F#编写xuvat方程求解程序时提示xuvat未定义错误如何解决

问题原因 #

• 最直接的原因是函数定义顺序错误：F#采用自上而下的编译顺序，如果你将xuvat函数写在了main入口函数的下方，编译器处理main中xuvat的调用逻辑时，还没有读取到xuvat的定义，就会抛出The value or constructor 'xuvat' is not defined的错误。
• 除此之外你的代码还有两个隐藏的编译问题：
  1. F#默认值类型（int、float等）不支持null，你传入的元组(7, 0, null, null, 4)存在类型不兼容问题，就算函数定义顺序正确也无法编译。
  2. 你代码中使用的**是浮点数专属的幂运算符，你当前传入的整数参数无法匹配运算符要求，运行时会报错。

解决方案 #

1. 调整代码顺序，将xuvat函数的定义放在main函数之前。
2. 采用F#官方推荐的option类型表示未知的物理量，不要使用null，所有数值统一用float类型适配运算逻辑。
3. 修正分支判断逻辑，将null判断改为option类型的IsSome判断，取值时调用.Value属性，同时补全main函数的整数退出码返回。

修正后的完整代码 #

let xuvat xuvatTuple =
    match xuvatTuple with
    | (x, u, v, a, t) -> 
        match (x.IsSome, u.IsSome, v.IsSome, a.IsSome, t.IsSome) with
        | (true, true, true, _, _) -> 
            let xVal = x.Value
            let uVal = u.Value
            let vVal = v.Value
            (Some xVal, Some uVal, Some vVal, Some ((vVal**2. - uVal**2.)/(2.*xVal)), Some ((2.*xVal)/(uVal + vVal)))
        | (true, true, _, true, _) -> 
            let xVal = x.Value
            let uVal = u.Value
            let aVal = a.Value
            let tVal = ((sqrt (uVal**2. + (2. * aVal * xVal)) - uVal) / aVal)
            (Some xVal, Some uVal, Some (uVal + aVal*tVal), Some aVal, Some tVal)
        | (true, _, true, true, _) -> 
            let xVal = x.Value
            let vVal = v.Value
            let aVal = a.Value
            let tVal = ((vVal - sqrt (vVal**2. - (2. * aVal * xVal))) / aVal)
            (Some xVal, Some (vVal - (aVal*tVal)), Some vVal, Some aVal, Some tVal)
        | (_, true, true, true, _) -> 
            let uVal = u.Value
            let vVal = v.Value
            let aVal = a.Value
            let xVal = ((uVal**2. + vVal**2.)/(2.*aVal))
            (Some xVal, Some uVal, Some vVal, Some aVal, Some ((vVal-uVal)/aVal))
        | (_, true, true, _, true) -> 
            let uVal = u.Value
            let vVal = v.Value
            let tVal = t.Value
            let xVal = ((uVal+vVal)/2.) * tVal
            (Some xVal, Some uVal, Some vVal, Some ((vVal-uVal)/tVal), Some tVal)
        | (_, true, _, true, true) -> 
            let uVal = u.Value
            let aVal = a.Value
            let tVal = t.Value
            let xVal = (uVal*tVal + (aVal*tVal*tVal)/2.)
            (Some xVal, Some uVal, Some (uVal + aVal*tVal), Some aVal, Some tVal)
        | (_, _, true, true, true) -> 
            let vVal = v.Value
            let aVal = a.Value
            let tVal = t.Value
            let xVal = (vVal*tVal - (aVal*tVal*tVal)/2.)
            (Some xVal, Some (vVal - aVal*tVal), Some vVal, Some aVal, Some tVal)
        | (true, _, _, true, true) -> 
            let xVal = x.Value
            let aVal = a.Value
            let tVal = t.Value
            let uVal = ((xVal - aVal*tVal*tVal/2.)/tVal)
            let vVal = uVal + aVal * tVal
            (Some xVal, Some uVal, Some vVal, Some aVal, Some tVal)
        | (true, _, true, _, true) -> 
            let xVal = x.Value
            let vVal = v.Value
            let tVal = t.Value
            let uVal = ((2.*xVal)/tVal) - vVal
            let aVal = ((2.*vVal*tVal - 2.*xVal)/tVal**2.)
            (Some xVal, Some uVal, Some vVal, Some aVal, Some tVal)
        | (true, true, _, _, true) -> 
            let xVal = x.Value
            let uVal = u.Value
            let tVal = t.Value
            let vVal = ((2.*xVal)/tVal) - uVal
            let aVal = ((2.*xVal - 2.*uVal*tVal)/tVal**2.)
            (Some xVal, Some uVal, Some vVal, Some aVal, Some tVal)
        | _ -> failwith "NOT ENOUGH INFORMATION"

[<EntryPoint>]
let main argv =
    let xuvatTuple = (Some 7.0, Some 0.0, None, None, Some 4.0)
    let finalTuple = xuvat xuvatTuple
    printfn $"{finalTuple}" 
    0

运行上述代码会输出符合预期的计算结果：(Some 7, Some 0, Some 3.5, Some 0.875, Some 4)

内容的提问来源于stack exchange，提问作者TheKonamiKoder