@djyde, thanks for your PR! By analyzing the history of the files in this pull request, we identified @simaQ, @benjycui and @RaoHai to be potential reviewers.

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Merging #7119 into master will increase coverage by <.01%.
The diff coverage is 100%.

@@            Coverage Diff             @@
##           master    #7119       /-   ##
==========================================
  Coverage   85.87%   85.87%    <.01%     
==========================================
  Files         231      231              
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  Misses        700      700

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LGTM.
But when you write options like below will have no effects:

const mapPropsToFields = (props: ArticleDetailOwnProps)  => props ...//
const onFieldsChange = (props, fields) => null

const ArticleDetailForm = Form.create<TOwnProps>({ onFieldsChange, mapPropsToFields })(ArticleDetail)

@BANG88 Of course, because when you define a function (or arrow function), the compiler has no idea where it will be, so the params is free to declare. But when you put the function to another function as a param (like Form.create), the compiler will check if the params type is correct.

Maybe not, only useful when you write function inside options, otherwise we don't know the type definitions.

@BANG88

你可能没有理解我的意思，我用中文表达吧。

你在定义一个函数的时候，参数的类型肯定是自由的。但是当这个函数是用作参数的话，就是有限制的了。

比如：

const mapPropsToFields = (props) => {
  return {
  }
}

export default Form.create<ICommentFormProps>({
  mapPropsToFields
})(CommentForm)

这样编译器不会认为错，是因为 props 默认就是 any.

但是如果你传一个不是 ICommentFormProps(也就是 TOwnProps) 类型的时候，编译器就会报错：

所以如果你需要把函数在外定义，编译器只能保证这个函数作为参数传递时它接受的参数的类型是安全的 (除了any)，至于 Intelisense, 就只能靠你手动声明类型了：

const mapPropsToFields = (props: TOwnProps) => {
  return {
  }
}

嗯 定义在外面需要手动添加类型。

conflict

resovled