为什么C预处理器认为枚举值相等? [英] Why does the C preprocessor consider enum values as equal?
问题描述
即使 A
和<$ c $,为什么以下代码中的 std :: cout
行也会运行c> B 不同吗?
Why does the std::cout
line in the following code run even though A
and B
are different?
#include <iostream>
enum T { A = 1, B = 2 };
// #define A 1
// #define B 2
int main() {
#if (A == B)
std::cout << A << B;
#endif
}
如果我使用#相反,定义
(如已注释掉),我没有得到预期的输出。
If I use #define
instead (as commented out), I get no output as I expect.
问题的原因:
我想为某些测试代码提供一个模式选择器,在其中可以通过在顶部注释/取消注释行来轻松更改模式:
I want to have a mode selector for some test code in which I can easily change modes by commenting/uncommenting lines on top:
enum T { MODE_RGB = 1, MODE_GREY = 2, MODE_CMYK = 3 };
// #define MODE MODE_RGB
#define MODE MODE_GREY
// #define MODE MODE_CMYK
int main() {
#if (MODE == MODE_RGB)
// do RGB stuff
#elif (MODE == MODE_GREY)
// do greyscale stuff
#else
// do CMYK stuff
#endif
// some common code
some_function(arg1, arg2,
#if (MODE == MODE_RGB)
// RGB calculation for arg3,
#elif (MODE == MODE_GREY)
// greyscale calculation for arg3,
#else
// CMYK calculation for arg3,
#endif
arg4, arg5);
}
我知道我可以使用数字值,例如
I know I can use numeric values e.g.
#define MODE 1 // RGB
...
#if (MODE == 1) // RGB
,但这会使代码的可读性降低。
but it makes the code less readable.
推荐答案
没有称为 A
的宏,或者 B
,因此在您的 #if
行, A
和 B
被替换为 0
,因此您实际上拥有:
There are no macros called A
or B
, so on your #if
line, A
and B
get replaced by 0
, so you actually have:
enum T { A = 1, B = 2 };
int main() {
#if (0 == 0)
std::cout << A << B;
#endif
}
预处理器在编译器对您的信息一无所知之前运行枚举
。预处理器只知道宏( #define
)。
The preprocessor runs before the compiler knows anything about your enum
. The preprocessor only knows about macros (#define
).
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