Swift不能与函数指针一起使用吗？-DoesSwiftnotworkwithfunctionpointers?

10  

Apple has made function pointers available as of beta 3, however they can only be referenced not called.

从beta 3开始，Apple已经提供了函数指针，但是它们只能被引用而不能被调用。

Using Swift with Cocoa and Objective-C

将Swift与Cocoa和Objective-C一起使用

Function Pointers

C function pointers are imported into Swift as CFunctionPointer, where Type is a Swift function type. For example, a function pointer that has the type int (*)(void) in C is imported into Swift as CFunctionPointer<() -> Int32>

C函数指针作为CFunctionPointer 导入Swift，其中Type是Swift函数类型。例如，在C中将类型为int（*）（void）的函数指针作为CFunctionPointer <（） - > Int32>导入到Swift中

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，可以被引用和传递。但是，您无法调用C函数指针或将闭包转换为C函数指针类型。

15  

This answer refers to an earlier version of the Swift language and may no longer be reliable.

这个答案涉及Swift语言的早期版本，可能不再可靠。

While C function pointers are not available 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swift闭包，它们作为块传递给C函数。

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。以下演示了它的工作原理。

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，您选择使用闭合装置，以及如何存储和调用它以供使用取决于您。但这应该给你一个开始。

2  

In the Apple documentation it is noted that C function pointers are not imported in Swift.

在Apple文档中，注意到在Swift中不导入C函数指针。