复杂的C演员解释
我想弄清楚以下 C 代码的作用是什么？
((void(*)())buf)();
其中“buf”是一个 char 数组。
I'm trying to figure out what the following code in C does?
((void(*)())buf)();
where 'buf' is a char array.
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让我们一步一步来。
这是一个指向函数的指针，该函数接受未指定的参数并且没有返回值。
只需将 buf 转换为该函数指针类型。最后
调用该函数。
所以整个语句是“将
buf解释为指向不带参数的void函数的指针，并调用该函数。”Let's take it one step at a time.
This is a pointer to a function that takes unspecified arguments and has no return value.
simply casts buf to this function pointer type. Finally,
calls this function.
So the entire statement is "interpret
bufas a pointer to avoidfunction without arguments, and call that function."它将
buf转换为void(*)()类型的函数指针（一个不返回任何内容/void 并接受未指定参数的函数）并调用它。ANSI 标准实际上并不允许将普通数据指针转换为函数指针，但您的平台可能允许。
It casts
bufto a function pointer of typevoid(*)()(A function returning nothing/void and taking unspecified arguments) and calls it.The ANSI standard does not really allow the casting of normal data pointers to function pointers, but your platform may allow it.
当我遇到令人难以置信的声明时，我倾向于使用“cdecl”命令。示例：
虽然在某些情况下我确实希望有一个工具可以解释“cdecl”的输出：/
I tend to use the "cdecl" command when I come across a mind boggling declaration. Example:
Although there are cases where I do wish that there's a tool out there that explains the output of "cdecl" :/
这会将
buf转换为void (*)()类型，这是一个指向接受未指定参数且不返回任何内容的函数的指针。然后它调用该地址（最右边的两个括号）处的函数。This casts
bufto the typevoid (*)(), a pointer to a function that takes unspecified parameters and returns nothing. Then it calls the function at that address (the two rightmost parentheses).((void(*)())buf)(); \------------/ cast `buf` to \---------/ type: pointer to function accepting a fixed but unspecified number of arguments and returning void \----------------/ and call that "function"((void(*)())buf)(); \------------/ cast `buf` to \---------/ type: pointer to function accepting a fixed but unspecified number of arguments and returning void \----------------/ and call that "function"它将 buf 转换为函数指针，该指针接受未指定的参数并调用它。
It casts
bufinto a function pointer, that takes unspecified arguments, and calls it.我猜想在很多情况下，它都会使机器崩溃。
否则，它将数组视为指向返回 void 的函数的指针并调用它。
I would guess that in many circumstances, it crashes the machine.
Otherwise, it treats the array as a pointer to a function that returns void and invokes it.
你可能会发现“专家c编程”是一本很好的读物——如果我没记错的话，解压这类东西在其中一章中。我已经很久没有读过这本书了，但我记得当时我认为这是值得的。
http://www.amazon.com/Expert-Programming-Peter-van-Linden/ dp/0131774298
You might find "expert c programming" a good read - unpacking this kind of thing is in one of the chapters, if I remember right. It's a long time since I read it, but I remember thinking it was worth the effort at the time.
http://www.amazon.com/Expert-Programming-Peter-van-Linden/dp/0131774298
lsc 提到有一个在线版本的“cdecl”工具，您可能会觉得有用：http://www.cdecl .org/
There is an online version of the 'cdecl' tool that lsc mentioned that you might find useful : http://www.cdecl.org/
调用函数指针。该函数没有参数。
函数指针 - 维基百科
calls a function pointer. the function has no arguments.
Function Pointer - Wikipedia