问题描述
函数指针定义的常用形式是:
The usual form of function pointer definitions is:
int function(int, int);
int (*ptr)(int, int);
但是我今天看到了一个我不理解的表格.有人可以解释一下吗?
but I saw a form today which I didn't understand. Can anyone explain this please?
int (*close) __P((struct __db *));
推荐答案
__P()宏通常用于支持从K& RC时代起没有原型(从C89引入C的时代)开始的C实现. ).逻辑基本上是
The __P() macro is usually used to support C implementations from the days of K&R C, when there were no prototypes (which were introduced to C with C89). Basically the logic is
#if SOME_LOGIC_TO_TEST_WHETHER_IMPLEMENTATION_SUPPORTS_PROTOTYPES
# define __P(argument_list) argument_list
#else
# define __P(argument_list) ()
#endif
应用于示例时，您能看到它的工作原理吗?请注意，要使此方法起作用并且不会引起语法错误，参数列表必须包含 函数调用 的括号，而不仅仅是 类似函数的宏 .因此，使用宏时会在双括号内.这可能就是它看起来不寻常的原因.
Can you see how this works when applied to your example? Note that for this to work and not cause a syntax error, the argument list must include the parentheses of the function call, not just the parentheses of the function-like macro. Hence the double parentheses when the macro is used. That's probably the reason why it looks unusual.
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