I’m trying to write a Java routine to evaluate math expressions from String
values like:
"5+3"
"1040"
"(1+10)*3"
I want to avoid a lot of ifthenelse statements.
How can I do this?
7
With JDK1.6, you can use the builtin Javascript engine.
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));
}
}
20
 60
It seems there’s a major problem there; It executes a script, not evaluates an expression. To be clear, engine.eval(“8;40+2”), outputs 42 ! If you want an expression parser that also check the syntax, I’ve just finished one (because I found nothing that suits my needs) : Javaluator.
Aug 29, 2012 at 12:33
 4
As a side note, if you need to use the result of this expression elsewhere in your code, you can typecast the result to a Double like so:
return (Double) engine.eval(foo);
Apr 2, 2014 at 22:44
 54
Security note: You should never use this in a server context with user input. The executed JavaScript can access all Java classes and thus hijack your application without limit.
– BoannSep 21, 2015 at 11:08
 5
@Boann, I request you to give me a reference about what you said.(to be sure 100%)
– parthoFeb 27, 2016 at 13:03
 28
@partho
new javax.script.ScriptEngineManager().getEngineByName("JavaScript") .eval("var f = new java.io.FileWriter('hello.txt'); f.write('UNLIMITED POWER!'); f.close();");
— will write a file via JavaScript into (by default) the program’s current directory– BoannFeb 27, 2016 at 13:37
I’ve written this eval
method for arithmetic expressions to answer this question. It does addition, subtraction, multiplication, division, exponentiation (using the ^
symbol), and a few basic functions like sqrt
. It supports grouping using (
…)
, and it gets the operator precedence and associativity rules correct.
public static double eval(final String str) {
return new Object() {
int pos = 1, ch;
void nextChar() {
ch = (++pos < str.length()) ? str.charAt(pos) : 1;
}
boolean eat(int charToEat) {
while (ch == ' ') nextChar();
if (ch == charToEat) {
nextChar();
return true;
}
return false;
}
double parse() {
nextChar();
double x = parseExpression();
if (pos < str.length()) throw new RuntimeException("Unexpected: " + (char)ch);
return x;
}
// Grammar:
// expression = term  expression `+` term  expression `` term
// term = factor  term `*` factor  term `/` factor
// factor = `+` factor  `` factor  `(` expression `)`  number
//  functionName `(` expression `)`  functionName factor
//  factor `^` factor
double parseExpression() {
double x = parseTerm();
for (;;) {
if (eat('+')) x += parseTerm(); // addition
else if (eat('')) x = parseTerm(); // subtraction
else return x;
}
}
double parseTerm() {
double x = parseFactor();
for (;;) {
if (eat('*')) x *= parseFactor(); // multiplication
else if (eat("https://stackoverflow.com/")) x /= parseFactor(); // division
else return x;
}
}
double parseFactor() {
if (eat('+')) return +parseFactor(); // unary plus
if (eat('')) return parseFactor(); // unary minus
double x;
int startPos = this.pos;
if (eat('(')) { // parentheses
x = parseExpression();
if (!eat(')')) throw new RuntimeException("Missing ')'");
} else if ((ch >= '0' && ch <= '9')  ch == '.') { // numbers
while ((ch >= '0' && ch <= '9')  ch == '.') nextChar();
x = Double.parseDouble(str.substring(startPos, this.pos));
} else if (ch >= 'a' && ch <= 'z') { // functions
while (ch >= 'a' && ch <= 'z') nextChar();
String func = str.substring(startPos, this.pos);
if (eat('(')) {
x = parseExpression();
if (!eat(')')) throw new RuntimeException("Missing ')' after argument to " + func);
} else {
x = parseFactor();
}
if (func.equals("sqrt")) x = Math.sqrt(x);
else if (func.equals("sin")) x = Math.sin(Math.toRadians(x));
else if (func.equals("cos")) x = Math.cos(Math.toRadians(x));
else if (func.equals("tan")) x = Math.tan(Math.toRadians(x));
else throw new RuntimeException("Unknown function: " + func);
} else {
throw new RuntimeException("Unexpected: " + (char)ch);
}
if (eat('^')) x = Math.pow(x, parseFactor()); // exponentiation
return x;
}
}.parse();
}
Example:
System.out.println(eval("((4  2^3 + 1) * sqrt(3*3+4*4)) / 2"));
Output: 7.5 (which is correct)
The parser is a recursive descent parser, so internally uses separate parse methods for each level of operator precedence in its grammar. I deliberately kept it short, but here are some ideas you might want to expand it with:
Variables:
The bit of the parser that reads the names for functions can easily be changed to handle custom variables too, by looking up names in a variable table passed to the
eval
method, such as aMap<String,Double> variables
.Separate compilation and evaluation:
What if, having added support for variables, you wanted to evaluate the same expression millions of times with changed variables, without parsing it every time? It’s possible. First define an interface to use to evaluate the precompiled expression:
@FunctionalInterface interface Expression { double eval(); }
Now to rework the original “eval” function into a “parse” function, change all the methods that return
double
s, so instead they return an instance of that interface. Java 8’s lambda syntax works well for this. Example of one of the changed methods:Expression parseExpression() { Expression x = parseTerm(); for (;;) { if (eat('+')) { // addition Expression a = x, b = parseTerm(); x = (() > a.eval() + b.eval()); } else if (eat('')) { // subtraction Expression a = x, b = parseTerm(); x = (() > a.eval()  b.eval()); } else { return x; } } }
That builds a recursive tree of
Expression
objects representing the compiled expression (an abstract syntax tree). Then you can compile it once and evaluate it repeatedly with different values:public static void main(String[] args) { Map<String,Double> variables = new HashMap<>(); Expression exp = parse("x^2  x + 2", variables); for (double x = 20; x <= +20; x++) { variables.put("x", x); System.out.println(x + " => " + exp.eval()); } }
Different datatypes:
Instead of
double
, you could change the evaluator to use something more powerful likeBigDecimal
, or a class that implements complex numbers, or rational numbers (fractions). You could even useObject
, allowing some mix of datatypes in expressions, just like a real programming language. 🙂
_{All code in this answer released to the public domain. Have fun!}
18
 3
Nice algorithm, starting from it I managed to impliment and logical operators. We created separate classes for functions to evaluate a function, so like your idea of variables, I create a map with functions and looking after the function name. Every function implements an interface with a method eval (T rightOperator , T leftOperator), so anytime we can add features without changing the algorithm code. And it is a good idea to make it work with generic types. Thanks you!
Jul 23, 2016 at 16:17
 1
 1
I try to give a description of what I understand from the code written by Boann, and examples described wiki.The logic of this algoritm starting from rules of operation orders. 1. operator sign  variable evaluation  function call  parenthesis (subexpressions); 2. exponentiation; 3. multiplication, division; 4. addition, subtraction;
Jul 24, 2016 at 15:05
 1
Algorithm methods are divided for each level of operations order as follows: parseFactor = 1. operator sign  variable evaluation  function call  parenthesis (subexpressions); 2. exponentiation; parseTerms = 3. multiplication, division; parseExpression = 4. addition, subtraction. The algorithm, call methods in reverse order (parseExpression > parseTerms > parseFactor > parseExpression (for subexpressions)), but every method to the first line call the method to the next level, so the entire execution order methods will be actually normal order of operations.
Jul 24, 2016 at 15:09
 1
For example the parseExpression method the
double x = parseTerm();
evaluate the left operator, after thisfor (;;) {...}
evaluate succesive operations of actual order level (addition, subtraction). The same logic are and in parseTerm method. The parseFactor does not have next level, so there are only evaluations of methods/ variables or in case of paranthesis – evaluate subexpression. Theboolean eat(int charToEat)
method check equality of the current cursor character with the charToEat character, if equal return true and move cursor to next character, I use name ‘accept’ for it.Jul 24, 2016 at 15:12
For my university project, I was looking for a parser / evaluator supporting both basic formulas and more complicated equations (especially iterated operators). I found very nice open source library for JAVA and .NET called mXparser. I will give a few examples to make some feeling on the syntax, for further instructions please visit project website (especially tutorial section).
https://mathparser.org/mxparsertutorial/
And few examples
1 – Simple furmula
Expression e = new Expression("( 2 + 3/4 + sin(pi) )/2");
double v = e.calculate()
2 – User defined arguments and constants
Argument x = new Argument("x = 10");
Constant a = new Constant("a = pi^2");
Expression e = new Expression("cos(a*x)", x, a);
double v = e.calculate()
3 – User defined functions
Function f = new Function("f(x, y, z) = sin(x) + cos(y*z)");
Expression e = new Expression("f(3,2,5)", f);
double v = e.calculate()
4 – Iteration
Expression e = new Expression("sum( i, 1, 100, sin(i) )");
double v = e.calculate()
Found recently – in case you would like to try the syntax (and see the advanced use case) you can download the Scalar Calculator app that is powered by mXparser.
4
 1
So far this is the best math library out there; simple to kickstart, easy to use and extendable. Definitely should be top answer.
Feb 6, 2019 at 7:00
 2
 1
I found mXparser can’t identify illegal formula, for example, ‘0/0’ will get a result as ‘0’. How can I solve this problem？
– lulijunMar 14, 2019 at 6:41
 2
I recently wrote a math expression parser called exp4j that was released under the apache license you can check it out here: objecthunter.net/exp4j
Dec 29, 2010 at 12:21
What kinds of expressions do you permit? Only single operator expressions? Are parentheses permitted?
Dec 10, 2014 at 13:20
Also take a look at Dijkstra’s twostack algorithm
Jan 21, 2016 at 15:33
Possible duplicate of Is there an eval() function in Java?
Sep 20, 2016 at 23:24
How can this possible be considered too broad? Dijkstra’s evaluation is the obvious solution here en.wikipedia.org/wiki/Shuntingyard_algorithm
Oct 13, 2018 at 9:43

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