Basic Expression Parsing
Click here for advanced expression parsing
When writing your own calculator it is necessary to build a converter that can transform an input mathematical expression such as ( 1 + 8 ) – ( ( 3 * 4 ) / 2 ), into a format that is more suited for evaluation by computers.
When evaluating expressions such as the one above (known as “infix notation“), that which appears simple and intuitive to us humans, is usually not so straightforward to implement in a programming language.
The shunting-yard algorithm is a method for parsing mathematical expressions written in infix notation to Reverse Polish Notation (RPN). The RPN notation is different to infix notation in that every operator (+, -, * etc) comes after the operands (numbers) and there are no parentheses (brackets).
So ( 3 * 4 ) for example becomes 3 4 *.
When given an input string in Reverse Polish Notation, it is then possible to employ a simple algorithm based around the use of a stack in order to evaluate the RPN expression and determine the mathematical result.
The Shunting-yard algorithm
This pseudocode shows how the shunting-yard algorithm converts an expression given in conventional infix notation into Reverse Polish Notation:
For each token
{
If (token is a number)
{
Add number to the output queue
}
If (token is an operator eg +,-,*...)
{
While (stack not empty AND
stack top element is an operator)
{
If ((token = left associative AND
precedence <= stack top element) OR
(token = right associative AND
precedence < stack top element))
{
Pop stack onto the output queue.
Exit while loop.
}
}
Push token onto stack
}
If (token is left bracket '(')
{
Push token on to stack
}
If (token is right bracket ')')
{
While (stack not empty AND
stack top element not a left bracket)
{
Pop the stack onto output queue
}
Pop the stack
}
}
While (stack not empty)
{
Pop stack onto output queue
}
Implementing the the shunting-yard algorithm in Java
The Java function to implement the shunting yard algorithm described above is as follows:
// Convert infix expression format into reverse Polish notation
public static String[] infixToRPN(String[] inputTokens)
{
ArrayList<String> out = new ArrayList<String>();
Stack<String> stack = new Stack<String>();
// For each tokens
for (String token : inputTokens)
{
// If token is an operator
if (isOperator(token))
{
// While stack not empty AND stack top element
// is an operator
while (!stack.empty() && isOperator(stack.peek()))
{
if ((isAssociative(token, LEFT_ASSOC) &&
cmpPrecedence(token, stack.peek()) <= 0) ||
(isAssociative(token, RIGHT_ASSOC) &&
cmpPrecedence(token, stack.peek()) < 0))
{
out.add(stack.pop());
continue;
}
break;
}
// Push the new operator on the stack
stack.push(token);
}
// If token is a left bracket '('
else if (token.equals("("))
{
stack.push(token); //
}
// If token is a right bracket ')'
else if (token.equals(")"))
{
while (!stack.empty() && !stack.peek().equals("("))
{
out.add(stack.pop());
}
stack.pop();
}
// If token is a number
else
{
out.add(token);
}
}
while (!stack.empty())
{
out.add(stack.pop());
}
String[] output = new String[out.size()];
return out.toArray(output);
}
Evaluating expressions in Reverse Polish Notation
Once we have obtained our input expression string in RPN format, we evaluate the expression using a simple algorithm, also based around the use of a stack. Pseudocode as follows:
For each token
{
If (token is a number)
{
Push value onto stack
}
If (token is an operator)
{
Pop 2 top values from the stack
Evaluate operator using popped values as args
Push result onto stack
}
}
The single value remaining on the stack is the evaluation. Java code for this is as follows:
public static double RPNtoDouble(String[] tokens)
{
Stack<String> stack = new Stack<String>();
// For each token
for (String token : tokens)
{
// If the token is a number push it onto the stack
if (!isOperator(token))
{
stack.push(token);
}
else
{
// Pop the two top entries
Double d2 = Double.valueOf( stack.pop() );
Double d1 = Double.valueOf( stack.pop() );
//Get the result
Double result = token.compareTo("+") == 0 ? d1 + d2 :
token.compareTo("-") == 0 ? d1 - d2 :
token.compareTo("*") == 0 ? d1 * d2 :
d1 / d2;
// Push result onto stack
stack.push( String.valueOf( result ));
}
}
return Double.valueOf(stack.pop());
}
Example
Consider the expression
( 1 + 2 ) * ( 3 / 4 ) - ( 5 + 6 ).
Passing this as a string argument to the infixToRPN function will yield this RPN format:
1 2 + 3 4 / * 5 6 + -
It is then simply a matter of evaluating the RPN format to give the desired mathematical result of -8.75.
Putting it all together: Java Implementation of Basic Expression Parser
Full Java code listing as follows:
package expressionparser;
import java.util.*;
public class ExpressionParser
{
// Associativity constants for operators
private static final int LEFT_ASSOC = 0;
private static final int RIGHT_ASSOC = 1;
// Operators
private static final Map<String, int[]> OPERATORS = new HashMap<String, int[]>();
static
{
// Map<"token", []{precendence, associativity}>
OPERATORS.put("+", new int[] { 0, LEFT_ASSOC });
OPERATORS.put("-", new int[] { 0, LEFT_ASSOC });
OPERATORS.put("*", new int[] { 5, LEFT_ASSOC });
OPERATORS.put("/", new int[] { 5, LEFT_ASSOC });
}
// Test if token is an operator
private static boolean isOperator(String token)
{
return OPERATORS.containsKey(token);
}
// Test associativity of operator token
private static boolean isAssociative(String token, int type)
{
if (!isOperator(token))
{
throw new IllegalArgumentException("Invalid token: " + token);
}
if (OPERATORS.get(token)[1] == type) {
return true;
}
return false;
}
// Compare precedence of operators.
private static final int cmpPrecedence(String token1, String token2)
{
if (!isOperator(token1) || !isOperator(token2))
{
throw new IllegalArgumentException("Invalid tokens: " + token1
+ " " + token2);
}
return OPERATORS.get(token1)[0] - OPERATORS.get(token2)[0];
}
// Convert infix expression format into reverse Polish notation
public static String[] infixToRPN(String[] inputTokens)
{
ArrayList<String> out = new ArrayList<String>();
Stack<String> stack = new Stack<String>();
// For each token
for (String token : inputTokens)
{
// If token is an operator
if (isOperator(token))
{
// While stack not empty AND stack top element
// is an operator
while (!stack.empty() && isOperator(stack.peek()))
{
if ((isAssociative(token, LEFT_ASSOC) &&
cmpPrecedence(token, stack.peek()) <= 0) ||
(isAssociative(token, RIGHT_ASSOC) &&
cmpPrecedence(token, stack.peek()) < 0))
{
out.add(stack.pop());
continue;
}
break;
}
// Push the new operator on the stack
stack.push(token);
}
// If token is a left bracket '('
else if (token.equals("("))
{
stack.push(token); //
}
// If token is a right bracket ')'
else if (token.equals(")"))
{
while (!stack.empty() && !stack.peek().equals("("))
{
out.add(stack.pop());
}
stack.pop();
}
// If token is a number
else
{
out.add(token);
}
}
while (!stack.empty())
{
out.add(stack.pop());
}
String[] output = new String[out.size()];
return out.toArray(output);
}
public static double RPNtoDouble(String[] tokens)
{
Stack<String> stack = new Stack<String>();
// For each token
for (String token : tokens)
{
// If the token is a value push it onto the stack
if (!isOperator(token))
{
stack.push(token);
}
else
{
// Token is an operator: pop top two entries
Double d2 = Double.valueOf( stack.pop() );
Double d1 = Double.valueOf( stack.pop() );
//Get the result
Double result = token.compareTo("+") == 0 ? d1 + d2 :
token.compareTo("-") == 0 ? d1 - d2 :
token.compareTo("*") == 0 ? d1 * d2 :
d1 / d2;
// Push result onto stack
stack.push( String.valueOf( result ));
}
}
return Double.valueOf(stack.pop());
}
public static void main(String[] args) {
String[] input = "( 1 + 2 ) * ( 3 / 4 ) - ( 5 + 6 )".split(" ");
String[] output = infixToRPN(input);
// Build output RPN string minus the commas
for (String token : output) {
System.out.print(token + " ");
}
// Feed the RPN string to RPNtoDouble to give result
Double result = RPNtoDouble( output );
}
}
C++ Implementation of Basic Expression Parser
#include <iostream>
#include <sstream>
#include <list>
#include <stack>
#include <map>
#include <string>
#include <vector>
#include <iterator>
#include <stdlib.h>
const int LEFT_ASSOC = 0;
const int RIGHT_ASSOC = 1;
// Map the different operators: +, -, *, / etc
typedef std::map< std::string, std::pair< int,int >> OpMap;
typedef std::vector<std::string>::const_iterator cv_iter;
typedef std::string::iterator s_iter;
const OpMap::value_type assocs[] =
{ OpMap::value_type( "+", std::make_pair<int,int>( 0, LEFT_ASSOC ) ),
OpMap::value_type( "-", std::make_pair<int,int>( 0, LEFT_ASSOC ) ),
OpMap::value_type( "*", std::make_pair<int,int>( 5, LEFT_ASSOC ) ),
OpMap::value_type( "/", std::make_pair<int,int>( 5, LEFT_ASSOC ) ) };
const OpMap opmap( assocs, assocs + sizeof( assocs ) / sizeof( assocs[ 0 ] ) );
// Test if token is an pathensesis
bool isParenthesis( const std::string& token)
{
return token == "(" || token == ")";
}
// Test if token is an operator
bool isOperator( const std::string& token)
{
return token == "+" || token == "-" ||
token == "*" || token == "/";
}
// Test associativity of operator token
bool isAssociative( const std::string& token, const int& type)
{
const std::pair<int,int> p = opmap.find( token )->second;
return p.second == type;
}
// Compare precedence of operators.
int cmpPrecedence( const std::string& token1, const std::string& token2 )
{
const std::pair<int,int> p1 = opmap.find( token1 )->second;
const std::pair<int,int> p2 = opmap.find( token2 )->second;
return p1.first - p2.first;
}
// Convert infix expression format into reverse Polish notation
bool infixToRPN( const std::vector<std::string>& inputTokens,
const int& size,
std::vector<std::string>& strArray )
{
bool success = true;
std::list<std::string> out;
std::stack<std::string> stack;
// While there are tokens to be read
for ( int i = 0; i < size; i++ )
{
// Read the token
const std::string token = inputTokens[ i ];
// If token is an operator
if ( isOperator( token ) )
{
// While there is an operator token, o2, at the top of the stack AND
// either o1 is left-associative AND its precedence is equal to that of o2,
// OR o1 has precedence less than that of o2,
const std::string o1 = token;
if ( !stack.empty() )
{
std::string o2 = stack.top();
while ( isOperator( o2 ) &&
( ( isAssociative( o1, LEFT_ASSOC ) && cmpPrecedence( o1, o2 ) == 0 ) ||
( cmpPrecedence( o1, o2 ) < 0 ) ) )
{
// pop o2 off the stack, onto the output queue;
stack.pop();
out.push_back( o2 );
if ( !stack.empty() )
o2 = stack.top();
else
break;
}
}
// push o1 onto the stack.
stack.push( o1 );
}
// If the token is a left parenthesis, then push it onto the stack.
else if ( token == "(" )
{
// Push token to top of the stack
stack.push( token );
}
// If token is a right bracket ')'
else if ( token == ")" )
{
// Until the token at the top of the stack is a left parenthesis,
// pop operators off the stack onto the output queue.
std::string topToken = stack.top();
while ( topToken != "(" )
{
out.push_back(topToken );
stack.pop();
if ( stack.empty() ) break;
topToken = stack.top();
}
// Pop the left parenthesis from the stack, but not onto the output queue.
if ( !stack.empty() ) stack.pop();
// If the stack runs out without finding a left parenthesis,
// then there are mismatched parentheses.
if ( topToken != "(" )
{
return false;
}
}
// If the token is a number, then add it to the output queue.
else
{
out.push_back( token );
}
}
// While there are still operator tokens in the stack:
while ( !stack.empty() )
{
const std::string stackToken = stack.top();
// If the operator token on the top of the stack is a parenthesis,
// then there are mismatched parentheses.
if ( isParenthesis( stackToken ) )
{
return false;
}
// Pop the operator onto the output queue./
out.push_back( stackToken );
stack.pop();
}
strArray.assign( out.begin(), out.end() );
return success;
}
double RPNtoDouble( std::vector<std::string> tokens )
{
std::stack<std::string> st;
// For each token
for ( int i = 0; i < (int) tokens.size(); ++i )
{
const std::string token = tokens[ i ];
// If the token is a value push it onto the stack
if ( !isOperator(token) )
{
st.push(token);
}
else
{
double result = 0.0;
// Token is an operator: pop top two entries
const std::string val2 = st.top();
st.pop();
const double d2 = strtod( val2.c_str(), NULL );
if ( !st.empty() )
{
const std::string val1 = st.top();
st.pop();
const double d1 = strtod( val1.c_str(), NULL );
//Get the result
result = token == "+" ? d1 + d2 :
token == "-" ? d1 - d2 :
token == "*" ? d1 * d2 :
d1 / d2;
}
else
{
if ( token == "-" )
result = d2 * -1;
else
result = d2;
}
// Push result onto stack
std::ostringstream s;
s << result;
st.push( s.str() );
}
}
return strtod( st.top().c_str(), NULL );
}
std::vector<std::string> getExpressionTokens( const std::string& expression )
{
std::vector<std::string> tokens;
std::string str = "";
for ( int i = 0; i < (int) expression.length(); ++i )
{
const std::string token( 1, expression[ i ] );
if ( isOperator( token ) || isParenthesis( token ) )
{
if ( !str.empty() )
{
tokens.push_back( str ) ;
}
str = "";
tokens.push_back( token );
}
else
{
// Append the numbers
if ( !token.empty() && token != " " )
{
str.append( token );
}
else
{
if ( str != "" )
{
tokens.push_back( str );
str = "";
}
}
}
}
return tokens;
}
// Print iterators in a generic way
template<typename T, typename InputIterator>
void Print( const std::string& message,
const InputIterator& itbegin,
const InputIterator& itend,
const std::string& delimiter)
{
std::cout << message << std::endl;
std::copy(itbegin,
itend,
std::ostream_iterator<T>(std::cout, delimiter.c_str()));
std::cout << std::endl;
}
int main()
{
std::string s = "( 1 + 2) * ( 3 / 4 )-(5+6)";
Print<char, s_iter>( "Input expression:", s.begin(), s.end(), "" );
// Tokenize input expression
std::vector<std::string> tokens = getExpressionTokens( s );
// Evaluate feasible expressions
std::vector<std::string> rpn;
if ( infixToRPN( tokens, tokens.size(), rpn ) )
{
double d = RPNtoDouble( rpn );
Print<std::string, cv_iter>( "RPN tokens: ", rpn.begin(), rpn.end(), " " );
std::cout << "Result = " << d << std::endl;
}
else
{
std::cout << "Mis-match in parentheses" << std::endl;
}
return 0;
}
So when running the above code as a simple Windows Console application using an example input expression of ( 1 + 2) * ( 3 / 4 )-(5+6) we get the following RPN tokens and mathematical result:
If an inappropriate input expression is used, one with a mis-match in the number of left/right parentheses such as ( 1 + 2 * ( 3 / 4 )-(5+6), then we get an error:
And if we insert an additional minus sign ‘-‘ so that the original expression used is now -( 1 + 2) * ( 3 / 4 )-(5+6), the algorithm recognizes that this should get treated as a unary minus operator, resulting in the following output:
UPDATE: 13 April 2013
Extending the algorithm to handle further mathematical expressions
So far our implementation has been applied to handle basic expressions based on standard mathematical operators +, -, *, / etc. The following downloadable C++ code makes a number of further improvements to handle additional mathematical operators sin, cos, tan, log, exp etc, as well as much more complicated subexpressions.
Additional sanity checks are included to make sure there are no mismatched numbers of parentheses, as well as some additional work on the tokenization of RPM strings to handle unary minus operators occuring at positions where an expression is expected eg -8 + 5 = -3 or 11 ^ -7 = 5.13158e-08.
All examples in this code have been verified using Google’s online calculator. To use Google’s built-in calculator simply enter your mathematical expression into the search box. For example:
Some examples are shown in the following table:
| Expression (infix) | RPN (postfix) | Result |
|---|---|---|
| exp( 1.11 ) | 1.11 exp | 3.034358 |
| sin( cos( 90 * pi / 180 ) ) | 90 pi * 180 / cos sin | 0.000001 |
| 34.5*(23+1.5)/2 | 34.5 23 1.5 + * 2 / | 422.625000 |
| 5 + ((1 + 2) * 4) – 3 | 5 1 2 + 4 * + 3 – | 14 |
| ( 1 + 2 ) * ( 3 / 4 ) ^ ( 5 + 6 ) | 1 2 + 3 4 / 5 6 + ^ * | 0.126705 |
| 3/2 + 4*(12+3) | 3 2 / 4 12 3 + * + | 61.5 |
| PI*pow(9/2,2) | PI 9 2 / 2 pow * | 63.617197 |
| ((2*(6-1))/2)*4 | 2 6 1 – * 2 / 4 * | 20 |
| ln(2)+3^5 | 2 ln 3 5 ^ + | 243.693147 |
| 11 ^ -7 | 11 -7 ^ | 5.13158e-08 |
| cos ( ( 1.3 + 1 ) ^ ( 1 / 3 ) ) – log ( -2 * 3 / -14 ) | 1.3 1 + 1 3 / ^ cos -2 3 * -14 / log – | 0.616143 |
| 1 * -sin( Pi / 2) | 1 Pi 2 / -sin * | -1 |
| -8 + 5 | -8 5 + | -3 |
| 1 – (-2^2) – 1 | 1 1 -2 2 ^ * – 1 – | 4 |
The file main.cpp is an example usage of applying the Tokenize method of the ExpressionParser class to first prime the input expression string into a suitable format, which is subsequently converted to Reverse Polish Notation using the InfixToRPN method.
If successful, the reverse Polish Notation is then evaluated to give the mathematical result.
Example usage shown here:
#include "ExpressionParser.h"
// Print iterators in a generic way
template<typename T, typename InputIterator>
void Print( const std::string& message,
const InputIterator& itbegin,
const InputIterator& itend,
const std::string& delimiter)
{
std::cout << message << std::endl;
std::copy(itbegin,
itend,
std::ostream_iterator<T>(std::cout, delimiter.c_str()));
std::cout << std::endl;
}
int main(int argc, char** argv)
{
std::string result;
//std::string originalInput = "-11 ^ -7";
//std::string originalInput = "-1*- sin( Pi / -2)";
//std::string originalInput = "-8 + 5";
//std::string originalInput = "5 + (-1 + 2 )";
//std::string originalInput = "cos ( ( -1.3 + -1 ) ^ ( 1 / -3 ) ) - log ( -2 * 3 / -14 )";
//std::string originalInput = "cos ( ( 1.3 + 1 ) ^ ( 1 / 3 ) )";
//std::string originalInput = "( 1.3 + -1 ) ^ ( 1 / 3 )";
//std::string originalInput = "cos(- 1.32001)";
//std::string originalInput = "-8+3";
//std::string originalInput = "ln(2)+3^5";
//std::string originalInput = "((2*(6-1))/2)*4";
//std::string originalInput = "PI*pow(9/2,2)";
//std::string originalInput = "PI*pow(9/-2,2)";
//std::string originalInput = "pow(2, 3)";
//std::string originalInput = "3/2 + 4*(12+3)";
//std::string originalInput = "( 1 + 2 ) * ( 3 / 4 ) ^ ( 5 + 6 )";
//std::string originalInput = "5 + ((1 + 2) * -4) - 3";
//std::string originalInput = "34.5*(23+1.5)/2";
//std::string originalInput = "sin( cos( 90 * pi / 180 ) )";
std::string originalInput = "exp( 1.11 )";
//std::string originalInput = "5 + ((1 + 2) * 4) - 3 ";
//std::string originalInput = "3 + 4 * 2 / ( 1 - 5 ) ^ 2 ^ 3";
//std::string originalInput = "2.5^3";
//std::string originalInput = "-cos(1/-20)+sin(1/30)+cos(1/50)";
//std::string originalInput = "--cos(1/-20)+-sin(1/30)";
//std::string originalInput = "--cos(1/-20)--sin(1/30)";
//std::string originalInput = "--cos(1/--20)";
//std::string originalInput = "SQRT(4 )";
//std::string originalInput = "5 + (-1 + 2 )";
//std::string originalInput = "-(2+3)*(4*-10-1)+100";
//std::string originalInput = "(2+3)*-(4*10-1)+100";
//std::string originalInput = "1 - (-2^2) - 1";
//std::string originalInput = "(4*-10-1)";
//std::string originalInput = "-(2+3)*(4*-10-1)";
//std::string originalInput = "-3/2 + 4*-( 12+3)";
//std::string originalInput = "-3/2 + 4*(-12-3)";
//std::string originalInput = "-3/2 + -4*(-12-3)";
Print<char, std::string::iterator>( "Input expression:", originalInput.begin(), originalInput.end(), "" );
ExpressionParser parser( originalInput );
if ( !parser.MatchingParetheses() )
{
std::cout << "Error: mismatched parentheses or empty input"
<< std::endl;
return 1;
}
std::vector<std::string> RPN;
if ( parser.InfixToRPN( RPN ) )
{
Print<std::string, std::vector<std::string>::const_iterator>( "RPN tokens: ",
RPN.begin(),
RPN.end(),
" " );
if ( parser.Evaluate( RPN, result ) )
{
std::cout << std::endl;
std::cout << "Result = " << result << std::endl;
}
}
else
{
std::cout << "Error: mismatched parentheses" << std::endl;
}
return 0;
}
Example outputs:
Download below. Please contact me if you need any help in using the software.
Comments
45 responses to “Mathematical Expression Parsers in Java and C++”
hi,
just want to say thanks, it was very useful and, u’ve made it easy.
P.S
i think, you should change this, like what u have in your java code :
If (token is right bracket ')')
{
While (stack not empty AND
stack top element not a left bracket)
{
Pop the stack onto output queue
Pop the stack
}
}
to this:
If (token is right bracket ')')
{
While (stack not empty AND
stack top element not a left bracket)
{
Pop the stack onto output queue
}
Pop the stack
}
Hi thanks for the feedback, glad you found it useful. Post has been updated accordingly.
Well spotted!
Hello,
Thanks for your post . I have one thing..
In the updated version, If while loop is ended , then stack is empty which means , you can’t pop from… after while loop
Hi Yasmin
I think I know what you mean. But the stack is not quite empty, the while loop keeps popping until it reaches a left parenthesis:
// Until the token at the top of the stack is a left parenthesis, pop operators
// off the stack onto the output queue.
while ( !stack.empty() && stack.top() != "(" )
{
// Add to end of list
outputQueue.push( stack.top() );
stack.pop();
}
If it is not left with a remaining “(” then the expression is invalid and the program exits.
I know, but in your original post the second pop statement from the stack is out of while loop .so if there is no opening ( and everything is just pop from the stack, it will release exception, so it should be handled in case user miswritting the opening bracket…
Ah, I see what you mean. I think this improvement got put into the in the downloadable, but the one shown here still relies on a valid input expression. I will update the code in due course. Thanks for pointing that out.
Hi I have updated the C++ code listing on this page to reflect this. You still need to enter each value separated by a whitespace though, the download has tokenizers etc to cope with these.
hello sir, could u tell me the java coding to use one expression for calculation which has been passed as string from another java file?
Hi Rakesh. I’m not sure I understand your question. Could you be more specific? It would help if you could express your problem in terms of what your current input is, and what your desired outcome would be. What do you mean by “passed as string from another java file?”
really cool, since easy to extract an abstract base class with generic token. thanx a lot!!
Thanks for your kind words. The intention was to keep this as simple as possible, so that others can use it and/or modify it in ways they see fit.
Hi,
the solution is really elegant…
Are there algorithms for inserting braces in a mathematical expression…..???
You mean braces in things like sin or log expressions?
Can your parser solve this questions?
Given a series of numbers as the input, the last one as the result. Use the rest numbers to calculate the result,only +, -, *, / are allowed. The order of the input cannot be changed. If there is an equation, print it; or print “no equation”. If more than one solution, any working equation is fine.
example:
input: 2, 3, 1, 4
output: 2+3-1 = 4.
Hi Andy. No this parser as it stands would not be able to do what describe. If I understand you correctly, would it not be a case of writing an additional routine to:
1. parse your original input string of “2,3,1,4”, separating it into an input part (“2,3,1”) and an expected output part (“4”).
2. replace each of the the commas in your input string with the arithmetic operators you prefer so that “2,3,1” becomes “2+3-1”
3. feed this new string “2+3-1” into the infixToRPN() routine and then the RPNtoDouble() routine.
4. compare the answer returned with the desired answer. If they are equal print the new string appended with the equals sign and the desired output string ie “2+3-1” + “=4” ; otherwise print “no equation”.
Hope this helps.
can you help me with this problem pleas.
Write a program that will evaluate mathematical equation and print its result.
ex. enter equation (3+4)/2*1
result 3.5
Hi have you tried using the expression “(3+4)/2*1” in the Java or C++ versions shown here?
For the C++ code, go the line that says
istringstream iss( “( 1 + 2 ) * ( 3 / 4 ) – ( 5 + 6 )” );
and replace this with
istringstream iss( “( 3 + 4 ) / 2 * 1” ).
Ditto the For the Java version, just simply use:
String[] input = “( 3 + 4 ) / 2 * 1″.split(” “);
Both yield the desired result of 3.5
Thanks for this great job!!!
I have a question how can i add a subexpression ???
example:
10 + 20 * sin ( 30+40 ) -cos ( 10*sin(10) )
Thanks
Will look into it one day…
The newest version is now able to evaluate expressions like this.
Thank you for sharing your code 🙂
You’re welcome.
How can I add sin, cos, tan, pi, square root and square to the code? Should they be added to own OPERATORS?
Newset version can now handle functions like these.
I think it may involve this. When I get a moment I will look into it. If you see any useful links please let me know.
Cool that you are going to look into other operators! Looking foward to it!
Done!
Thanks,
But how can I tell the difference between subtraction(-) and negative (-) ?
such as 8-3 is subtraction but -8+3 is negative.
Now covered in newest version!
Really nice introduction into the shunting yard algorithm. If you are interested in a versatile and extendable expression parser using an LL(1) grammar for Java follow this link.
http://cogitolearning.co.uk/?p=523
Very slick. Nice one!
[…] I know it might be awkward to ask this type of question but really this bug is causing a lot of problem for me. I used the cpp mathematical parsing code from this link.C++ Implementation of Basic Expression Parser https://www.technical-recipes.com/2011/a-mathematical-expression-parser-in-java-and-cpp/ […]
Hi, Andy. Can I use this Java code for my own project?
Yes.
Hi Andy,
We are currently working on a RPN Calculator and your algorithm help us a lot.
Thank you very much for sharing your code,
Théophile
No really sure why but this “5+6”
outputs->
Input expression:
5+6
RPN tokens:
5 +
Result = 5
No really sure why but this with parentheses “(5+6)”
outputs->
Input expression:
(5+6)
RPN tokens:
5 6 +
Result = 11
Hi Anthony which language version are you using the C++ or Java?
I believe its a bug in the getExpressionTokens()
If you put in the string “5 + 6 ” with the extra space character after the 6, it should correctly output 11.
It’s because the free code (the c++ provided above, not the payed code) does not allow for the final character to get pushed to the token stack, if the final character is a number.
Hi,
Can I use abs function with your code ?? For example 2 + abs(-1), work with your code or not ??
Regards
Hi :D, can you show me how to input from a file (Java code) method? Tks
Hi. I tried to run the zip file but it says in expressionparser.cpp in line 263 ‘prev’ is not a member of ‘std’. How do i fix this?
std::prev is part of C++11. If you don’t have this there is documentation online on how to write your own.
One possible implementation:
// copy-pasted from cppreference::difference_type n = 1)
template
BidirIt prev(BidirIt it,
typename std::iterator_traits
{
std::advance(it, -n);
return it;
}
Hi,
I’m trying to run this on my machine but am getting this error:
“ExpressionParser::InfixToRPN(std::__1::vector<std::__1::basic_string<char, std::__1::char_traits, std::__1::allocator >, std::__1::allocator<std::__1::basic_string<char, std::__1::char_traits, std::__1::allocator > > >&)”, referenced from:
_main in main-1b71a2.o
“ExpressionParser::MatchingParetheses()”, referenced from:
_main in main-1b71a2.o
“ExpressionParser::Evaluate(std::__1::vector<std::__1::basic_string<char, std::__1::char_traits, std::__1::allocator >, std::__1::allocator<std::__1::basic_string<char, std::__1::char_traits, std::__1::allocator > > > const&, std::__1::basic_string<char, std::__1::char_traits, std::__1::allocator >&)”, referenced from:
_main in main-1b71a2.o
“ExpressionParser::ExpressionParser(std::__1::basic_string<char, std::__1::char_traits, std::__1::allocator > const&)”, referenced from:
_main in main-1b71a2.o
ld: symbol(s) not found for architecture x86_64
clang: error: linker command failed with exit code 1 (use -v to see invocation)
any ideas?
failed to aprse this expression : -( 1 + 2) * -( 3 / 4 )-(5+6)
*parse