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

Question

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 magpie th 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.

Answer 1

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

Answer 2

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, ie, [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.