我从一段代码中得到一个错误。我只会显示一行代码，至少我认为是错误报告中导致它的那一行。这是：
b = temp(temp.length-1).toInt; //temp is an ArrayBuffer[String]
错误是：
For input string: "z"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
at java.lang.Integer.parseInt(Integer.java:449)
at java.lang.Integer.parseInt(Integer.java:499)
at scala.collection.immutable.StringLike$class.toInt(StringLike.scala:231)
at scala.collection.immutable.StringOps.toInt(StringOps.scala:31)
at Driver$.stringParse$1(Driver.scala:59)
at Driver$.main(Driver.scala:86)
at Driver.main(Driver.scala)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25)
at java.lang.reflect.Method.invoke(Method.java:597)
at scala.tools.nsc.util.ScalaClassLoader$$anonfun$run$1.apply(ScalaClassLoader.scala:78)
at scala.tools.nsc.util.ScalaClassLoader$class.asContext(ScalaClassLoader.scala:24)
at scala.tools.nsc.util.ScalaClassLoader$URLClassLoader.asContext(ScalaClassLoader.scala:88)
at scala.tools.nsc.util.ScalaClassLoader$class.run(ScalaClassLoader.scala:78)
at scala.tools.nsc.util.ScalaClassLoader$URLClassLoader.run(ScalaClassLoader.scala:101)
at scala.tools.nsc.ObjectRunner$.run(ObjectRunner.scala:33)
at scala.tools.nsc.ObjectRunner$.runAndCatch(ObjectRunner.scala:40)
at scala.tools.nsc.MainGenericRunner.runTarget$1(MainGenericRunner.scala:56)
at scala.tools.nsc.MainGenericRunner.process(MainGenericRunner.scala:80)
at scala.tools.nsc.MainGenericRunner$.main(MainGenericRunner.scala:89)
at scala.tools.nsc.MainGenericRunner.main(MainGenericRunner.scala)
据我所知，这导致了这个问题。由于它是不可变的，我知道它无法更改。但我不确定。我以此为基础
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
一旦我做了类似上面唯一的代码的事情，它会改变整个对象吗？Temp 是一个 ArrayBuffer[String]。所以我试图访问一个数字的字符串表示形式，并将其转换。但是这样做是否会改变现状并阻止我做任何事情？
如果您认为放置我所有的代码会有所帮助，请告诉我编辑它，但它很多，我不想惹恼任何人。我感谢任何可以帮助我理解这一点的人！
*编辑：我的代码（仅在这里帮助我找出我的错误，但没有必要查看。我只是看不到它在哪里给了我这个错误）。
我的代码的重点是解析顶部的任何一个字符串。它将组合成一个字符串，然后读取其他两个符号以配合它。它可以很好地解析 str，但是当它在 str2 中读取“z”和在 str3 中读取“y”时发现了问题。可以看出，问题在于递归之后的第二个字符串。同样重要的是要注意字符串必须采用这种形式。所以它只能像“（和x（和yz））”那样解析，但不能以任何其他更方便的方式解析。
val str = "(and x y)";
val str2 = "(and x (and y z))"; //case with expression on th right side
val str3 = "(and (and x y) z)"; //case with expression ont he left side
var i = 0; //just counter used to loop through the finished parsed array to make a list
//var position = 0; //this is used for when passing it in the parser to start off at zero
var hold = new ArrayBuffer[String]();//finished array should be here
def stringParse ( exp: String, expreshHolder: ArrayBuffer[String] ): ArrayBuffer[String] = { //takes two arguments, string, arraybuffer
var b = 0; //position of where in the expression String I am currently in
var temp = expreshHolder; //holder of expressions without parens
var arrayCounter = 0;
if(temp.length == 0)
b = 0;
else {
b = temp(temp.length-1).toInt;
temp.remove(temp.length-1);
arrayCounter = temp.length;
} //this sets the position of wherever the string was read last plus removes that check from the end of the ArrayBuffer
//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') {
//first create the 'and', 'or', 'not' expression types to figure out
temp += exp(b).toString;
b = b+1;
temp(arrayCounter) = temp(arrayCounter) + exp(b).toString; //concatenates the second letter
b = b+1;
temp(arrayCounter) = temp(arrayCounter) + exp(b).toString; //concatenates the last letter for the expression type
//arrayCounter+=1;
//this part now takes the symbols and puts them in an array
b+=1;
while(exp(b) == ' ') {b+=1;} //just skips any spaces until it reaches the FIRST symbol
if(exp(b) == '(') {
temp += b.toString;
temp = stringParse(exp, temp);
b = temp(temp.length-1).toInt;
temp.remove(temp.length-1);
arrayCounter = temp.length-1
} else {
temp += exp(b).toString;
arrayCounter+=1; b+=1; }
while(exp(b) == ' ') {b+=1;} //just skips any spaces until it reaches the SECOND symbol
if(exp(b) == '(') {
temp += b.toString;
temp = stringParse(exp, temp);
b = temp(temp.length-1).toInt;
temp.remove(temp.length-1);
arrayCounter = temp.length-1
} else {
temp += exp(b).toString;
arrayCounter+=1;
b+=1;
}
temp;
} else { var fail = new ArrayBuffer[String]; fail +="failed"; fail;}
}
hold = stringParse(str2, ho );
for(test <- hold) println(test);