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
你可以看看Symja框架:
ExprEvaluator util = new ExprEvaluator();
IExpr result = util.evaluate("10-40");
System.out.println(result.toString()); // -> "-30"
请注意,可以计算更复杂的表达式:
// D(...) gives the derivative of the function Sin(x)*Cos(x)
IAST function = D(Times(Sin(x), Cos(x)), x);
IExpr result = util.evaluate(function);
// print: Cos(x)^2-Sin(x)^2
其他回答
import javax.script.ScriptEngine;
import javax.script.ScriptEngineManager;
import javax.script.ScriptException;
public class test2 {
public static void main(String[] args) throws ScriptException {
String s = "10+2";
ScriptEngineManager mn = new ScriptEngineManager();
ScriptEngine en = mn.getEngineByName("js");
Object result = en.eval(s);
System.out.println(result);
}
}
还有一个选择:https://github.com/stefanhaustein/expressionparser
我已经实现了一个简单而灵活的选项,以允许两者:
即时处理(Calculator.java, SetDemo.java) 构建和处理解析树(TreeBuilder.java)
上面链接的TreeBuilder是进行符号推导的CAS演示包的一部分。还有一个BASIC解释器的例子,我已经开始使用它来构建一个TypeScript解释器。
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.
对于JDK1.6,您可以使用内置的Javascript引擎。
import javax.script.ScriptEngineManager;
import javax.script.ScriptEngine;
import javax.script.ScriptException;
public class Test {
public static void main(String[] args) throws ScriptException {
ScriptEngineManager mgr = new ScriptEngineManager();
ScriptEngine engine = mgr.getEngineByName("JavaScript");
String foo = "40+2";
System.out.println(engine.eval(foo));
}
}
使用带有代码注入处理的JDK1.6 Javascript引擎尝试下面的示例代码。
import javax.script.ScriptEngine;
import javax.script.ScriptEngineManager;
public class EvalUtil {
private static ScriptEngine engine = new ScriptEngineManager().getEngineByName("JavaScript");
public static void main(String[] args) {
try {
System.out.println((new EvalUtil()).eval("(((5+5)/2) > 5) || 5 >3 "));
System.out.println((new EvalUtil()).eval("(((5+5)/2) > 5) || true"));
} catch (Exception e) {
e.printStackTrace();
}
}
public Object eval(String input) throws Exception{
try {
if(input.matches(".*[a-zA-Z;~`#$_{}\\[\\]:\\\\;\"',\\.\\?]+.*")) {
throw new Exception("Invalid expression : " + input );
}
return engine.eval(input);
} catch (Exception e) {
e.printStackTrace();
throw e;
}
}
}