F# 此表达式的类型应为string,但此处的类型为Gen<;字符串>;
我有这样的记录:F# 此表达式的类型应为string,但此处的类型为Gen<;字符串>;,f#,fscheck,F#,Fscheck,我有这样的记录: type Client = { Name : string; Income : int ; YearsInJob : int UsesCreditCard : bool; CriminalRecord : bool } 为了创建此记录的生成器,我尝试了以下代码: let chooseFromList xs = gen { let! idx = Gen.choose(0, List.length xs - 1) ret
type Client =
{ Name : string; Income : int ; YearsInJob : int
UsesCreditCard : bool; CriminalRecord : bool }
为了创建此记录的生成器,我尝试了以下代码:
let chooseFromList xs =
gen {
let! idx = Gen.choose(0, List.length xs - 1)
return (List.nth xs idx)
}
let generateName = (Gen.oneof [ gen {return "A"};gen {return "B"};gen {return "C"}])
let generateIncome= (chooseFromList [0..5000])
let generate_YearsInJob= chooseFromList [0..45]
let generate_UsesCreditCard = (Gen.oneof [ gen { return true }; gen { return false } ])
let generate_UsesCriminalRecord= (Gen.oneof [ gen { return true }; gen { return false } ])
let genertate_Client =
{
Name= generateName;
Income=generateIncome;
YearsInJob=generate_YearsInJob
UsesCreditCard=generate_UsesCreditCard
CriminalRecord=generate_UsesCriminalRecord
}
问题在第Name=generateName行代码>我遇到了erorr:
This expression was expected to have type string,but here has type Gen<string>
此表达式应具有string类型,但此处具有Gen类型
最后一行存在此错误。当您想要为客户端生成生成器时,必须首先为字段数据生成示例-因此gen{…}
语法非常方便:
let genertate_Client =
gen {
let! name = generateName
let! income = generateIncome
let! yearsInJob = generate_YearsInJob
let! creditCard = genergenerate_UsesCreditCard
let! criminal = generate_UsesCriminalRecord
return {
Name = name;
Income = income;
YearsInJob = yearsInJob;
UsesCreditCard = creditCard;
CriminalRecord = criminal }
}
当然,这里您可以使用let代码>计算中的语法修复名称的类型?或者问它第一个值?还是把所有的东西都串起来?或者可能是其中一个字符串(而不是生成器,每个生成器生成一个不同的字符串)?错误是它所说的:生成器不是字符串。