swift - 二元运算符 "=="不能应用于 (Int, Int, Int, Int) -> Int 类型的操作数

Question

var a11: Int = 0
var a12: Int = 0
var a21: Int = 0
var a22: Int = 0

var valueDeterminant = a11 * a12 * a21 + a22

func calculateDeterminant(a11: Int, a12: Int, a21: Int, a22: Int) -> Int {
    return valueDeterminant
}


if calculateDeterminant == 0 {
    print("The trasformation is irreversable, due to the fact that the determinant is \(calculateDeterminant)")
 } else if calculateDeterminant == 1 {
    print("The determinant is\(calculateDeterminant) \nThe trasformation is reverable, and it keeps the area")
 } else {
    print("The determinant is \(calculateDeterminant) \nThe transformation is reversable")
}

该错误表示二元运算符 == 不能用于 (Int, Int, Int, Int) 类型。为什么？我该如何解决这个问题？另外，如果我将返回值更改为 Bool，则会出现另一个错误，指出我无法输入返回 Bool 值的 Int 值。

Answer 1

函数是一流的值。您正在将函数 calculateDeterminant 与 Int 进行比较。不支持。调用该函数，然后比较整数值。

var a11: Int = 0
var a12: Int = 0
var a21: Int = 0
var a22: Int = 0

var valueDeterminant = a11 * a12 * a21 + a22

func calculateDeterminant() -> Int {
  return a11 * a12 * a21 * a22
}


if calculateDeterminant() == 0 {
  print("The trasformation is irreversable, due to the fact that the determinant is \(calculateDeterminant())")
} else if calculateDeterminant() == 1 {
  print("The determinant is\(calculateDeterminant()) \nThe trasformation is reverable, and it keeps the area")
} else {
  print("The determinant is \(calculateDeterminant()) \nThe transformation is reversable")
}

或者像这样定义它......

func calculateDeterminant(a11: Int, a12: Int, a21: Int, a22: Int) -> Int {
    return a11 * a12 * a21 * a22
}

...并这样调用它。

if (calculateDeterminant(a11: a11, a12: a12, a21: a21, a22: a22) == 0 {...}