c - 在 Swift 中使用将函数作为参数的 C 函数

Question

我正在围绕 C 数学库编写一个包装器。每个函数都接受一个或两个函数作为参数。但是，这些子函数(以及父函数)的参数不是 Swifty - 因此是包装器。

我已经清理了示例代码，只显示了三个主要部分:c 库函数，将传递给包装器的所需 Swift 函数(主体未显示，但围绕着 c 库函数) , 以及所需的 C 函数形式。

//C library function, that calls the passed function dozens, hundreds or thousands of times, each time it changes the data provided in p, and uses the output from x
//The Swift arrays are passed as pointers, and the length of the and x array are m and n respectively
returnValue = cLibraryFunc(passedFunc, &p, &x, Int32(m), Int32(n), Int32(itmax), &opts, &info, &work, &covar, &adata)


//I would like to create a Swift function that would look like this (internals could be any myriad of things that takes inputs p and adata and returns data in x:
func desiredSwifty(p: inout [Double], x: inout [Double], m: Int, n: Int, adata: inout [Double]) {
    //very simple example
    //this example knows the length of p (so m as well)
    //and assumes that adata length is the same as the x length (n)
    //obviously, it could ifer m and n from p.count and x.count

    for i in 0..<n {
        x[i] = p[0] + p[1]*adata[i]  + p[2]*pow(adata[i], 2)
    }
}


//And the wrapper would "convert" it -internally- into the form that the C library function requires:
func requiredC(p: UnsafeMutablePointer<Double>?, x: UnsafeMutablePointer<Double>?, m: Int32, n: Int32, adata: UnsafeMutablePointer<Void>?) {
    //same thing, but using pointers, and uglier

    //first, have to bitcast the void back to a double
    let adataDouble : UnsafeMutablePointer<Double> = unsafeBitCast(adata, to: UnsafeMutablePointer<Double>.self)

    for i in 0..<Int(n) {
        x![i] = p![0] + p![1]*adataDouble[i]  + p![2]*pow(adataDouble[i], 2)
    }
}

添加

我应该补充一点，我可以访问 c 源代码，所以我可以添加一些虚拟参数(可能是为了找到一种方法来传递上下文)。但是考虑到文档似乎表明不能使用 c 函数指针获取上下文，这可能没有用。

Answer 1

(注意:以下示例在 Xcode 8 beta 2 上使用 Swift 3。)

您的问题是关于将另一个 C 函数作为参数的 C 函数，因此让我们将问题简化为该问题。这是一个简单的 C 函数，它接受一个参数，它是又是一个 C 函数，它接受一个指向 double 组的指针和一个整数计数:

// cfunction.h:
void cFunc(void (*func)(double *values, int count));

// cfunction.c:
void cFunc(void (*func)(double *values, int count)) {
    double x[] = { 1.2, 3.4, 5,6 };
    func(x, 3);
}

这个函数被导入到 Swift 中作为

func cFunc(_ func: (@convention(c) (UnsafeMutablePointer<Double>?, Int32) -> Swift.Void)!)

这里 @convention(c) 声明 block 具有 C 风格的调用惯例。特别是，在 Swift 中，您只能传递一个全局函数或一个不捕获任何上下文的闭包。

Swift 包装器的一个简单示例是

func swiftyFunc(passedFunc: (@convention(c) (UnsafeMutablePointer<Double>?, Int32) -> Void)) {
    cFunc(passedFunc) 
}

你可以这样使用:

func functionToPass(values: UnsafeMutablePointer<Double>?, count: Int32) {
    let bufPtr = UnsafeBufferPointer(start: values, count: Int(count))
    for elem in bufPtr { print(elem) }
}

swiftyFunc(passedFunc: functionToPass)

或者使用闭包参数:

swiftyFunc { (values, count) in
    let bufPtr = UnsafeBufferPointer(start: values, count: Int(count))
    for elem in bufPtr { print(elem) }
}