如何在Nim中声明可为空的字段？

问题描述

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: "")