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How do Golang generics affect function type inference?

Apr 16, 2024 am 11:45 AM
golang Generics

Go泛型通过引入类型参数化改变了函数类型推断:参数类型:[]A,其中A是满足number接口的实际类型参数,在调用时具体化。返回值类型:与参数类型相同,推断为A。

How do Golang generics affect function type inference?

Go 泛型如何影响函数类型推断

Go 语言中的泛型引入了一种新的类型推断机制,它影响着函数类型推断的方式。了解这些变化对于正确使用 Go 泛型非常重要。

传统类型推断

在 Go 泛型之前,函数类型推断是基于上下文中的类型信息。例如,以下函数:

func SumNumbers(numbers []int) int {
  sum := 0
  for _, number := range numbers {
    sum += number
  }
  return sum
}
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会推断出 SumNumbers 的参数类型为 []int,返回值类型为 int。这是因为 range 语句中的 number 变量具有类型 int,而 sum 变量是类型的 int

泛型中的类型推断

Go 泛型引入了一层额外的类型参数化,这改变了函数类型推断的方式。考虑以下泛型函数:

func Sum[T number](numbers []T) T {
  sum := zero[T]()
  for _, number := range numbers {
    sum += number
  }
  return sum
}
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现在,Sum 函数具有一个类型参数 T,指示类型 T 应该是数字类型,也就是说,它必须实现 number 接口。在这种情况下,函数类型推断如下:

  • 函数参数类型:[]A,其中 A 是满足 number 接口的实际类型参数(在调用时具体化)。
  • 函数返回值类型:A,与函数参数类型相同。

实战案例

以下是使用 Sum 泛型函数的示例:

// 求和整型数组
sum := Sum([]int{1, 2, 3}) // 推断 T 为 int

// 求和浮点型数组
sumF := Sum([]float64{1.2, 2.3, 3.4}) // 推断 T 为 float64
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正如你所看到的,Sum 泛型函数可以根据调用时的实际类型参数进行类型推断。

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