为什么我不能将可选的Swift String传递给允许NULL指针的C函数? [英] Why can't I pass an optional Swift String to C function that allows NULL pointers?

问题描述

我有一个处理C字符串的C函数.该函数实际上允许字符串为NULL指针.声明如下:

I have a C function that deals with C strings. The function actually allows strings to be NULL pointers. The declaration is like follows:

size_t embeddedSize ( const char *_Nullable string );

在C语言中使用此功能没问题:

It is no problem to use this function like this in C:

size_t s1 = embeddedSize("Test");
size_t s2 = embeddedSize(NULL); // s2 will be 0

现在我正在尝试从Swift中使用它.以下代码有效

Now I'm trying to use it from Swift. The following code works

let s1 = embeddedSize("Test")
let s2 = embeddedSize(nil) // s2 will be 0

但是什么都行不通呢?是给它传递一个可选的字符串！该代码将无法编译:

But what doesn't work is passing an optional string to it! This code will not compile:

let someString: String? = "Some String"
let s2 = embeddedSize(someString)

编译器抛出一个关于未解开可选选项的错误，Xcode询问我是否忘记添加!或?.但是，为什么我要拆开它? NULL或nil是要馈入函数的有效值.参见上文，我直接将nil传递给它，并且编译得很好，并返回了预期的结果.在我的真实代码中，字符串是从外部馈送的，它是可选的，我不能强行解开字符串，如果字符串为nil，则字符串将断开.那我怎么用一个可选的字符串来调用那个函数呢?

The compiler throws an error about the optional not being unwrapped and Xcode asks me if I maybe forgot to add ! or ?. However, why would I want to unwrap it? NULL or nil are valid values to be fed to the function. See above, I just passed nil to it directly and that compiled just well and returned the expected result. In my real code the string is fed from external and it is optional, I cannot force unwrap it, that will break if the string was nil. So how can I call that function with an optional string?

推荐答案

最可能的答案是，虽然字符串文字可转换为UnsafePointer<CChar>，而nil可转换为UnsafePointer<CChar>，而String也是，String?可能不在Swift 2中.

The most likely answer is that while string literals are convertible to UnsafePointer<CChar>, and nil is convertible to UnsafePointer<CChar>, and String also is, String? might not be in Swift 2.

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