java - 计算器无法正常计算三位数

Question

所以我的计算器可以完美地处理任何涉及两位数字(包括 C 按钮)的计算，但是如果我尝试任何三位数计算，这实际上不起作用，例如 1 + 2 + 3 = 5 有人可以告诉我为什么会这样，无论如何我可以修复它吗？

public class Calculator {  

  private long currentInput;          //current input
  private long previousInput;         // previous input
  private long result;            // result of calculation
  private String lastOperator = "";  // keeps track of the last operator entered


  /* New digit entered as integer value i - moves currentInput 1 decimal place to the left and adds i in "one's column" */
  public void inDigit(long i) {
    currentInput = (currentInput * 10) + i;
  }


  /* Operator entered  + - or *   */
  public void inOperator(String op) {
    previousInput = currentInput;      // save the new input as previous to get ready for next input
    currentInput = 0;
    lastOperator = op;                 // remember which operator was entered
  } 


   /* Equals operation sets result to previousInput + - or * currentInput (depending on lastOperator) */
  public void inEquals() {
    if (lastOperator.equals("+")) {
      result = previousInput + currentInput;
    } else if (lastOperator.equals("-")) { 
      result = previousInput - currentInput;
    } else if (lastOperator.equals("*"))  {
      result = previousInput * currentInput;
    } 
    lastOperator = "";       // reset last operator to "nothing"
  }


  /* Clear operation */
  public void inClear() {
    currentInput = 0;
    previousInput = 0;
    result = 0;
    lastOperator = "";
  } 

  /* returns the current result */
  public String getResult() {  
    return Long.toString(result);  //converts int to String
  }

  /* returns the previous input value */
  public String getPreviousInput() {
    return Long.toString(previousInput);
  }
  /* returns the current input value */
  public String getCurrentInput() {
    return Long.toString(currentInput);
  }

Answer 1

好吧，您只存储最后两个操作数，因此对于 1 + 2 + 3 = 5 的示例，当您输入第二个 + 符号时，1 会丢失，而 2 + 3 = 5。