F#使用and递归绑定出现null问题，求修复方法

原问题代码 #

type QueryInfo = 
    {   Title: string
        Check: Client -> bool
        Positive: Decision
        Negative: Decision
    }
and Decision = 
    | Result of string
    | Query of QueryInfo
let rec tree1 = 
    Query({Title=null;Check=(fun cl -> false);Positive=aaa; Negative=aaa})
and aaa = Result("SOME")

问题原因 #

F#处理相互递归的直接值绑定时，对于引用类型（比如Decision联合类型，底层为引用类型），会先为变量分配内存并初始化为null，再后续填充实际值。当tree1初始化时，aaa还未完成实际值的填充，因此Positive和Negative会被赋值为null。

修复方案 #

方案1：用函数延迟求值

将递归值改为函数，利用函数调用的延迟特性，确保求值时依赖项已完成初始化：

type QueryInfo = 
    {   Title: string
        Check: Client -> bool
        Positive: Decision
        Negative: Decision
    }
and Decision = 
    | Result of string
    | Query of QueryInfo

let rec tree1 () = 
    Query({Title=null;Check=(fun cl -> false);Positive=aaa(); Negative=aaa()})
and aaa () = Result("SOME")

使用时调用tree1()即可获取完整的Decision值。

方案2：用Lazy<T>延迟初始化

通过lazy包装递归值，在需要时再强制求值，避免提前引用未初始化的变量：

type QueryInfo = 
    {   Title: string
        Check: Client -> bool
        Positive: Decision
        Negative: Decision
    }
and Decision = 
    | Result of string
    | Query of QueryInfo

let rec tree1 = lazy (
    Query({Title=null;Check=(fun cl -> false);Positive=aaa.Value; Negative=aaa.Value})
)
and aaa = lazy (Result("SOME"))

使用时通过tree1.Value获取实际的Decision值。

方案3：自引用场景优化（若tree1需指向自身）

如果实际需求是让Positive和Negative指向tree1自身，可用as关键字简化：

type QueryInfo = 
    {   Title: string
        Check: Client -> bool
        Positive: Decision
        Negative: Decision
    }
and Decision = 
    | Result of string
    | Query of QueryInfo

let rec tree1 = 
    let rec info = {
        Title = null
        Check = (fun cl -> false)
        Positive = Query info
        Negative = Query info
    }
    Query info

内容的提问来源于stack exchange，提问作者dondublon