具有类型约束的 Go 泛型：不能在赋值中使用 false（无类型布尔常量）作为 T 值

Question

我有以下代码：

func createGrid[T int | bool](size int, forFlooded bool) [][]T {

    var tempGrid [][]T

    for i := 0; i <= size+1; i++ {

        for j := 0; j <= size+1; j++ {

            if forFlooded {

                tempGrid[i][j] = false

            } else {

                tempGrid[i][j] = -1

            }

        }

    }

    return tempGrid

}

我在分配的行中收到以下错误tempGrid[i][j]：

cannot use false (untyped bool constant) as T value in assignment

cannot use -1 (untyped int constant) as T value in assignment

这就是我使用它的方式：

var grid [][]int;

grid = createGrid(n, false);

var flooded [][]bool;

flooded = createGrid(n, true);

在这里，我得到CannotInferTypeArgs( cannot infer T) 错误。

我也试过将该类型约束提取到接口中。我究竟做错了什么？

Answer 1

泛型是编译器为不同类型创建同一函数的多个实例的快捷方式。让我们尝试手动完成编译器的工作。您将获得这两个函数 - 一个用于 int：func createGridInt(size int, forFlooded bool) [][]int {&nbsp; &nbsp; var tempGrid [][]int&nbsp; &nbsp; for i := 0; i <= size+1; i++ {&nbsp; &nbsp; &nbsp; &nbsp; for j := 0; j <= size+1; j++ {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if forFlooded {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; tempGrid[i][j] = false&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; } else {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; tempGrid[i][j] = -1&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }&nbsp; &nbsp; return tempGrid}一个用于布尔：func createGridBool(size int, forFlooded bool) [][]bool {&nbsp; &nbsp; var tempGrid [][]bool&nbsp; &nbsp; for i := 0; i <= size+1; i++ {&nbsp; &nbsp; &nbsp; &nbsp; for j := 0; j <= size+1; j++ {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if forFlooded {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; tempGrid[i][j] = false&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; } else {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; tempGrid[i][j] = -1&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }&nbsp; &nbsp; return tempGrid}突然间，这不再有意义了，因为您不能在第一个函数中分配false给intas，而在第二个函数中则相反。要解决这个问题，您需要使用Tas 您正在处理的类型。这也解决了将布尔标志传递给函数的代码味道：func createGrid[T int | bool](size int, fillWith T) [][]T {&nbsp; &nbsp; var tempGrid [][]T&nbsp; &nbsp; for i := 0; i <= size+1; i++ {&nbsp; &nbsp; &nbsp; &nbsp; for j := 0; j <= size+1; j++ {&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; tempGrid[i][j] = fillWith&nbsp; &nbsp; &nbsp; &nbsp; }&nbsp; &nbsp; }&nbsp; &nbsp; return tempGrid}你可以这样称呼它：createGrid(n, false)createGrid(n, -1)