无C-的malloc（）在雨燕'D内存？

问题描述

我使用的是斯威夫特编译器的桥接报头的功能调用，使用的malloc分配内存的C函数（）。然后，它返回一个指向该内存。该函数的原型是这样的：

I'm using the Swift compiler's Bridging Header feature to call a C function that allocates memory using malloc(). It then returns a pointer to that memory. The function prototype is something like:

char *the_function(const char *);

在斯威夫特，我用它是这样的：

In Swift, I use it like this:

var ret = the_function(("something" as NSString).UTF8String)

let val = String.fromCString(ret)!

请原谅我关于雨燕，但通常在C，如果the_function（）被malloc'ing内存还，别人需要释放（），它在某个时候。

Forgive my ignorance concerning Swift but normally in C, if the_function() is malloc'ing memory and returning it, somebody else needs to free() it at some point.

这是由雨燕莫名其妙地处理还是我在这个例子中内存泄露？

Is this being handled by Swift somehow or am I leaking memory in this example?

先谢谢了。

推荐答案

斯威夫特不管理与分配的内存的malloc（），最终你必须要释放内存

Swift does not manage memory that is allocated with malloc(), you have to free the memory eventually:

let ret = the_function("something") // returns pointer to malloc'ed memory
let str = String.fromCString(ret)!  // creates Swift String by *copying* the data
free(ret) // releases the memory

println(str) // `str` is still valid (managed by Swift)

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请注意，一个斯威夫特字符串被自动转换为UTF-8
当传递给C函数取为const char * 参数字符串
在String价值UnsafePointer＆LT;＆UINT8 GT;功能参数的行为。
这就是为什么

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Note that a Swift String is automatically converted to a UTF-8string when passed to a C function taking a const char * parameteras described in String value to UnsafePointer<UInt8> function parameter behavior.That's why

let ret = the_function(("something" as NSString).UTF8String)

可以简化为

let ret = the_function("something")

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