我试图写一个Java例程来计算数学表达式从字符串值,如:
"5 + 3" "10-4 * 5" "(1 + 10) * 3"
我想避免很多如果-then-else语句。 我该怎么做呢?
我试图写一个Java例程来计算数学表达式从字符串值,如:
"5 + 3" "10-4 * 5" "(1 + 10) * 3"
我想避免很多如果-then-else语句。 我该怎么做呢?
当前回答
你也可以试试BeanShell解释器:
Interpreter interpreter = new Interpreter();
interpreter.eval("result = (7+21*6)/(32-27)");
System.out.println(interpreter.get("result"));
其他回答
我已经使用迭代解析和分流码算法,我真的很喜欢开发表达式求值器,你可以在这里找到所有的代码
https://github.com/nagaraj200788/JavaExpressionEvaluator
有73个测试用例,甚至工作于大整数,大小数
支持所有关系,算术表达式和两者的组合。 甚至支持三元运算符。
增加了增强,以支持有符号的数字,如-100+89,这是有趣的,详细信息请检查TokenReader.isUnaryOperator()方法,我已经更新了上面链接中的代码
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.
像RHINO或NASHORN这样的外部库可以用来运行javascript。javascript可以计算简单的公式,而不需要对字符串进行分割。如果代码写得好,也不会对性能造成影响。 下面是一个使用RHINO -的示例
public class RhinoApp {
private String simpleAdd = "(12+13+2-2)*2+(12+13+2-2)*2";
public void runJavaScript() {
Context jsCx = Context.enter();
Context.getCurrentContext().setOptimizationLevel(-1);
ScriptableObject scope = jsCx.initStandardObjects();
Object result = jsCx.evaluateString(scope, simpleAdd , "formula", 0, null);
Context.exit();
System.out.println(result);
}
另一种方法是使用Spring表达式语言或SpEL,它在计算数学表达式时做了更多的工作,因此可能会有点过度。您不必使用Spring框架来使用这个表达式库,因为它是独立的。从SpEL文档中复制示例:
ExpressionParser parser = new SpelExpressionParser();
int two = parser.parseExpression("1 + 1").getValue(Integer.class); // 2
double twentyFour = parser.parseExpression("2.0 * 3e0 * 4").getValue(Double.class); //24.0
解决这个问题的正确方法是使用词法分析器和解析器。您可以自己编写这些页面的简单版本,或者这些页面还包含指向Java词法分析器和解析器的链接。
创建递归下降解析器是非常好的学习练习。