function - 为什么我的Scala函数返回类型Unit而不是最后一行？

Question

我试图找出问题所在，并尝试了我在Scala上阅读过的各种样式，但是它们都不起作用。我的代码是:

....

val str = "(and x y)";

def stringParse ( exp: String, pos: Int, expreshHolder: ArrayBuffer[String], follow: Int )  

    var b = pos; //position of where in the expression String I am currently in
    val temp = expreshHolder; //holder of expressions without parens
    var arrayCounter = follow; //just counts to make sure an empty spot in the array is there to put in the strings

    if(exp(b) == '(') {
        b = b + 1;

        while(exp(b) == ' '){b = b + 1} //point of this is to just skip any spaces between paren and start of expression type

        if(exp(b) == 'a') {
               temp(arrayCounter) = exp(b).toString; 
               b = b+1; 
               temp(arrayCounter)+exp(b).toString; b = b+1; 
               temp(arrayCounter) + exp(b).toString; arrayCounter+=1}
               temp;

         }

}

val hold: ArrayBuffer[String] = stringParse(str, 0, new ArrayBuffer[String], 0);
for(test <- hold) println(test);

我的错误是:

Driver.scala:35: error: type mismatch;
found   : Unit
 required: scala.collection.mutable.ArrayBuffer[String]
ho = stringParse(str, 0, ho, 0);
                ^one error found

当我在方法声明中的参数之后添加等号时，如下所示:

def stringParse ( exp: String, pos: Int, expreshHolder: ArrayBuffer[String], follow: Int )  ={....}

它将其更改为“任何”。我对这是如何工作感到困惑。有任何想法吗？非常感激。

Answer 1

如果要返回值，则必须添加等号。现在，函数返回值为Any的原因是，您有2条控制路径，每条返回的路径的类型不同-1是满足if条件的条件(返回值为temp)，而另一条是if的条件不是(返回值将为b = b + 1或递增后的b)。