如何访问和控制Swift中的char ***

Question

我的c API如下所示：

int enumDataFormat(char ***SampleNames);

Swift代码：

var formats =  UnsafeMutablePointer<UnsafeMutablePointer<Int8>>.alloc(1)
let count = enumDataFormat(&formats)
for index in 0..<count {
    //Question in here: I don't known how to access memory in Swift
}

Objective-C代码：

工作得非常好！

char **formats;
int count = enumDataFormat(&formats);
if (count > 0) {
    for (int i = 0; i < count; i++) {

        //obtain the string from memory
        char *formatString = *(formats + i);
        NSLog(@"%s", formatString);
    }
}

  

问题：在Swift中，如何访问内存中的字符串？

Answer 1

你可以使用UnsafePointers在Swift中进行指针运算：

let oldPointer = UnsafePointer<UInt8>()
let newPointer = oldPointer + 10

它创建10个字节的偏移量。

Answer 2

你不需要alloc()，只需要一个类型＆＃34;指向的变量 指向Int8＆＃34;的指针，类似于（Objective-）C代码：

var formats : UnsafeMutablePointer<UnsafeMutablePointer<Int8>> = nil
let count = Int(enumDataFormat(&formats))

您只需使用下标符号即可访问第i个C字符串指针...

for i in 0 ..< count {

    let formatString = formats[i]

...并将其转换为带

的Swift字符串

    if let format = String.fromCString(formatString) {
        println(format)
    } else {
        // Not a valid UTF-8 string.
    }
}