'try' 表达式中没有调用抛出函数答案

【问题标题】：No calls to throwing functions occur within 'try' expression'try' 表达式中没有调用抛出函数
【发布时间】：2022-01-05 07:06:24
【问题描述】：

`枚举outofBound：错误 { 案例负数案例超出限制 }

func sqareRoot(Number: Int) throws -> String
{
    if Number < 10
    {
        throw outofBound.NegativeNumber
    }
     if  Number > 100
    {
        throw outofBound.OutOfLimit
    }
    else
    {
        let number1 = Double (Number / 4)
        print(number1)
        let number2 = Double (number1 / 2)
        print(number2)
        return "number2"
    }
    
}
do
{
 let Result = try sqareRoot(Number: 3600)
}
catch outofBound.NegativeNumber
{
    print("Please provide positive number")
}
catch outofBound.OutOfLimit
{
    print("please provide less than 1000 number")
}
catch
{
    print("unkonw errors occured")
}
print(squareRoot(Number: 2500))
print(squareRoot(Number: 3600))
print(squareRoot(Number: 4))

它给出了这个建议，我没有得到它，请帮助我。

{ 'try' 表达式中没有调用抛出函数 'catch' 块无法访问，因为在 'do' 块中没有抛出错误 }

【问题讨论】：

标签： error-handling swiftui try-catch

【解决方案1】：

试试这样的：

 func test() {
    do {
        let result = try sqareRoot(Number: 3600)
         print("result: \(result)")
    }
    catch (let error) {
        switch error {
        case outofBound.NegativeNumber:
            print("Please provide >= 10 number")
        case outofBound.OutOfLimit:
            print("please provide <= 100 number")
        default:
            print("error: \(error)")
        }
    }
     // test print
     print(try? sqareRoot(Number: 25))
  }
     
    func sqareRoot(Number: Int) throws -> String {
        if Number < 10 {
            throw outofBound.NegativeNumber
        }
        if Number > 100 {
            throw outofBound.OutOfLimit
        }
        else {
            let number1 = Double (Number / 4)
            print("number1 \(number1)")
            let number2 = Double (number1 / 2)
            print("number2 \(number2)")
            return String(number2)
        }
    }

【讨论】：