t *data, size_t n);

int main() {
    struct data_t data[] = {
        {3, func_add, "+"},
        {3, func_multiply, "*"},
        {1, func_subtract, "-"}
    };
    
    permutations(17, data, sizeof(data)/sizeof(*data));
    
    getchar();
    return 0;
}

void permutations(int x, struct data_t *data, size_t n) {
    if(!n) return;
    
    permutations(x, data + 1, n - 1);
    printf("&#37;d %s %d = ", x, data[0].name, data[0].n);
    x = (*data[0].func)(x, data[0].n);
    printf("%d [%d]\n", x, 3-n);
    permutations(x, data + 1, n - 1);
    
    putchar('\n');
}

[edit] Wait, I think my algorithm is flawed . . . [/edit]

[edit=2] This simpler program works better:

Code:

#include <stdio.h>

void permutations(int *data, size_t n, const char *prev);

int main() {
    int data[] = {1, 2, 3};
    
    permutations(data, sizeof(data)/sizeof(*data), "combination:");
    
    getchar();
    return 0;
}

void permutations(int *data, size_t n, const char *prev) {
    char stack[100];
    
    if(!n) return;

    permutations(data + 1, n - 1, prev);

    sprintf(stack, "%s %d", prev, *data);
    printf("%s %d\n", prev, *data);

    permutations(data + 1, n - 1, stack);
}

[/edit]

[edit=3] There you go, this one's much easier to understand:

Code:

#include <stdio.h>

struct data_t {
    int n;
    int (*func)(int x, int n);
    const char *name;
};

int func_add(int x, int n) { return x + n; }
int func_multiply(int x, int n) { return x * n; }
int func_subtract(int x, int n) { return x - n; }

void permutations(int x, struct data_t *data, size_t n, const char *prev);

int main() {
    struct data_t data[] = {
        {3, func_add, "+"},
        {3, func_multiply, "*"},
        {1, func_subtract, "-"}
    };
    
    permutations(17, data, sizeof(data)/sizeof(*data), "combination:");
    
    getchar();
    return 0;
}

void permutations(int x, struct data_t *data, size_t n, const char *prev) {
    char stack[BUFSIZ];
    int t;
    
    if(!n) return;
    
    permutations(x, data + 1, n - 1, prev);
    
    t = (*data[0].func)(x, data[0].n);
    sprintf(stack, "%s [%d %s %d = %d]", prev, x, data[0].name, data[0].n, t);
    puts(stack);
    
    permutations(t, data + 1, n - 1, stack);
}

Have you ever programmed a solution to the Towers of Hanoi? It's quite similar to this. [/edit]

[edit=4] Here's the Wikipedia entry for "Towers of Hanoi": http://en.wikipedia.org/wiki/Tower_of_Hanoi
It includes a recursive solution written in C. [/edit]

I think that last post has enough edits, time for a new one.

In case you can't run it, here's the output from the last program in the previous post:

Code:

combination: [17 - 1 = 16]
combination: [17 * 3 = 51]
combination: [17 * 3 = 51] [51 - 1 = 50]
combination: [17 + 3 = 20]
combination: [17 + 3 = 20] [20 - 1 = 19]
combination: [17 + 3 = 20] [20 * 3 = 60]
combination: [17 + 3 = 20] [20 * 3 = 60] [60 - 1 = 59]

As you can see, though the order is reversed, the program calculates all possible combinations, "permutations" if you will.

My 2D game made in August of 2006 has over 200 options. In it, I used a bunch of if statements nested inside one another and it wasn't unusual to see 6 levels of nested ifs for several dozen options/settings. Basically, it went like this:

Code:

if (OptionSelected[0] == 1) // first option
{
	if (OptionSelected[1] == 1) // first option of the submenu
	{
		// do stuff that the option refers to
	}
	
	else if (OptionSelected[1] == 2) // second option of the submenu
	{
		if (OptionSelected[2] == 1) // another submenu's first option
		...
	}
	
	else // last option
	{
		OptionSelected[0] = 0; // allows for going up a menu level for going back
	}
}

else if (OptionSelected[0] == 2) // second option on the main menu
{
	// do stuff
}

Perhaps something similar could be used.

This whole thread is about how to avoid that sort of code! From the original post:

A massive function full of nested if statements doesn't sound good to me.

The code I provided, a recursive Towers-of-Hanoi type function, will find every permutation of its arguments, which is a much cleaner solution than a bunch of nested if statements.