algorithm question - selecting bits

[divideby0]

I've got defines representing option choices where the result can be none, all, or any combination. I'm using an if / else if / else structure to test which options are set, but I wondering if there is a more logical / efficient way of doing this?

So far, I've got this

#define OPT_NONE  0
#define OPT_1     1
#define OPT_2     2
      ...
#define OPT_7    64
#define OPT_8   128
#define OPT_ALL 255
 
unsigned int result = OPT_NONE;
 

The user can then select which options he or she wants, and I build a query string based the selection(s).

switch(user_entered_choice)
{
   case OPT_NONE : result = OPT_NONE; break;
   case OPT_1    : result += OPT_1;   break;
       ...
   case OPT_8    : result += OPT_8;   break;
}
 
if(result == OPT_NONE)
   // default option selected
else if(result == OPT_ALL)
   // all options selected
else if(result == OPT_1 ... result == OPT_8) // not valid code
   // a single option 1 of 8 was selected
else if(((result & OPT_1) == OPT_1) &&
        ((result & OPT_2) == OPT_2) ... 
 

The last else if bit of code seems way too clumsy, but I'm at a loss trying to figure out how to combine these selections. Would I even want to use this approach?

TIA

[weaknessforcats]

I wouldn't worry about efficiency until the program works.

One time I wrote:

int main()
{
     while(1)
     {
        featureAdd(choice);
        featureRevise(choice);
        featureDelete(choice);
        etc...
     }
}

where each function:

void featureAdd()
{
    if (choice != ADD)
    return;
 
    etc...to do the add
}

because no one could tell me what the menu was or when the add feature may be called. As a result my feature functions inside the loop can be in any order without changing the logic of the program.

Efficient? not really.

Practical? an absolute lifesaver.

[divideby0]

Thank you for the reply. I *think* that I've got it worked out... well it seems to return what I expect.

If there's a silver lining, it's that I don't have a time constraint to worry about or someone to answer to, so I can putz around with it when I feel up to the madness. :)

I cannot thank you, Banfa, and Rabbit enough for all of the help that each of you has given.

Cheers.