Question

我在Swift中创建了一个函数来解决并给出二次函数的解。我不知道如何调整我的功能，以便它提供想象的解决方案而不是打印，＆＃34;没有真正的解决方案。＆＃34;

我对编程比较陌生，可以使用一些帮助。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

由于您的功能可以返回一些不同的选项，因此我们要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。

您的函数现在可以返回此枚举的实例：

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)

然后我们可以在其上使用switch语句。如果它是积极的，你有两个真正的根源。如果它为零，则您有一个真实（重复）根。如果它是否定的，那么你有两个非真实的根源：

    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。我建议您在应用中嵌入swift-pons，而不是重新发明轮子。

完成后，您可以更改TwoNonReal枚举，以返回两个Complex个数字：

    case TwoNonReal(firstSolution: Complex, secondSolution: Complex)

然后你可以像这样计算它们：

    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)
    }

如何调整我的Swift代码以给出二次方的假想解？

1 个答案: