ANSI C: ordinal position of enum

[xyz000]

Hi,

I'm looking for an algorithm to efficiently determine the ordinal position of a non-linear enum. Suppose I have:

typedef enum NewTypeT
{
    ABC = 5,
    DEF,
    GHI,
    JKL = 73,
    MNO,
    PQR
} NewType;

Then I'm after a function (or macro?) to evaluate the position of an enum's declaration. Something like:

int ordinal(NewType n)
{
    /* work out ordinal position of n */
}

void foo(void)
{
    int x = ordinal(MNO); /* x should be 5 because "MNO" is the fifth element */
}

Any ideas?

[Rubberman]

Sorry, but there is no way to do this without you providing some sort of means to associate the enum (which is basically a global variable) and its position in the set, such as using an array to hold the values. Example:

NewType NTordinal[] = { ABC, DEF, GHI, JKL, MNO, PQR };

Now, you can determine the position of each in the array and since each entry is kept in sync with the enum list itself, there you are. You just have to remember to update the array when you add/remove/change the enums. Your ordinal function could also use a switch statement to do the same and return the values. IE,

switch (NewType t)
{
    case ABC: return 0;
    .
    .
    .
    case PQR: return 5;
    default: return -1;
}

[Roxoff]

Rather than do this by algorithm, you may find it easier to just add a simple array to assist with this - you give up some memory for convenience, e.g.

typedef enum NewTypeT
{
    ABC = 5,
    DEF,
    GHI,
    JKL = 73,
    MNO,
    PQR
} NewType;

const  NewType NewTypeOrdinals[] = {ABC, DEF, GHI, JKL, MNO, PQR};

Don't forget to put a comment into your enum so that if someone changes it they update the array. Then when you want an ordinal, just search the array, the index where you find it is the ordinal you want. If you do this, you might want to include an 'end of list' marker, to make the searching easier to detect the end.

You could even do this kind of thing with an stl container, such as a map, if you want to do this without even having to write the search code.

[xyz000]

Thanks for your answers. This is a great head start to solving my problem.