我如何调整我的 Swift 代码来给出二次方程的假想解？

Question

我在 Swift 中创建了一个函数来求解并给出二次函数的解。我不知道如何调整我的功能，以便它给出假想的解决方案而不是打印，"There are no real solutions."

我对编程还比较陌生，需要一些帮助。这是我的代码：

func quadraticFormulaSolver(variableA a: Double, variableB b: Double, variableC c: Double) -> (Double, Double) {
let firstSolution: Double = (-b + sqrt((b * b) + (-4.0 * a * c))) / 2.0
let secondSolution: Double = (-b - sqrt((b * b) + (-4.0 * a * c))) / 2.0
let checkSolution: Double = sqrt((b * b) + (-4.0 * a * c))

if checkSolution > 0 {
    print("There are two real solutions and they are \(firstSolution) and \(secondSolution)")
    return(firstSolution, secondSolution) }

guard firstSolution != 0.0 else {
    print("There is one real solution and it is \(firstSolution)")
   return(firstSolution, secondSolution) }


guard checkSolution < 0 else {
print("There are no real solutions")
    return(firstSolution, secondSolution) }

    return(firstSolution, secondSolution)
}

Answer 1

由于您的函数可以 return 几个不同的选择，让我们做一个 Enum 来表示选项：

enum QuadraticSolution {
    case TwoReal(firstSolution: Double, secondSolution: Double)
    case OneReal(solution: Double)
    case TwoNonReal
}

我们稍后会回到 TwoNonReal。

您的函数现在可以 return 这个枚举的一个实例：

func quadraticFormulaSolver(variableA a: Double, variableB b: Double, variableC c: Double) -> QuadraticSolution {

为了使代码更具可读性，让我们过滤掉判别式：

    let discriminant = (b * b) - (4.0 * a * c)

然后我们可以在上面使用switch语句。如果它是积极的，你有两个真正的根源。如果它为零，则您有一个真正的（重复的）根。如果它是负数，你有两个非实根：

    switch discriminant {
    case _ where discriminant > 0:
        let firstSolution = (-b + sqrt(discriminant)) / (2.0 * a)
        let secondSolution = (-b - sqrt(discriminant)) / (2.0 * a)
        return .TwoReal(firstSolution: firstSolution, secondSolution: secondSolution)
    case _ where discriminant == 0:
        let solution = (-b) / (2.0 * a)
        return .OneReal(solution: solution)
    default: // discriminant is negative
        return .TwoNonReal
    }
}

Swift 没有非实数的内置类型。我建议您将 swift-pons 嵌入到您的应用程序中，而不是重新发明轮子。

完成后，您可以将 TwoNonReal 枚举更改为 return 两个 Complex 数字：

    case TwoNonReal(firstSolution: Complex, secondSolution: Complex)

然后你可以这样计算它们：

    default: // discriminant is negative
        let base = (-b) / (2.0 * a)
        let firstSolution = base + (Complex.sqrt(-1.0 * discriminant)) / (2.0 * a)
        let secondSolution = base - (Complex.sqrt(-1.0 * discriminant)) / (2.0 * a)
        return .TwoNonReal(firstSolution: firstSolution, secondSolution: secondSolution)
    }

我如何调整我的 Swift 代码来给出二次方程的假想解？

How can I adapt my Swift code to give imaginary solutions to a quadratic?

math

quadratic

ios

swift