I'm trying to write a Java routine to evaluate math expressions from String values like:
"5+3"
"10-4*5"
"(1+10)*3"
I want to avoid a lot of if-then-else statements.
How can I do this?
With JDK1.6, you can use the built-in 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));
}
}
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('/')) 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 a Map<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 doubles, 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 like BigDecimal, or a class that implements complex numbers, or rational numbers (fractions). You could even use Object, 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!
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/
https://mathparser.org/mxparser-tutorial/
https://mathparser.org/api/
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.
The correct way to solve this is with a lexer and a parser. You can write simple versions of these yourself, or those pages also have links to Java lexers and parsers.
Creating a recursive descent parser is a really good learning exercise.
HERE is another open source library on GitHub named EvalEx.
Unlike the JavaScript engine this library is focused in evaluating mathematical expressions only. Moreover, the library is extensible and supports use of boolean operators as well as parentheses.
You can evaluate expressions easily if your Java application already accesses a database, without using any other JARs.
Some databases require you to use a dummy table (eg, Oracle's "dual" table) and others will allow you to evaluate expressions without "selecting" from any table.
For example, in Sql Server or Sqlite
select (((12.10 +12.0))/ 233.0) amount
and in Oracle
select (((12.10 +12.0))/ 233.0) amount from dual;
The advantage of using a DB is that you can evaluate many expressions at the same time. Also most DB's will allow you to use highly complex expressions and will also have a number of extra functions that can be called as necessary.
However performance may suffer if many single expressions need to be evaluated individually, particularly when the DB is located on a network server.
The following addresses the performance problem to some extent, by using a Sqlite in-memory database.
Here's a full working example in Java
Class. forName("org.sqlite.JDBC");
Connection conn = DriverManager.getConnection("jdbc:sqlite::memory:");
Statement stat = conn.createStatement();
ResultSet rs = stat.executeQuery( "select (1+10)/20.0 amount");
rs.next();
System.out.println(rs.getBigDecimal(1));
stat.close();
conn.close();
Of course you could extend the above code to handle multiple calculations at the same time.
ResultSet rs = stat.executeQuery( "select (1+10)/20.0 amount, (1+100)/20.0 amount2");
You can also try the BeanShell interpreter:
Interpreter interpreter = new Interpreter();
interpreter.eval("result = (7+21*6)/(32-27)");
System.out.println(interpreter.get("result"));
Another way is to use the Spring Expression Language or SpEL which does a whole lot more along with evaluating mathematical expressions, therefore maybe slightly overkill. You do not have to be using Spring framework to use this expression library as it is stand-alone. Copying examples from SpEL's documentation:
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
This article discusses various approaches. Here are the 2 key approaches mentioned in the article:
JEXL from Apache
Allows for scripts that include references to java objects.
// Create or retrieve a JexlEngine
JexlEngine jexl = new JexlEngine();
// Create an expression object
String jexlExp = "foo.innerFoo.bar()";
Expression e = jexl.createExpression( jexlExp );
// Create a context and add data
JexlContext jctx = new MapContext();
jctx.set("foo", new Foo() );
// Now evaluate the expression, getting the result
Object o = e.evaluate(jctx);
Use the javascript engine embedded in the JDK:
private static void jsEvalWithVariable()
{
List<String> namesList = new ArrayList<String>();
namesList.add("Jill");
namesList.add("Bob");
namesList.add("Laureen");
namesList.add("Ed");
ScriptEngineManager mgr = new ScriptEngineManager();
ScriptEngine jsEngine = mgr.getEngineByName("JavaScript");
jsEngine.put("namesListKey", namesList);
System.out.println("Executing in script environment...");
try
{
jsEngine.eval("var x;" +
"var names = namesListKey.toArray();" +
"for(x in names) {" +
" println(names[x]);" +
"}" +
"namesListKey.add(\"Dana\");");
}
catch (ScriptException ex)
{
ex.printStackTrace();
}
}
if we are going to implement it then we can can use the below algorithm :--
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.
This is another interesting alternative
https://github.com/Shy-Ta/expression-evaluator-demo
The usage is very simple and gets the job done, for example:
ExpressionsEvaluator evalExpr = ExpressionsFactory.create("2+3*4-6/2");
assertEquals(BigDecimal.valueOf(11), evalExpr.eval());
It seems like JEP should do the job
It's too late to answer but I came across same situation to evaluate expression in java, it might help someone
MVEL does runtime evaluation of expressions, we can write a java code in String to get it evaluated in this.
String expressionStr = "x+y";
Map<String, Object> vars = new HashMap<String, Object>();
vars.put("x", 10);
vars.put("y", 20);
ExecutableStatement statement = (ExecutableStatement) MVEL.compileExpression(expressionStr);
Object result = MVEL.executeExpression(statement, vars);
Try the following sample code using JDK1.6's Javascript engine with code injection handling.
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;
}
}
}
This is actually complementing the answer given by #Boann. It has a slight bug which causes "-2 ^ 2" to give an erroneous result of -4.0. The problem for that is the point at which the exponentiation is evaluated in his. Just move the exponentiation to the block of parseTerm(), and you'll be all fine. Have a look at the below, which is #Boann's answer slightly modified. Modification is in the comments.
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 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('/')) x /= parseFactor(); // division
else if (eat('^')) x = Math.pow(x, parseFactor()); //exponentiation -> Moved in to here. So the problem is fixed
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();
eat(')');
} 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);
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 -> This is causing a bit of problem
return x;
}
}.parse();
}
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; }
//.........
}
I think what ever way you do this it's going to involve a lot of conditional statements. But for single operations like in your examples you could limit it to 4 if statements with something like
String math = "1+4";
if (math.split("+").length == 2) {
//do calculation
} else if (math.split("-").length == 2) {
//do calculation
} ...
It gets a whole lot more complicated when you want to deal with multiple operations like "4+5*6".
If you are trying to build a calculator then I'd surgest passing each section of the calculation separatly (each number or operator) rather than as a single string.
You might have a look at the Symja framework:
ExprEvaluator util = new ExprEvaluator();
IExpr result = util.evaluate("10-40");
System.out.println(result.toString()); // -> "-30"
Take note that definitively more complex expressions can be evaluated:
// 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
package ExpressionCalculator.expressioncalculator;
import java.text.DecimalFormat;
import java.util.Scanner;
public class ExpressionCalculator {
private static String addSpaces(String exp){
//Add space padding to operands.
//https://regex101.com/r/sJ9gM7/73
exp = exp.replaceAll("(?<=[0-9()])[\\/]", " / ");
exp = exp.replaceAll("(?<=[0-9()])[\\^]", " ^ ");
exp = exp.replaceAll("(?<=[0-9()])[\\*]", " * ");
exp = exp.replaceAll("(?<=[0-9()])[+]", " + ");
exp = exp.replaceAll("(?<=[0-9()])[-]", " - ");
//Keep replacing double spaces with single spaces until your string is properly formatted
/*while(exp.indexOf(" ") != -1){
exp = exp.replace(" ", " ");
}*/
exp = exp.replaceAll(" {2,}", " ");
return exp;
}
public static Double evaluate(String expr){
DecimalFormat df = new DecimalFormat("#.####");
//Format the expression properly before performing operations
String expression = addSpaces(expr);
try {
//We will evaluate using rule BDMAS, i.e. brackets, division, power, multiplication, addition and
//subtraction will be processed in following order
int indexClose = expression.indexOf(")");
int indexOpen = -1;
if (indexClose != -1) {
String substring = expression.substring(0, indexClose);
indexOpen = substring.lastIndexOf("(");
substring = substring.substring(indexOpen + 1).trim();
if(indexOpen != -1 && indexClose != -1) {
Double result = evaluate(substring);
expression = expression.substring(0, indexOpen).trim() + " " + result + " " + expression.substring(indexClose + 1).trim();
return evaluate(expression.trim());
}
}
String operation = "";
if(expression.indexOf(" / ") != -1){
operation = "/";
}else if(expression.indexOf(" ^ ") != -1){
operation = "^";
} else if(expression.indexOf(" * ") != -1){
operation = "*";
} else if(expression.indexOf(" + ") != -1){
operation = "+";
} else if(expression.indexOf(" - ") != -1){ //Avoid negative numbers
operation = "-";
} else{
return Double.parseDouble(expression);
}
int index = expression.indexOf(operation);
if(index != -1){
indexOpen = expression.lastIndexOf(" ", index - 2);
indexOpen = (indexOpen == -1)?0:indexOpen;
indexClose = expression.indexOf(" ", index + 2);
indexClose = (indexClose == -1)?expression.length():indexClose;
if(indexOpen != -1 && indexClose != -1) {
Double lhs = Double.parseDouble(expression.substring(indexOpen, index));
Double rhs = Double.parseDouble(expression.substring(index + 2, indexClose));
Double result = null;
switch (operation){
case "/":
//Prevent divide by 0 exception.
if(rhs == 0){
return null;
}
result = lhs / rhs;
break;
case "^":
result = Math.pow(lhs, rhs);
break;
case "*":
result = lhs * rhs;
break;
case "-":
result = lhs - rhs;
break;
case "+":
result = lhs + rhs;
break;
default:
break;
}
if(indexClose == expression.length()){
expression = expression.substring(0, indexOpen) + " " + result + " " + expression.substring(indexClose);
}else{
expression = expression.substring(0, indexOpen) + " " + result + " " + expression.substring(indexClose + 1);
}
return Double.valueOf(df.format(evaluate(expression.trim())));
}
}
}catch(Exception exp){
exp.printStackTrace();
}
return 0.0;
}
public static void main(String args[]){
Scanner scanner = new Scanner(System.in);
System.out.print("Enter an Mathematical Expression to Evaluate: ");
String input = scanner.nextLine();
System.out.println(evaluate(input));
}
}
A Java class that can evaluate mathematical expressions:
package test;
public class Calculator {
public static Double calculate(String expression){
if (expression == null || expression.length() == 0) {
return null;
}
return calc(expression.replace(" ", ""));
}
public static Double calc(String expression) {
String[] containerArr = new String[]{expression};
double leftVal = getNextOperand(containerArr);
expression = containerArr[0];
if (expression.length() == 0) {
return leftVal;
}
char operator = expression.charAt(0);
expression = expression.substring(1);
while (operator == '*' || operator == '/') {
containerArr[0] = expression;
double rightVal = getNextOperand(containerArr);
expression = containerArr[0];
if (operator == '*') {
leftVal = leftVal * rightVal;
} else {
leftVal = leftVal / rightVal;
}
if (expression.length() > 0) {
operator = expression.charAt(0);
expression = expression.substring(1);
} else {
return leftVal;
}
}
if (operator == '+') {
return leftVal + calc(expression);
} else {
return leftVal - calc(expression);
}
}
private static double getNextOperand(String[] exp){
double res;
if (exp[0].startsWith("(")) {
int open = 1;
int i = 1;
while (open != 0) {
if (exp[0].charAt(i) == '(') {
open++;
} else if (exp[0].charAt(i) == ')') {
open--;
}
i++;
}
res = calc(exp[0].substring(1, i - 1));
exp[0] = exp[0].substring(i);
} else {
int i = 1;
if (exp[0].charAt(0) == '-') {
i++;
}
while (exp[0].length() > i && isNumber((int) exp[0].charAt(i))) {
i++;
}
res = Double.parseDouble(exp[0].substring(0, i));
exp[0] = exp[0].substring(i);
}
return res;
}
private static boolean isNumber(int c) {
int zero = (int) '0';
int nine = (int) '9';
return (c >= zero && c <= nine) || c =='.';
}
public static void main(String[] args) {
System.out.println(calculate("(((( -6 )))) * 9 * -1"));
System.out.println(calc("(-5.2+-5*-5*((5/4+2)))"));
}
}
How about something like this:
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);
}
}
and do the similar thing for every other mathematical operator accordingly ..
It is possible to convert any expression string in infix notation to a postfix notation using Djikstra's shunting-yard algorithm. The result of the algorithm can then serve as input to the postfix algorithm with returns the result of the expression.
I wrote an article about it here, with an implementation in java
Yet another option: https://github.com/stefanhaustein/expressionparser
I have implemented this to have a simple but flexible option to permit both:
Immediate processing (Calculator.java, SetDemo.java)
Building and processing a parse tree (TreeBuilder.java)
The TreeBuilder linked above is part of a CAS demo package that does symbolic derivation. There is also a BASIC interpreter example and I have started to build a TypeScript interpreter using it.
External library like RHINO or NASHORN can be used to run javascript. And javascript can evaluate simple formula without parcing the string. No performance impact as well if code is written well.
Below is an example with RHINO -
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);
}
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);
}
}
I have done using iterative parsing and shunting Yard algorithm and i have really enjoyed developing the expression evaluator ,you can find all the code here
https://github.com/nagaraj200788/JavaExpressionEvaluator
Has 73 test cases and even works for Bigintegers,Bigdecimals
supports all relational, arithmetic expression and also combination of both .
even supports ternary operator .
Added enhancement to support signed numbers like -100+89 it was intresting, for details check TokenReader.isUnaryOperator() method and i have updated code in above Link
I have read other threads on pset5 Valgrind memory errors, but that didn't help me. I get 0 leaks, but this instead:
==1917== Conditional jump or move depends on uninitialised value(s)
Looks like you're trying to use a variable that might not have a value? Take a closer look at line 34 of dictionary.c.
The error refers to line 34 which is this: lower[i] = tolower(word[i]);
To supply context, the code below attempts to check if a word exists in the dictionary that has been uploaded to a hash table. I am attempting to convert the wanted word to lowercase because all the dictionary words are also lowercase and so that their hashes would be identical. The program successfully completes all tasks, but then stumbles upon these memory errors.
Any hints as to why Valgrind is mad at me? Thank you!
// Returns true if word is in dictionary else false
bool check(const char *word)
{
char lower[LENGTH + 1];
//Converts word to lower so the hashes of the dictionary entry and searched word would match
for (int i = 0; i < LENGTH + 1; i++)
{
lower[i] = tolower(word[i]);
}
// Creates node from the given bucket
node *tmp = table[hash(lower)];
// Traverses the linked list
while (tmp != NULL)
{
if (strcasecmp(word, tmp->word) == 0)
{
return true;
}
tmp = tmp->next;
}
return false;
}
Below is the whole dictionary.c file:
// Implements a dictionary's functionality
#include <string.h>
#include <strings.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include "dictionary.h"
// Represents a node in a hash table
typedef struct node
{
char word[LENGTH + 1];
struct node *next;
}
node;
// Number of buckets in hash table 26^3
const unsigned int N = 17576;
// Hash table
node *table[N];
int count = 0;
// Returns true if word is in dictionary else false
bool check(const char *word)
{
char lower[LENGTH + 1];
//Converts word to lower so the hashes of the dictionary entry and searched word would match
for (int i = 0; i < LENGTH + 1; i++)
{
lower[i] = tolower(word[i]);
}
// Creates node from the given bucket
node *tmp = table[hash(lower)];
// Traverses the linked list
while (tmp != NULL)
{
if (strcasecmp(word, tmp->word) == 0)
{
return true;
}
tmp = tmp->next;
}
return false;
}
// Hashes word to a number
unsigned int hash(const char *word)
{
// Modified hash function by Dan Berstein taken from http://www.cse.yorku.ca/~oz/hash.html
unsigned int hash = 5381;
int c;
while ((c = *word++))
{
hash = (((hash << 5) + hash) + c) % N; /* hash * 33 + c */
}
return hash;
}
// Loads dictionary into memory, returning true if successful else false
bool load(const char *dictionary)
{
FILE *inptr = fopen(dictionary, "r");
if (dictionary == NULL)
{
printf("Could not load %s\n.", dictionary);
return false;
}
// Create a char array to temporarily hold the new word (r stands for read)
char r_word[N+1];
// Until the end of file
while (fscanf(inptr, "%s", r_word) != EOF)
{
// Increments count
count++;
// Create a node
node *new_node = malloc(sizeof(node));
if (new_node == NULL)
{
unload();
return false;
}
strcpy(new_node->word, r_word);
// Hash the node
int index = hash(new_node->word);
// Places the node at the right index
new_node->next = table[index];
table[index] = new_node;
}
fclose(inptr);
return true;
}
// Returns number of words in dictionary if loaded else 0 if not yet loaded
unsigned int size(void)
{
if (&load == false)
{
return '0';
}
else
{
return count;
}
}
// Unloads dictionary from memory, returning true if successful else false
bool unload(void)
{
// Interates over the array
for (int i = 0; i < N; i++)
{
node *head = table[i];
while (head != NULL)
{
node *tmp = head;
head = head->next;
free(tmp);
}
}
return true;
}
This loop iterates through the maximum length of word-
for (int i = 0; i < LENGTH + 1; i++)
{
lower[i] = tolower(word[i]);
}
Except if you look at how word is created-
while (fscanf(inptr, "%s", r_word) != EOF)
{
// Increments count
count++;
// Create a node
node *new_node = malloc(sizeof(node));
if (new_node == NULL)
{
unload();
return false;
}
strcpy(new_node->word, r_word);
Notice, the variable r_word, may not be exactly of length LENGTH + 1. So what you really have in word is N number of characters, where N is not necessarily LENGTH + 1, it could be less.
So looping over the entire 0 -> LENGTH + 1 becomes problematic for words that are shorter than LENGTH + 1. You're going over array slots that do not have a value, they have garbage values.
What's the solution? This is precisely why c strings have \0-
for (int i = 0; word[i] != '\0'; i++)
{
lower[i] = tolower(word[i]);
}
This will stop the loop as soon as the NULL character is reached, which, you must have already learnt, marks the end of a string - aka a char array.
There may still be more errors in your code. But for your particular question - reading out of bounds is the answer.
I got a question to answer with the best complexity we can think about.
We got one sorted array (int) and X value. All we need to do is to find how many places in the array equals the X value.
This is my solution to this situation, as i don't know much about complexity.
All i know is that better methods are without for loops :X
class Question
{
public static int mount (int [] a, int x)
{
int first=0, last=a.length-1, count=0, pointer=0;
boolean found=false, finish=false;
if (x < a[0] || x > a[a.length-1])
return 0;
while (! found) **//Searching any place in the array that equals to x value;**
{
if ( a[(first+last)/2] > x)
last = (first+last)/2;
else if ( a[(first+last)/2] < x)
first = (first+last)/2;
else
{
pointer = (first+last)/2;
count = 1;
found = true; break;
}
if (Math.abs(last-first) == 1)
{
if (a[first] == x)
{
pointer = first;
count = 1;
found = true;
}
else if (a[last] == x)
{
pointer = last;
count = 1;
found = true;
}
else
return 0;
}
if (first == last)
{
if (a[first] == x)
{
pointer = first;
count = 1;
found = true;
}
else
return 0;
}
}
int backPointer=pointer, forwardPointer=pointer;
boolean stop1=false, stop2= false;
while (!finish) **//Counting the number of places the X value is near our pointer.**
{
if (backPointer-1 >= 0)
if (a[backPointer-1] == x)
{
count++;
backPointer--;
}
else
stop1 = true;
if (forwardPointer+1 <= a.length-1)
if (a[forwardPointer+1] == x)
{
count++;
forwardPointer++;
}
else
stop2 = true;
if (stop1 && stop2)
finish=true;
}
return count;
}
public static void main (String [] args)
{
int [] a = {-25,0,5,11,11,99};
System.out.println(mount(a, 11));
}
}
The print command count it right and prints "2".
I just want to know if anyone can think about better complexity for this method.
Moreover, how can i know what is the time/space-complexity of the method?
All i know about time/space-complexity is that for loop is O(n). I don't know how to calculate my method complexity.
Thank a lot!
Editing:
This is the second while loop after changing:
while (!stop1 || !stop2) //Counting the number of places the X value is near our pointer.
{
if (!stop1)
{
if ( a[last] == x )
{
stop1 = true;
count += (last-pointer);
}
else if ( a[(last+forwardPointer)/2] == x )
{
if (last-forwardPointer == 1)
{
stop1 = true;
count += (forwardPointer-pointer);
}
else
forwardPointer = (last + forwardPointer) / 2;
}
else
last = ((last + forwardPointer) / 2) - 1;
}
if (!stop2)
{
if (a[first] == x)
{
stop2 = true;
count += (pointer - first);
}
else if ( a[(first+backPointer)/2] == x )
{
if (backPointer - first == 1)
{
stop2 = true;
count += (pointer-backPointer);
}
else
backPointer = (first + backPointer) / 2;
}
else
first = ((first + backPointer) / 2) + 1;
}
}
What do you think about the changing? I think it would change the time complexity to O(long(n)).
First let's examine your code:
The code could be heavily refactored and cleaned (which would also result in more efficient implementation, yet without improving time or space complexity), but the algorithm itself is pretty good.
What it does is use standard binary search to find an item with the required value, then scans to the back and to the front to find all other occurrences of the value.
In terms of time complexity, the algorithm is O(N). The worst case is when the entire array is the same value and you end up iterating all of it in the 2nd phase (the binary search will only take 1 iteration). Space complexity is O(1). The memory usage (space) is unaffected by growth in input size.
You could improve the worst case time complexity if you keep using binary search on the 2 sub-arrays (back & front) and increase the "match range" logarithmically this way. The time complexity will become O(log(N)). Space complexity will remain O(1) for the same reason as before.
However, the average complexity for a real-world scenario (where the array contains various values) would be very close and might even lean towards your own version.
I'm very new to Arduino and C programming.
I'm making a GPS speedo and I'm trying to read in some serial, store a value from a substring and echo it back via serial.
At the moment I'm having problems storing the substring.
I've gotten to the point where I'm able to get some data between < and >.
But the data doesn't come in like that. It's a NMEA data stream and the data I want is between ,N, and ,K,.
So I've been trying to replace ,N, with < and ,K, with > .
Just can't get it to work. I get error: request for member 'replace' in 'c', which is of non-class type 'char'
Here's my code so far....
int indata = 0;
int scrubdata = 0;
char inString[32];
int stringPos = 0;
boolean startRead = false; // is reading?
void setup() {
Serial.begin(4800);
}
void loop() {
String pageValue = readPage();
Serial.print(pageValue);
}
String readPage(){
//read the page, and capture & return everything between '<' and '>'
stringPos = 0;
memset( &inString, 0, 32 ); //clear inString memory
while(true){
if (Serial.available() > 0) {
char c = Serial.read();
c.replace(",N,", "<");
c.replace(",K,", ">");
if (c == '<' ) { //'<' is our begining character
startRead = true; //Ready to start reading the part
}
else if(startRead){
if(c != '>'){ //'>' is our ending character
inString[stringPos] = c;
stringPos ++;
}
else{
//got what we need here! We can disconnect now
startRead = false;
return inString;
}
}
}
}
}
By Default:
Serial.read() returns an int if you must process the data this way, try casting it to char with:
char c = (char) Serial.read();
Another way to do this:
Would be to seek your beginning string (discarding un-needed data) using Serial.find() then reading data until you met your end character ",K," with Serial.readBytesUntil()
Something like this would work quite well:
char inData[64]; //adjust for your data size
Serial.setTimeout(2000); //Defaults to 1000 msecs set if necessary
Serial.find(",N,"); //Start of Data
int bRead = Serial.readBytesUntil(",K,", inData, 64); //Read until end of data
inData[bRead] = 0x00; //Zero terminate if using this as a string
return inData;