xcode Swift 不能使用函数指针吗？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/24088312/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Does Swift not work with function pointers?
提问by Mike
I'm trying to use a C library in Swift, and I'm having trouble calling any function that takes a function pointer as one of it's arguments. For example, part of the lua.h file that I'm trying to use in Swift looks like this:
我正在尝试在 Swift 中使用 C 库，但在调用任何将函数指针作为参数之一的函数时遇到问题。例如，我尝试在 Swift 中使用的 lua.h 文件的一部分如下所示：
LUA_API void (lua_setuservalue) (lua_State *L, int idx);
typedef int (*lua_CFunction) (lua_State *L);
LUA_API void (lua_callk) (lua_State *L, int nargs, int nresults, int ctx,
lua_CFunction k);
I use the bridging header to get access to the library, and from my Swift code I can call lua_setuservaluewithout any trouble. But if I try to call lua_callkI get "use of unresolved identifier 'lua_callk'". If I remove the function pointer from the declaration for lua_callk, I no longer get this error. Any help is quite appreciated.
我使用桥接头来访问库，并且从我的 Swift 代码中，我可以毫无困难地调用lua_setuservalue。但是，如果我尝试调用lua_callk，我会得到“使用未解析的标识符‘lua_callk’”。如果我从lua_callk的声明中删除函数指针，则不会再出现此错误。非常感谢任何帮助。
回答by Jim Hayes
This answer refers to an earlier version of the Swift language and may no longer be reliable.
这个答案指的是早期版本的 Swift 语言，可能不再可靠。
While C function pointers are notavailable in Swift, you can still use swift closures which are passed to C functions as blocks.
虽然 C 函数指针在 Swift中不可用，但您仍然可以使用作为块传递给 C 函数的 swift 闭包。
Doing so requires a few "shim" routines in C to take the block and wrap it in a C function. The following demonstrates how it works.
这样做需要在 C 中使用一些“填充”例程来获取块并将其包装在 C 函数中。下面演示了它是如何工作的。
Swift:
迅速：
func foo(myInt: CInt) -> CInt {
return myInt
}
var closure: (CInt) -> CInt = foo;
my_c_function(closure)
C:
C：
void my_c_function(int (^closure)(int))
{
int x = closure(10);
printf("x is %d\n", x);
}
Of course what you choose to do with the closure, and how you store and recall it for use is up to you. But this should give you a start.
当然，您选择如何处理封盖，以及如何储存和召回以供使用取决于您。但这应该给你一个开始。
回答by TiLogic
Apple has made function pointers available as of beta 3, however they can only be referenced not called.
Apple 从 beta 3 开始提供函数指针，但是它们只能被引用不能被调用。
Using Swift with Cocoa and Objective-C
将 Swift 与 Cocoa 和 Objective-C 结合使用
Function Pointers
C function pointers are imported into Swift as
CFunctionPointer<Type>, whereTypeis a Swift function type. For example, a function pointer that has the typeint (*)(void)in C is imported into Swift asCFunctionPointer<() -> Int32>
函数指针
C 函数指针作为 导入到 Swift 中
CFunctionPointer<Type>，其中Type是 Swift 函数类型。例如，一个具有int (*)(void)C 中类型的函数指针被导入到 Swift 中作为CFunctionPointer<() -> Int32>
Beta 3 Release Notes(PDF)
Beta 3 发行说明(PDF)
Function pointers are also imported now, and can be referenced and passed around. However, you cannot call a C function pointer or convert a closure to C function pointer type.
函数指针现在也被导入，并且可以被引用和传递。但是，您不能调用 C 函数指针或将闭包转换为 C 函数指针类型。
回答by Antarr Byrd
In the Apple documentationit is noted that C function pointers are not imported in Swift.
在 Apple文档中，注意到C function pointers are not imported in Swift.
回答by Kamil.S
Since Swift 3.1an arbitrary address being a function pointer can be called like this:
从Swift 3.1 开始，可以像这样调用任意地址作为函数指针：
let fakeIMP = unsafeBitCast(0x1e233d1d0, to: IMP.self)
unsafeBitCast(fakeIMP,to:(@convention(c)()->Void).self)()
Assuming for this^ particular call signature:
假设对于这个^特定的调用签名：
void cFunction();
