声明一个枚举变量是这样的:

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

但是，你可以使用typedef来缩短变量声明，如下所示:

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

有一个命名约定来区分类型和变量是一个好主意:

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

当你说

enum {RANDOM, IMMEDIATE, SEARCH} strategy;

创建一个名为“strategy”的实例变量。这不是一件非常有用的事情——你需要一个类型定义:

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

值得一提的是，在c++中，你可以使用“enum”来定义一个新类型，而不需要typedef语句。

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

我发现这种方法更友好。

[编辑-澄清c++状态-我原来有这个，然后删除它!]］

如前所述，您的代码没有任何问题。你确定你没做过类似的事吗

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

错误消息指向哪些行?当它说“以前的‘战略’宣言在这里”时，“这里”是什么?它显示了什么?

值得指出的是，您不需要类型定义。你可以像下面这样做

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

这是一个风格问题，你是否喜欢typedef。如果没有它，如果要引用枚举类型，则需要使用枚举策略。有了它，你可以说策略。

两种方法都有优缺点。一种方法比较冗长，但将类型标识符保留在标记名称空间中，在那里它们不会与普通标识符冲突(想想struct stat和stat函数:它们也不冲突)，并且在那里您立即看到它是一个类型。另一个较短，但将类型标识符引入普通名称空间。

如果为枚举声明名称，则不会发生错误。

如果没有声明，你必须使用类型定义:

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

它不会显示错误…

你尝试两次声明策略，这就是为什么你会得到上面的错误。以下工作没有任何抱怨(编译gcc -ansi - petic -Wall):

#include <stdio.h>

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

int main(int argc, char** argv){
    printf("strategy: %d\n", strategy);

    return 0;
}

如果将第二行改为:

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

从警告中，你可以很容易地发现你的错误:

enums.c:5:1: warning: data definition has no type or storage class [enabled by default]
enums.c:5:1: warning: type defaults to ‘int’ in declaration of ‘strategy’ [-Wimplicit-int]
enums.c:5:1: error: conflicting types for ‘strategy’
enums.c:4:36: note: previous declaration of ‘strategy’ was here

因此，编译器使用strategy = IMMEDIATE来声明一个默认类型为int的名为strategy的变量，但之前已经声明了一个具有此名称的变量。

然而，如果你把赋值放在main()函数中，它将是一个有效的代码:

#include <stdio.h>

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

int main(int argc, char** argv){
    strategy=SEARCH;
    printf("strategy: %d\n", strategy);

    return 0;
}

@ThoAppelsin in his comment to question posted is right. The code snippet posted in the question it is valid and with no errors. The error you have must be because other bad syntax in any other place of your c source file. enum{a,b,c}; defines three symbolic constants (a, b and c) which are integers with values 0,1 and 2 respectively, but when we use enum it is because we don't usually care about the specific integer value, we care more about the meaning of the symbolic constant name. This means you can have this:

#include <stdio.h>
enum {a,b,c};
int main(){
  printf("%d\n",b);
  return 0;
}

这个输出是1。

这也是有效的:

#include <stdio.h>
enum {a,b,c};
int bb=b;
int main(){
  printf("%d\n",bb);
  return 0;
}

并将输出与之前相同的结果。

如果你这样做:

enum {a,b,c};
enum {a,b,c};

你会有一个错误，但是如果你这样做:

enum alfa{a,b,c};
enum alfa;

你不会有任何错误。

你可以这样做:

enum {a,b,c};
int aa=a;

aa是一个值为0的整型变量。但你也可以这样做:

enum {a,b,c} aa= a;

和将具有相同的效果(即aa是一个值为0的int)。

你还可以这样做:

enum {a,b,c} aa= a;
aa= 7;

aa将是int，值为7。

因为你不能使用enum重复符号常量定义，就像我之前说过的，如果你想使用enum声明int变量，你必须使用标签:

enum tag1 {a,b,c};
enum tag1 var1= a;
enum tag1 var2= b;

typedef的使用是为了保证你不用每次都用枚举tag1来定义变量。使用typedef，你可以输入Tag1:

typedef enum {a,b,c} Tag1;
Tag1 var1= a;
Tag1 var2= b;

你还可以有:

typedef enum tag1{a,b,c}Tag1;
Tag1 var1= a;
enum tag1 var2= b;

最后要说的是，因为我们谈论的是定义的符号常量，所以在使用enum时最好使用大写字母，例如:

enum {A,B,C};

而不是

enum {a,b,c};

这份声明似乎有些混乱。

如下所示，当strategy出现在{RANDOM, IMMEDIATE, SEARCH}之前时，

enum strategy {RANDOM, IMMEDIATE, SEARCH};

您正在创建一个名为enum策略的新类型。但是，在声明变量时，需要使用枚举策略本身。你不能只使用策略。所以下面的内容无效。

enum strategy {RANDOM, IMMEDIATE, SEARCH};
strategy a;

但是，以下是有效的

enum strategy {RANDOM, IMMEDIATE, SEARCH};

enum strategy queen = RANDOM;
enum strategy king = SEARCH;
enum strategy pawn[100];

---

当strategy出现在{RANDOM, IMMEDIATE, SEARCH}后面时，您正在创建一个匿名枚举，然后将strategy声明为该类型的变量。

现在，你可以这样做

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

但是，您不能声明任何其他类型为enum {RANDOM, IMMEDIATE, SEARCH}的变量，因为您从未命名过它。所以下面的内容无效

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

---

你也可以把这两个定义结合起来

enum strategy {RANDOM, IMMEDIATE, SEARCH} a, b;

a = RANDOM;
b = SEARCH;
enum strategy c = IMMEDIATE;

---

前面提到的Typedef用于创建更短的变量声明。

typedef enum {RANDOM, IMMEDIATE, SEARCH} strategy;

现在你已经告诉编译器enum {RANDOM, IMMEDIATE, SEARCH}是策略的同义词。所以现在你可以自由地使用策略作为变量类型。你不再需要输入enum strategy。以下是现在有效的

strategy x = RANDOM;

---

你也可以结合Typedef和枚举名称来获取

typedef enum strategyName {RANDOM, IMMEDIATE, SEARCH} strategy;

使用这个方法除了可以互换使用strategy和enum strategyName之外没有太多的好处。

typedef enum strategyName {RANDOM, IMMEDIATE, SEARCH} strategy;

enum strategyName a = RANDOM;
strategy b = SEARCH;

我尝试了gcc，并提出了我的需要，我被迫使用最后一种选择，编译出来的错误。

类型定义enum状态{a = 0, b = 1, c = 2}状态;

typedef enum state {a = 0, b = 1, c = 2} state;

typedef enum state old; // New type, alias of the state type.
typedef enum state new; // New type, alias of the state type.

new now     = a;
old before  = b;

printf("State   now = %d \n", now);
printf("Sate before = %d \n\n", before);

Tarc的答案是最好的。

全会的许多讨论都是在转移话题。

比较以下代码片段:-

int strategy;
strategy = 1;   
void some_function(void) 
{
}

这给了

error C2501: 'strategy' : missing storage-class or type specifiers
error C2086: 'strategy' : redefinition

用这个编译没有问题。

int strategy;
void some_function(void) 
{
    strategy = 1;   
}

变量策略需要在声明中或在函数中设置。您不能在全局范围内编写任意软件——特别是赋值软件。

他使用enum {RANDOM, IMMEDIATE, SEARCH}而不是int这一事实只是在一定程度上让那些无法超越它的人感到困惑。 问题中的重新定义错误消息表明这是作者做错的地方。

现在你应该可以看出为什么下面的第一个例子是错误的，而其他三个例子是正确的。

例1。错了!

enum {RANDOM, IMMEDIATE, SEARCH} strategy;
strategy = IMMEDIATE;
void some_function(void) 
{
}

例2。正确的。

enum {RANDOM, IMMEDIATE, SEARCH} strategy = IMMEDIATE;
void some_function(void) 
{
}

例3。正确的。

enum {RANDOM, IMMEDIATE, SEARCH} strategy;
void some_function(void) 
{
    strategy = IMMEDIATE;
}

例4。正确的。

void some_function(void) 
{
    enum {RANDOM, IMMEDIATE, SEARCH} strategy;
    strategy = IMMEDIATE;
}

如果你有一个工作的程序，你应该能够将这些代码段粘贴到你的程序中，并看到一些编译和一些不编译。

我最喜欢和唯一使用的结构是:

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

我相信这会解决你的问题。在我看来，使用新字体是正确的选择。

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.

enum bool类型和常量在C中遵循以下模式:

+--------------+-----+-----+-----+
|   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 |
+-----------+-----+---------------+------+

这在C中编译很好:

#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++

在c++中，枚举可以有类型

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

在这种情况下，常量和标识符都具有相同的类型bool，如果一个数字不能用该类型表示，就会发生错误。也许= 2，这不是bool类型。此外，True、False和Bool不能小写，否则它们将与语言关键字冲突。枚举也不能有指针类型。

在c++中枚举的规则是不同的。

#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;
}

c++中的enum变量不再只是无符号整数等，它们也是enum类型，只能在enum中赋值常量。然而，这是可以抛弃的。

#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
}

枚举类

Enum struct与Enum类相同

#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
}

范围解析操作符仍然可以用于非范围枚举。

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

但是因为w不能被定义为作用域中的其他东西，所以::w和::a::w之间没有区别