这是另一个有趣的选择 https://github.com/Shy-Ta/expression-evaluator-demo

它的用法非常简单，可以完成工作，例如:

  ExpressionsEvaluator evalExpr = ExpressionsFactory.create("2+3*4-6/2");  
  assertEquals(BigDecimal.valueOf(11), evalExpr.eval()); 

这样怎么样:

String st = "10+3";
int result;
for(int i=0;i<st.length();i++)
{
  if(st.charAt(i)=='+')
  {
    result=Integer.parseInt(st.substring(0, i))+Integer.parseInt(st.substring(i+1, st.length()));
    System.out.print(result);
  }         
}

并相应地对其他数学运算符做类似的事情。

下面是GitHub上另一个名为EvalEx的开源库。

与JavaScript引擎不同，这个库只专注于计算数学表达式。此外，该库是可扩展的，支持使用布尔运算符和圆括号。

这是另一个有趣的选择 https://github.com/Shy-Ta/expression-evaluator-demo

它的用法非常简单，可以完成工作，例如:

  ExpressionsEvaluator evalExpr = ExpressionsFactory.create("2+3*4-6/2");  
  assertEquals(BigDecimal.valueOf(11), evalExpr.eval()); 

import java.util.*;

public class check { 
   int ans;
   String str="7 + 5";
   StringTokenizer st=new StringTokenizer(str);

   int v1=Integer.parseInt(st.nextToken());
   String op=st.nextToken();
   int v2=Integer.parseInt(st.nextToken());

   if(op.equals("+")) { ans= v1 + v2; }
   if(op.equals("-")) { ans= v1 - v2; }
   //.........
}

如果我们要实现它，那么我们可以使用下面的算法

While there are still tokens to be read in, 1.1 Get the next token. 1.2 If the token is: 1.2.1 A number: push it onto the value stack. 1.2.2 A variable: get its value, and push onto the value stack. 1.2.3 A left parenthesis: push it onto the operator stack. 1.2.4 A right parenthesis: 1 While the thing on top of the operator stack is not a left parenthesis, 1 Pop the operator from the operator stack. 2 Pop the value stack twice, getting two operands. 3 Apply the operator to the operands, in the correct order. 4 Push the result onto the value stack. 2 Pop the left parenthesis from the operator stack, and discard it. 1.2.5 An operator (call it thisOp): 1 While the operator stack is not empty, and the top thing on the operator stack has the same or greater precedence as thisOp, 1 Pop the operator from the operator stack. 2 Pop the value stack twice, getting two operands. 3 Apply the operator to the operands, in the correct order. 4 Push the result onto the value stack. 2 Push thisOp onto the operator stack. While the operator stack is not empty, 1 Pop the operator from the operator stack. 2 Pop the value stack twice, getting two operands. 3 Apply the operator to the operands, in the correct order. 4 Push the result onto the value stack. At this point the operator stack should be empty, and the value stack should have only one value in it, which is the final result.