如何检查函数的类型参数是否静态解析? [英] How to check if a function's type parameters are statically resolved?
问题描述
非常简单的示例:
-
let myfun x = x
这里在智能感知中说"x:'a->'a".在FSI中，它显示为"x:'a->'a" -
let inline myfun x = x
这里在智能感知中说"x:'a->'a".在FSI中，它说"x:'a->'a"<<<<为什么不^a? -
let inline myfun (x: 'b) = x
这里在智能感知中说"x:'b->'b".在FSI中，它显示为"x:'b->'b" -
let inline myfun (x: ^b) = x
这里在智能感知中说"x:'b->'b".在FSI中，它表示"x:^ b-> ^ b"不同
let myfun x = x
Here in the intellisense it says "x: 'a -> 'a". In the FSI it says "x: 'a -> 'a"let inline myfun x = x
Here in the intellisense it says "x: 'a -> 'a". In the FSI it says "x: 'a -> 'a" <<<< why not^a?let inline myfun (x: 'b) = x
Here in the intellisense it says "x: 'b -> 'b". In the FSI it says "x: 'b -> 'b"let inline myfun (x: ^b) = x
Here in the intellisense it says "x: 'b -> 'b". In the FSI it says "x: ^b -> ^b" <<<< different
既然智能感知器从不显示^b，那么我应该寻找^b作为FSI中静态解析"的指标吗?
Since the intellisense never shows ^b, should I look for ^b as an indicator of "statically resolved" in FSI?
inline是否保证静态解析"?
Does inline guarantee "statically resolved"?
推荐答案
内联确实允许但不强制使用静态解析类型，这就是为什么在情况2中它与情况1相同.
我认为在大多数情况下，即使您未指定^，类型推断也足够聪明，可以猜测是否应该静态解析类型.
Inline does allow but does not force statically resolved types, that's why in case 2. it remains the same as in case 1.
I think in most cases type inference is smart enough to guess if the type should really be statically resolved, even if you don't specify the ^.
例如，如果在情况3中将函数主体更改为sqrt x，则会得到
For example if you change your function body to sqrt x in case 3. you'll get
> let inline myfun (x: 'b) = sqrt x;;
val inline myfun : ^b -> ^a when ^b : (static member Sqrt : ^b -> ^a)
我个人总是一开始尝试不明确地指定类型，然后检查是否对推断感到满意，如果不满意，则尝试添加内联而不是帽子类型.
I personally always try not to specify types explicitly at first try, then I check if I'm happy with the inference, if I'm not then I try adding inline, but not the hat types.
为什么智能感知有时会有所不同?那可能是一个小错误.
Why intellisense shows sometimes something different? that's probably a small bug.
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