将C字符串数组转换为Swift字符串数组 [英] Converting array of C strings to Swift string array
问题描述
在Swift 3中,签名为const char *f()
的C函数在导入时被映射为UnsafePointer<Int8>! f()
.结果可以转换为Swift字符串,如下所示:
In Swift 3, C function with signature const char *f()
is mapped to UnsafePointer<Int8>! f()
on import. It's result can be converted to a Swift string as:
let swiftString = String(cString: f())
问题是,如何将以NULL结尾的C字符串C数组映射到Swift字符串数组?
The question is, how a NULL terminated C array of C strings can be mapped to Swift array of strings?
原始的C签名:
const char **f()
导入的Swift签名:
Imported Swift signature:
UnsafeMutablePointer<UnsafePointer<Int8>?>! f()
快速字符串数组:
let stringArray: [String] = ???
推荐答案
据我所知,没有内置方法.
您必须遍历返回的指针数组,将C字符串转换为Swift String
,直到找到nil
指针:
There is no built-in method as far as I know.
You have to iterate over the returned pointer array, converting C strings to Swift String
s, until a nil
pointer is found:
if var ptr = f() {
var strings: [String] = []
while let s = ptr.pointee {
strings.append(String(cString: s))
ptr += 1
}
// Now p.pointee == nil.
print(strings)
}
备注: Swift 3使用可选的指针作为可以为nil
的指针.
在您的情况下,f()
返回一个隐式展开的可选内容,因为
头文件未经过审核":编译器不知道是否
该函数是否可以返回NULL
.
Remark: Swift 3 uses optional pointers for pointers that can be nil
.
In your case, f()
returns an implicitly unwrapped optional because
the header file is not "audited": The compiler does not know whether
the function can return NULL
or not.
使用可空性注释"可以提供该信息 到Swift编译器:
Using the "nullability annotations" you can provide that information to the Swift compiler:
const char * _Nullable * _Nullable f(void);
// Imported to Swift as
public func f() -> UnsafeMutablePointer<UnsafePointer<Int8>?>?
如果函数可以返回NULL
和
const char * _Nullable * _Nonnull f(void);
// Imported to Swift as
public func f() -> UnsafeMutablePointer<UnsafePointer<Int8>?>
如果保证f()
返回非NULL结果.
if f()
is guaranteed to return a non-NULL result.
有关可空性注释的更多信息,请参见例如 在Swift博客中可空性和Objective-C .
For more information about the nullability annotations, see for example Nullability and Objective-C in the Swift blog.
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