如何使用Swift调用需要参数类型(int *)的Objective-C方法? [英] How can I use Swift to call an objective-C method that expects an argument of type (int *)?

问题描述

现有的Objective-C方法具有以下签名:

An existing objective-C method has the following signature:

-(BOOL)barcodeSetScanBeep:(BOOL)enabled volume:(int)volume beepData:(int *)data length:(int)length error:(NSError **)error;

请注意，beepData:应该是(int *).

Note that beepData: expects (int *).

可以通过传入C数组在Objective-C中使用此方法:

This method can be used from objective-C by passing in a C array:

int beepData[] = {1200,100};

如何从Swift调用相同的方法?我最好的尝试let beepData: [Int] = [1200, 100]无法编译.

How can I call the same method from Swift? My best attempt, let beepData: [Int] = [1200, 100], doesn't compile.

推荐答案

int是一个32位C整数，并映射为Int32到Swift.

int is a C 32-bit integer and mapped to Swift as Int32.

一个int *参数映射为UnsafeMutablePointer<Int32>到Swift， 并且您可以通过&将变量数组作为输入参数"传递.

A int * parameter is mapped to Swift as UnsafeMutablePointer<Int32>, and you can pass a variable array as "inout parameter" with &.

因此它应该大致如下:

var beepData : [ Int32 ] = [ 1200, 100 ]
var error : NSError?
if !DTDevices.sharedDevice().barcodeSetScanBeep(true, volume: Int32(100),
                beepData: &beepData, length: Int32(beepData.count),
                error: &error) {
    println(error!)
}

Swift还定义了类型别名

Swift defines also a type alias

/// The C 'int' type.
typealias CInt = Int32

，因此，如果您想强调一下，可以在上面的代码中用CInt替换Int32 您正在使用C整数.

so you could replace Int32 by CInt in above code if you want to emphasize that you are working with C integers.

有关更多信息，请参见与C API交互" "文档.

For more information, see "Interacting with C APIs" in the "Using Swift with Cocoa and Objective-C" documentation.

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