枚举问题。 [英] enum question.
问题描述
考虑以下代码：
#include< stdio.h>
#define SOMEVAL 1234
enum tree_types {PINE = 10，BIRCH，LARCH，OAK = 100，MAPLE，ELM，WILLOW};
void print_val（enum tree_types tree）
{
printf（树是：％d \ n，树）;
}
int main （）
{
enum tree_types trees;
trees = PINE;
print_val（树）;
trees = OAK;
print_val（树）;
树= ELM;
print_val（树）;
trees = SOMEVAL;
print_val（树）;
返回0;
}
为什么允许使用SOMEVAL的assignmnt？我认为枚举类型只允许在枚举列表中指定值的
。为什么编译器
甚至没有警告它？我错了吗？
谢谢。
-
电子邮件：句柄，（点分隔），在gmail dot com。
At_sea_with_C写道：
请考虑以下代码：
#include< stdio.h>
#define SOMEVAL 1234
enum tree_types {PINE = 10 ，BIRCH，LARCH，OAK = 100，MAPLE，ELM，WILLOW};
void print_val（enum tree_types tree）
{
printf（" tree is：％d \ n"，tree）;
}
int main（）
{
enum tree_types trees;
trees = PINE;
print_val（树）;
trees = OAK;
print_val（树）;
trees = ELM;
print_val（树）;
trees = SOMEVAL;
print_val（树）;
返回0;
}
为什么是允许使用SOMEVAL的assignmnt？我认为枚举类型只允许在枚举列表中指定值的
。为什么编译器
甚至没有警告它？我错了吗？
这是允许的，因为你的想法是什么是错的。在C中，
enum只是某种整数的伪装（编译器
选择哪种），加上一些命名常量的引入
（所有类型`int''，不接受替代）。
这个自由的一个半实用的应用是当enum
值时不仅仅是任意数字代码，而是可以以有意义的方式组合
：
枚举CarStatus {
ALL_GREEN = 0x0，
DOOR_OPEN = 0x1，
BELTS_FASTENED = 0x2，
HEADLIGHTS_ON = 0x4
}状态;
...
status = DOOR_OPEN | HEADLIGHTS_ON;
结果是C'的枚举主要用于文档，
不用于执行。此外，一些调试器能够将
枚举值转换为名称;在你的例子中，调试器可能是
能够显示PINE而不是10。
-
Eric Sosman
es*****@acm-dot-org.inva 盖子
3月10日上午8：24，At_sea_with_C< blindal ... @ dev.null.invalidwrote：
（剪掉）
我认为枚举类型只允许在枚举列表中指定值的
。为什么编译器
甚至没有警告它？我错了吗？
谢谢。
摘自C标准草案（n869.pdf）：
"附件I
（资料性）
常见警告
1实施可能会产生警告在许多情况下，
这些都没有被指定为本国际标准的一部分。
以下是一些比较常见的情况。
....
通过赋值枚举常量给予枚举类型除
之外的对象的值
类型的成员，
或具有相同类型的枚举变量，
或返回相同的函数的值枚举类型
（6.7.2.2）"
所以，我想一些实现可能会产生你预期的警告类型，但是
它不是必需的。
-
希望这有帮助，
Ste ven
At_sea_with_C写道：
请考虑以下代码：
#include< stdio.h>
#define SOMEVAL 1234
enum tree_types {PINE = 10，BIRCH，LARCH，OAK = 100，MAPLE，ELM，WILLOW};
void print_val（enum tree_types tree）
{
printf（"树是：％d \ n"，树）;
}
int main（）
{
enum tree_types trees;
trees = PINE;
print_val（树）;
trees = OAK;
print_val（树）;
trees = ELM;
print_val（树）;
trees = SOMEVAL;
print_val（树）;
返回0;
}
为什么允许使用SOMEVAL的assignmnt？我认为枚举类型只允许在枚举列表中指定值的
。为什么编译器
甚至没有警告它？我错了吗？
因为C'的枚举被破坏了。也许你在想C ++哪里
这是非法的？
-
Ian Collins。
Consider the following code:
#include <stdio.h>
#define SOMEVAL 1234
enum tree_types { PINE = 10, BIRCH, LARCH, OAK = 100, MAPLE, ELM, WILLOW };
void print_val(enum tree_types tree)
{
printf("tree is: %d\n", tree);
}
int main()
{
enum tree_types trees;
trees = PINE;
print_val(trees);
trees = OAK;
print_val(trees);
trees = ELM;
print_val(trees);
trees = SOMEVAL;
print_val(trees);
return 0;
}
Why is the assignmnt with SOMEVAL allowed? I thought an enum type was only
allowed to have values specified in the enumeration list. Why is compiler
not even warning about it? Am I wrong?
Thanks.
--
Email: The handle, (dot seperated), at gmail dot com.
At_sea_with_C wrote:Consider the following code:
#include <stdio.h>
#define SOMEVAL 1234
enum tree_types { PINE = 10, BIRCH, LARCH, OAK = 100, MAPLE, ELM, WILLOW };
void print_val(enum tree_types tree)
{
printf("tree is: %d\n", tree);
}
int main()
{
enum tree_types trees;
trees = PINE;
print_val(trees);
trees = OAK;
print_val(trees);
trees = ELM;
print_val(trees);
trees = SOMEVAL;
print_val(trees);
return 0;
}
Why is the assignmnt with SOMEVAL allowed? I thought an enum type was only
allowed to have values specified in the enumeration list. Why is compiler
not even warning about it? Am I wrong?It''s allowed because "what you thought" is wrong. In C, an
enum is merely a disguise for some kind of integer (the compiler
chooses which kind), plus the introduction of some named constants
(all of type `int'', accept no substitutes).
One semi-useful application of this freedom is when the enum
values aren''t just arbitrary numeric codes, but can be combined
in meaningful ways:
enum CarStatus {
ALL_GREEN = 0x0,
DOOR_OPEN = 0x1,
BELTS_FASTENED = 0x2,
HEADLIGHTS_ON = 0x4
} status;
...
status = DOOR_OPEN | HEADLIGHTS_ON;
The upshot is that C''s enum is mostly useful for documentation,
not for enforcement. Also, some debuggers are able to convert
enum values to the names; in your example a debugger might be
able to show you PINE instead of 10.
--
Eric Sosman
es*****@acm-dot-org.invalid
On Mar 10, 8:24 am, At_sea_with_C <blindal...@dev.null.invalidwrote:
(snipped)
I thought an enum type was only
allowed to have values specified in the enumeration list. Why is compiler
not even warning about it? Am I wrong?
Thanks.
An excerpt from a draft of the C standard (n869.pdf):
"Annex I
(informative)
Common warnings
1 An implementation may generate warnings in many situations,
none of which are specified as part of this International Standard.
The following are a few of the more common situations.
....
A value is given to an object of an enumeration type other than
by assignment of an enumeration constant that is a member of that
type,
or an enumeration variable that has the same type,
or the value of a function that returns the same enumeration type
(6.7.2.2)"
So, I suppose some implementations may
generate the type of warning you had expected, but
it''s not required.
--
Hope this helps,
Steven
At_sea_with_C wrote:Consider the following code:
#include <stdio.h>
#define SOMEVAL 1234
enum tree_types { PINE = 10, BIRCH, LARCH, OAK = 100, MAPLE, ELM, WILLOW };
void print_val(enum tree_types tree)
{
printf("tree is: %d\n", tree);
}
int main()
{
enum tree_types trees;
trees = PINE;
print_val(trees);
trees = OAK;
print_val(trees);
trees = ELM;
print_val(trees);
trees = SOMEVAL;
print_val(trees);
return 0;
}
Why is the assignmnt with SOMEVAL allowed? I thought an enum type was only
allowed to have values specified in the enumeration list. Why is compiler
not even warning about it? Am I wrong?
Because C''s enums are broken. Maybe you where thinking of C++ where
this is illegal?
--
Ian Collins.
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