为什么 F# 突然从函数返回值作为 @+行号？

发布于 2024-12-10 14:34:57 字数 415 浏览 0 评论 0原文

我在 F# 中有一些代码工作正常，但现在有任何函数返回与 @ + 行号绑定的 let 名称吗？所以之前当我做这样的事情时：

let sqlData = GetSummary 2011

我会获取数据。现在我得到 sqlData@160 作为返回类型，这会破坏其他需要特定类型的函数。任何关于为什么会突然发生这种情况的帮助将不胜感激。

更新：所以我将 GetSummary 代码更改为这样，只是为了看看会发生什么：

 let GetSummary fiscalYear =
    let g = 22
    g

没有任何改变。我得到了完全相同的结果，即使我有一个断点，它也不会中断函数。它也不会进入 GetSummary。为什么它会忽略某些功能而不忽略其他功能？

原文

I have some code in F# that was working fine, but now any function returns the name of the let binding with @ + line number? So before when I did something like this:

let sqlData = GetSummary 2011

I would get the data. Now I get sqlData@160 as the return type, which break other functions expecting a specific type. Any help as to why this suddenly started happening would be appreciated.

UPDATE:
So I changed the GetSummary code to this just to see what would happen:

 let GetSummary fiscalYear =
    let g = 22
    g

And nothing changed. I got the exact same results and it would not break in the function even though I have a breakpoint. It also would not step into GetSummary either. Why would it ignore the some functions and not others?

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

与往事干杯 2024-12-17 14:34:57

我的猜测是您已经更改了 GetSummary 函数中的某些内容，以便它实际上返回一个函数而不是返回数据（当您给它一个参数时）。例如，此代码片段的行为有点相似 - 它返回一些内部内容，并且不进入 bar 函数内部：

> let bar n = printfn "inside bar"; n + 1;;
val bar : int -> int

> let foo n = bar;; // This takes 'n' and returns a function..
val foo : 'a -> (int -> int)

> printfn "%O" (foo 10);; // Here, we call 'foo', but the result is a function!
FSI_0016+it@15-3

My guess is that you've changed something in the GetSummary function, so that it actually returns a function instead of returning the data (when you give it a single argument). For example, this snippet behaves a bit similarly - it returns some internal things and it doesn't step inside the bar function:

> let bar n = printfn "inside bar"; n + 1;;
val bar : int -> int

> let foo n = bar;; // This takes 'n' and returns a function..
val foo : 'a -> (int -> int)

> printfn "%O" (foo 10);; // Here, we call 'foo', but the result is a function!
FSI_0016+it@15-3

回复收藏 0 原文