How to define an enumerated type (enum) in C?

It's worth pointing out that you don't need a typedef. You can just do it like the following

enum strategy { RANDOM, IMMEDIATE, SEARCH };
enum strategy my_strategy = IMMEDIATE;

It's a style question whether you prefer typedef. Without it, if you want to refer to the enumeration type, you need to use enum strategy. With it, you can just say strategy.

Both ways have their pro and cons. The one is more wordy, but keeps type identifiers into the tag-namespace where they won't conflict with ordinary identifiers (think of struct stat and the stat function: these don't conflict either), and where you immediately see that it's a type. The other is shorter, but brings type identifiers into the ordinary namespace.

Declaring an enum variable is done like this:

enum strategy {RANDOM, IMMEDIATE, SEARCH};
enum strategy my_strategy = IMMEDIATE;

However, you can use a typedef to shorten the variable declarations, like so:

typedef enum {RANDOM, IMMEDIATE, SEARCH} strategy;
strategy my_strategy = IMMEDIATE;

Having a naming convention to distinguish between types and variables is a good idea:

typedef enum {RANDOM, IMMEDIATE, SEARCH} strategy_type;
strategy_type my_strategy = IMMEDIATE;

When you say

enum {RANDOM, IMMEDIATE, SEARCH} strategy;

you create a single instance variable, called 'strategy' of a nameless enum. This is not a very useful thing to do - you need a typedef:

typedef enum {RANDOM, IMMEDIATE, SEARCH} StrategyType; 
StrategyType strategy = IMMEDIATE;

As written, there's nothing wrong with your code. Are you sure you haven't done something like

int strategy;
...
enum {RANDOM, IMMEDIATE, SEARCH} strategy;

What lines do the error messages point to? When it says "previous declaration of 'strategy' was here", what's "here" and what does it show?

It's worth mentioning that in C++ you can use "enum" to define a new type without needing a typedef statement.

enum Strategy {RANDOM, IMMEDIATE, SEARCH};
...
Strategy myStrategy = IMMEDIATE;

I find this approach a lot more friendly.

[edit - clarified C++ status - I had this in originally, then removed it!]

My favorite and only used construction always was:

typedef enum MyBestEnum
{
    /* good enough */
    GOOD = 0,
    /* even better */
    BETTER,
    /* divine */
    BEST
};

I believe that this will remove your problem you have. Using new type is from my point of view right option.

C

enum stuff q;
enum stuff {a, b=-4, c, d=-2, e, f=-3, g} s;

Declaration which acts as a tentative definition of a signed integer s with complete type and declaration which acts as a tentative definition of signed integer q with incomplete type in the scope (which resolves to the complete type in the scope because the type definition is present anywhere in the scope) (like any tentative definition, the identifiers q and s can be redeclared with the incomplete or complete version of the same type int or enum stuff multiple times but only defined once in the scope i.e. int q = 3; and can only be redefined in a subscope, and only usable after the definition). Also you can only use the complete type of enum stuff once in the scope because it acts as a type definition.

A compiler enumeration type definition for enum stuff is also made present at file scope (usable before and below) as well as a forward type declaration (the type enum stuff can have multiple declarations but only one definition/completion in the scope and can be redefined in a subscope). It also acts as a compiler directive to substitute a with rvalue 0, b with -4, c with 5, d with -2, e with -3, f with -1 and g with -2 in the current scope. The enumeration constants now apply after the definition until the next redefinition in a different enum which cannot be on the same scope level.

typedef enum bool {false, true} bool;

//this is the same as 
enum bool {false, true};
typedef enum bool bool;

//or
enum bool {false, true};
typedef unsigned int bool;

//remember though, bool is an alias for _Bool if you include stdbool.h. 
//and casting to a bool is the same as the !! operator 

The tag namespace shared by enum, struct and union is separate and must be prefixed by the type keyword (enum, struct or union) in C i.e. after enum a {a} b, enum a c must be used and not a c. Because the tag namespace is separate to the identifier namespace, enum a {a} b is allowed but enum a {a, b} b is not because the constants are in the same namespace as the variable identifiers, the identifier namespace. typedef enum a {a,b} b is also not allowed because typedef-names are part of the identifier namespace.

The type of enum bool and the constants follow the following pattern in C:

+--------------+-----+-----+-----+
|   enum bool  | a=1 |b='a'| c=3 |  
+--------------+-----+-----+-----+
| unsigned int | int | int | int |  
+--------------+-----+-----+-----+

+--------------+-----+-----+-----+
|   enum bool  | a=1 | b=-2| c=3 |  
+--------------+-----+-----+-----+
|      int     | int | int | int |  
+--------------+-----+-----+-----+

+--------------+-----+---------------+-----+
|   enum bool  | a=1 |b=(-)0x80000000| c=2 |
+--------------+-----+---------------+-----+
| unsigned int | int |  unsigned int | int |
+--------------+-----+---------------+-----+

+--------------+-----+---------------+-----+
|   enum bool  | a=1 |b=(-)2147483648| c=2 |
+--------------+-----+---------------+-----+
| unsigned int | int |  unsigned int | int |
+--------------+-----+---------------+-----+

+-----------+-----+---------------+------+
| enum bool | a=1 |b=(-)0x80000000| c=-2 |
+-----------+-----+---------------+------+
|    long   | int |      long     |  int |
+-----------+-----+---------------+------+

+-----------+-----+---------------+------+
| enum bool | a=1 | b=2147483648  | c=-2 |
+-----------+-----+---------------+------+
|    long   | int |      long     |  int |
+-----------+-----+---------------+------+

+-----------+-----+---------------+------+
| enum bool | a=1 | b=-2147483648 | c=-2 |
+-----------+-----+---------------+------+
|    int    | int |      int      |  int |
+-----------+-----+---------------+------+

+---------------+-----+---------------+-----+
|   enum bool   | a=1 | b=99999999999 | c=1 |
+---------------+-----+---------------+-----+
| unsigned long | int | unsigned long | int |
+---------------+-----+---------------+-----+

+-----------+-----+---------------+------+
| enum bool | a=1 | b=99999999999 | c=-1 |
+-----------+-----+---------------+------+
|    long   | int |      long     |  int |
+-----------+-----+---------------+------+

This compiles fine in C:

#include <stdio.h>
enum c j;
enum c{f, m} p;
typedef int d;
typedef int c;
enum c j;
enum m {n} ;
int main() {
  enum c j;
  enum d{l};
  enum d q; 
  enum m y; 
  printf("%llu", j);
}

C++

In C++, enums can have a type

enum Bool: bool {True, False} Bool;
enum Bool: bool {True, False, maybe} Bool; //error

In this situation, the constants and the identifier all have the same type, bool, and an error will occur if a number cannot be represented by that type. Maybe = 2, which isn't a bool. Also, True, False and Bool cannot be lower case otherwise they will clash with language keywords. An enum also cannot have a pointer type.

The rules for enums are different in C++.

#include <iostream>
c j; //not allowed, unknown type name c before enum c{f} p; line
enum c j; //not allowed, forward declaration of enum type not allowed and variable can have an incomplete type but not when it's still a forward declaration in C++ unlike C
enum c{f, m} p;
typedef int d;
typedef int c; // not allowed in C++ as it clashes with enum c, but if just int c were used then the below usages of c j; would have to be enum c j;
[enum] c j;
enum m {n} ;
int main() {
  [enum] c j;
  enum d{l}; //not allowed in same scope as typedef but allowed here 
  d q;
  m y; //simple type specifier not allowed, need elaborated type specifier enum m to refer to enum m here
  p v; // not allowed, need enum p to refer to enum p
  std::cout << j;
}

Enums variables in C++ are no longer just unsigned integers etc, they're also of enum type and can only be assigned constants in the enum. This can however be cast away.

#include <stdio.h>
enum a {l} c;
enum d {f} ;
int main() {
  c=0; // not allowed;
  c=l;
  c=(a)1;
  c=(enum a)4;
  printf("%llu", c); //4
}

Enum classes

enum struct is identical to enum class

#include <stdio.h>
enum class a {b} c;
int main() {
  printf("%llu", a::b<1) ; //not allowed
  printf("%llu", (int)a::b<1) ;
  printf("%llu", a::b<(a)1) ;
  printf("%llu", a::b<(enum a)1);
  printf("%llu", a::b<(enum class a)1) ; //not allowed 
  printf("%llu", b<(enum a)1); //not allowed
}

The scope resolution operator can still be used for non-scoped enums.

#include <stdio.h>
enum a: bool {l, w} ;
int main() {
  enum a: bool {w, l} f;
  printf("%llu", ::a::w);
}

But because w cannot be defined as something else in the scope, there is no difference between ::w and ::a::w