peekCString 和 peekCStringLen 是懒惰的吗？

发布于 2024-08-11 16:46:30 字数 459 浏览 5 评论 0原文

我有一个 C 函数，它创建一个以 null 结尾的字符串并返回指向它的指针，还有相应的释放函数。

foreign import ccall unsafe "get_str" getStr :: IO CString
foreign import ccall unsafe "free_str" freeStr :: CString -> IO ()

我想从返回的 CString 创建一个 Haskell String，并尽快释放 CString。

do cStr <- getStr
   str <- peekCString cStr
   freeStr cStr
   -- here str is used

在使用 str 之前释放 cStr 安全吗？换句话说，peekCString是一次性创建Haskell String，还是延迟创建？

原文

I have a C function that creates a null terminated string and returns a pointer to it, there is also corresponding deallocation function.

foreign import ccall unsafe "get_str" getStr :: IO CString
foreign import ccall unsafe "free_str" freeStr :: CString -> IO ()

I want to create a Haskell String from the returned CString, and free CString as soon as possible.

do cStr <- getStr
   str <- peekCString cStr
   freeStr cStr
   -- here str is used

Is it safe to free cStr before str is used? In other words, does peekCString create Haskell String all at once, or is it created lazily?

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往昔成烟 2024-08-18 16:46:30

peekCString 是严格的——例如，它不会通过 unsafeInterleaveIO 暂停循环，所以一旦你有了字符串的头部，你肯定已经计算了尾部。这是实现：

peekCAString cp = do
  l <- lengthArray0 nUL cp
  if l <= 0 then return "" else loop "" (l-1)
  where
    loop s i = do
        xval <- peekElemOff cp i
        let val = castCCharToChar xval
        val `seq` if i <= 0 then return (val:s) else loop (val:s) (i-1)

peekCString is strict -- it doesn't suspend the loop via unsafeInterleaveIO, for example, so once you have the head of the string, you've definitely already computed the tail. Here's the implementation:

peekCAString cp = do
  l <- lengthArray0 nUL cp
  if l <= 0 then return "" else loop "" (l-1)
  where
    loop s i = do
        xval <- peekElemOff cp i
        let val = castCCharToChar xval
        val `seq` if i <= 0 then return (val:s) else loop (val:s) (i-1)

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