What does the following mean with respect to enumerations in C programming?

user_17469998 0
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In a book I'm reading, enumeration constants have been introduced before arrays. Use of arrays has been demonstrated only through a couple examples. Following is stated :

enum corvid { magpie, raven, jay, corvid_num};

char const * const bird [ corvid_num ] = 
{
  [raven] = "raven",
  [magpie] = "magpie",
  [jay] = "jay",
};

for ( unsigned i = 0; i < corvid_num ; ++i)
  printf ("Corvid %u is the %s\n", i, bird[i]);

This declares a new integer type enum corvid for which we know four different values.

Takeaway - Enumeration constants have either an explicit or a positional value

As you might have guessed, positional values start from 0 onward, so in our example we have raven with value 0, magpie with 1, jay with 2, and corvid_num with 3. This last 3 is obviously the 3 we are interested in.

Question 1:

Does [magpie] = "magpie" imply that magpieth position refers to value "magpie".

Question 2:

According to loop, how is bird[0] equal to "raven" , since this is explicit and not a positional value. Also after the first iteration is [i + 1] gonna be equal to [magpie]. In all why is loop variable's type unsigned and not corvid or enum corvid?

I think I have misunderstood enumeration constants.

Also from,

As you might have guessed, positional values start from 0 onward, so in our example we have raven with value 0, magpie with 1, jay with 2, and corvid_num with 3. This last 3 is obviously the 3 we are interested in.

Is author right to say that raven should have value 0 and not magpie, if this is a typo, entirety of my confusion will dispense.

enums c types
2 Answers

In the declaration char const * const bird [ corvid_num ] the item between [ ] is the size of the array. In the initializer list, [ raven ] = " raven ", the item between [ ] is a so-called designated initializer, used to initialize a particular item in the array.

Question 1 : Does ' [ magpie ] = " magpie "' imply that 'magpie'TH position refers to value "magpie".

Yes.

Question 2 : According to loop, how is bird [0] equal to "raven" , since this is explicit and not a positional value.

Because enums are named constants. If an enumeration constant (raven etc) isn't given a number explicitly, it is given one implicitly, starting at 0. raven in your example is the second item in the enum, so it gets value 1 and whenever you type raven in the source it will be equivalent to typing 1.

As you might have guessed, positional values start from 0 onward, so in our example we have raven with value 0, magpie with 1, jay with 2, and corvid_num with 3. This last 3 is obviously the 3 we are interested in.

This is an error in Gustedt - Modern C. Errata might exist? The output from the example code is:

Corvid 0 is the magpie
Corvid 1 is the raven
Corvid 2 is the jay

Notably, if we had not used designated initializers but just done this:

char const * const bird [ corvid_num ] = 
{
    "raven",
    "magpie",
    "jay",
};

Then the output would instead have become

Corvid 0 is the raven
Corvid 1 is the magpie
Corvid 2 is the jay
user_5845180

Each enum case gets the value of the previous case plus one, unless explicitly overridden, and the first case gets the value zero. So here magpie is 0, raven is 1, etc. (The quoted part saying otherwise is wrong.) The final case corvid_num is the count of other cases in the array, since the + 1 of that case accounts for the zero case in the beginning.

The array initialisation is using C99 designated initializers, which allows assigning values to specific indices and in any order, i.e., [raven] = "raven" initialises the index raven (here 1). It would be possible to just traditionally list each value in order, but doing it this way allows reordering the enum cases without having to sync the changes with the array.

user_2048830
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