如何解决如何在Nim中声明可为空的字段?
stock_symbol
是可选的,它仅对某些公司存在,在Nim中进行声明的正确方法是什么?
我必须使用ref
还是其他方法?
type
Company = object
name: string
stock_symbol: string
echo Company(name: "Microsoft",stock_symbol: "MSFT")
echo Company(name: "Kinetic",stock_symbol: nil)
类似的复合类型问题,通常您需要同时了解证券交易所和交易代码
type
SymbolWithExchange = object
exchange: string
symbol: string
Company2 = object
name: string
stock_symbol: SymbolWithExchange
echo Company2(
name: "Microsoft",stock_symbol: SymbolWithExchange("NYSE","MSFT")
)
echo Company2(name: "Kinetic",stock_symbol: nil)
解决方法
应该使用类似Option[T]
的
import options
type
Company = object
name: string
stock_symbol: Option[string]
echo Company(name: "Microsoft",stock_symbol: some("MSFT"))
echo Company(name: "Kinetic",stock_symbol: none(string))
和
import options
type
SymbolWithExchange = object
exchange: string
symbol: string
Company2 = object
name: string
stock_symbol: Option[SymbolWithExchange]
echo Company2(
name: "Microsoft",stock_symbol: some(SymbolWithExchange(exchange: "NYSE",symbol: "MSFT"))
)
echo Company2(name: "Kinetic",stock_symbol: none(SymbolWithExchange))
,
可以使用'options'模块,但是我发现它引起的问题比它要解决的更多,因为如果您正在构建库,或者您打算将它以some()开头,由其他人使用,该要求可能会很大。
在您的情况下,建议使用构造函数初始化值。
type
Company = object
name: string
stock_symbol: string
proc company_constructor(name: string,stock_symbol: string = ""): Company =
return Company(name:name,stock_symbol:stock_symbol)
let
company_a = company_constructor("Microsoft","MSFT")
company_b = company_constructor("Kinetic")
echo company_a
echo company_b
输出:
(name: "Microsoft",stock_symbol: "MSFT")
(name: "Kinetic",stock_symbol: "")
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