c - 在 C 中使用 Swift 中的 Char 数组

Question

目前我有三个文件

主.swift

var dog = Dog()

dog.age = 12
dog.name = "H" // This is the non workable code

CTester.c

#include <stdio.h>

struct Dog {
    int age;
    char name[10];
} Dog;

和 C 测试桥接 header

#import "CTester.c"

我试图在 Swift 中使用 C 结构，但是，Char 数组在 Swift 中显示为 10 Int8 的数组。如何获取 Swift 字符串并将其分配给 char 数组？

Answer 1

因此，您可以像这样为您的Dog 编写一些扩展:

swift 2

extension Dog {
    var Name: String {
        mutating get {
            return withUnsafePointer(&self.name) {namePtr in
                let charPtr = UnsafePointer<CChar>(namePtr)
                //Sorry, this code may crash here...
                return String(CString: charPtr, encoding: NSUTF8StringEncoding)!
            }
        }
        set {
            withUnsafeMutablePointer(&self.name) {namePtr in
                let charPtr = UnsafeMutablePointer<CChar>(namePtr)
                let size = sizeofValue(self.name)
                strncpy(charPtr, newValue, size - 1)
                charPtr[size - 1] = 0
            }
        }
    }
}

Swift 3(使用 Xcode 8 beta 4 测试)

extension Dog {
    var Name: String {
        mutating get {
            return withUnsafePointer(&self.name) {namePtr in
                let charPtr = UnsafePointer<CChar>(namePtr)
                //The result may contain the Unicode replacement character ("\u{FFFD}")
                return String(cString: charPtr)
            }
        }
        set {
            withUnsafeMutablePointer(&self.name) {namePtr in
                let charPtr = UnsafeMutablePointer<CChar>(namePtr)
                let size = sizeofValue(self.name)
                strncpy(charPtr, newValue, size - 1)
                charPtr[size - 1] = 0
            }
        }
    }
}

看到the linked thread在 Martin R 的评论中，可能有一些改进的空间......(特别是使用 strlcpy 减少你的代码大小和更好的安全性。请检查它。)

但无论如何，它的作用是:

dog.Name = "H"
print(dog.Name) //->H

---

我发现 Name 的 getter 可能会使您的应用程序崩溃...在您仅存储 ASCII 字符的限制下“有效”。