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@mycodeschool
Created November 29, 2013 07:02
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/*
Evaluation Of postfix Expression in C++
Input Postfix expression must be in a desired format.
Operands must be integers and there should be space in between two operands.
Only '+' , '-' , '*' and '/' operators are expected.
*/
#include<iostream>
#include<stack>
#include<string>
using namespace std;
// Function to evaluate Postfix expression and return output
int EvaluatePostfix(string expression);
// Function to perform an operation and return output.
int PerformOperation(char operation, int operand1, int operand2);
// Function to verify whether a character is operator symbol or not.
bool IsOperator(char C);
// Function to verify whether a character is numeric digit.
bool IsNumericDigit(char C);
int main()
{
string expression;
cout<<"Enter Postfix Expression \n";
getline(cin,expression);
int result = EvaluatePostfix(expression);
cout<<"Output = "<<result<<"\n";
}
// Function to evaluate Postfix expression and return output
int EvaluatePostfix(string expression)
{
// Declaring a Stack from Standard template library in C++.
stack<int> S;
for(int i = 0;i< expression.length();i++) {
// Scanning each character from left.
// If character is a delimitter, move on.
if(expression[i] == ' ' || expression[i] == ',') continue;
// If character is operator, pop two elements from stack, perform operation and push the result back.
else if(IsOperator(expression[i])) {
// Pop two operands.
int operand2 = S.top(); S.pop();
int operand1 = S.top(); S.pop();
// Perform operation
int result = PerformOperation(expression[i], operand1, operand2);
//Push back result of operation on stack.
S.push(result);
}
else if(IsNumericDigit(expression[i])){
// Extract the numeric operand from the string
// Keep incrementing i as long as you are getting a numeric digit.
int operand = 0;
while(i<expression.length() && IsNumericDigit(expression[i])) {
// For a number with more than one digits, as we are scanning from left to right.
// Everytime , we get a digit towards right, we can multiply current total in operand by 10
// and add the new digit.
operand = (operand*10) + (expression[i] - '0');
i++;
}
// Finally, you will come out of while loop with i set to a non-numeric character or end of string
// decrement i because it will be incremented in increment section of loop once again.
// We do not want to skip the non-numeric character by incrementing i twice.
i--;
// Push operand on stack.
S.push(operand);
}
}
// If expression is in correct format, Stack will finally have one element. This will be the output.
return S.top();
}
// Function to verify whether a character is numeric digit.
bool IsNumericDigit(char C)
{
if(C >= '0' && C <= '9') return true;
return false;
}
// Function to verify whether a character is operator symbol or not.
bool IsOperator(char C)
{
if(C == '+' || C == '-' || C == '*' || C == '/')
return true;
return false;
}
// Function to perform an operation and return output.
int PerformOperation(char operation, int operand1, int operand2)
{
if(operation == '+') return operand1 +operand2;
else if(operation == '-') return operand1 - operand2;
else if(operation == '*') return operand1 * operand2;
else if(operation == '/') return operand1 / operand2;
else cout<<"Unexpected Error \n";
return -1;
}
@arsharaj
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arsharaj commented Jun 13, 2020 via email

@vaibhavs017
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vaibhavs017 commented Jul 5, 2020

Why -'0' in line 64

Because 0 in ASCII table is referred as 48 in decimal and 9 as 54 in decimal notation form.
Hence when evaluating the char to get integers this is very nice way of doing .

@abhisheknaw
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@arsharaj what function/work does this portion of your code do?
if(mychar==','){ // Delimiter comma handling
if(num!=(-1)){
temp_stack.push(num);
num=-1;
}
plzz can u explain?

@GrezzyTap
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The fact that '0' is 48, is it like a built in value? so if I write a program like:
char c1='2';
char c2='4';
then would c1+c2=6?

int x=c1+c2;
above expression will result in 102
so to get 6 do the following,
int x=c1+c2-2*'0';

@Malikhaidar
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evaluate expression to postfix() method 12 38 + 5 * with algorithm.
solution?

@kesava-karri
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sir this code is not debuging

Don't forget to give the input in post fix form...

@kesava-karri
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kesava-karri commented Oct 31, 2020

#include <iostream>
#include <stack>
#include <string>

using namespace std;

int do_operation(int op2, int op1, char opr);
bool is_operator(char C);
bool is_operand(char C);

int evaluatePostfix(string exp) {
    stack<int> S;
    for(int i=0; i<exp.length(); i++) {
        if(exp[i] == ' ' || exp[i] ==',') {
            continue;
        }
        else if(is_operand(exp[i])) {
            int operand = 0;
            while(is_operand(exp[i])) {
                operand = operand*10 + exp[i] - '0' ;
                i++;
            }
            i--;
            S.push(operand);
        }

        else if(is_operator(exp[i])) {
            int op2 = S.top(); 
            S.pop();
            int op1 = S.top();
            S.pop();
            int result = do_operation(op2, op1, exp[i]);
            S.push(result);
        }
    }
    return S.top();
}

int do_operation(int op2, int op1, char opr) {
    if(opr == '/') return op1/op2;
    else if(opr == '*') return op1*op2;
    else if(opr == '+') return op1+op2;
    else if(opr == '-') return op1-op2;
    else {
        return -1;
    }
}

bool is_operator(char C) {
    if(C == '/' or '*' or '+' or '-') {
        return true;
    }
    else 
        return false;
}

bool is_operand(char C) {
    if(C>='0' && C<='9') {
        return true;
    }
    else 
        return false;
}

int main() {
    cout << evaluatePostfix("12 38 + 5 *"); // The input should be in postfix form
}

evaluate expression to postfix() method 12 38 + 5 * with algorithm.
solution?

@Malikhaidar

@akshittrivedi007
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Why -'0' in line 64

Because of ASCII value , reference with 0

@soumyabag
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Hey can you leave the C code for it please?

#include<stdio.h>
#include<string.h>
#define MAX 30
struct stack
{
char str[MAX];
int top;
}s;

int checkoperand(char ch)
{
if(ch>='0' && ch<='9')
return 1;
return 0;
}

int operator(char ch)
{
if(ch=='*' || ch=='+' || ch=='/' || ch=='-')
return 1;
return 0;
}

void push(int data)
{
s.top++;
s.str[s.top]=data;
}

char peek()
{
return s.str[s.top];
}

void pop()
{
s.top--;
}

int operation(int x, int y, char op)
{
if(op == '+')
return y+x;
else if(op == '-')
return y-x;
else if(op == '')
return y
x;
else if(op == '/')
return y/x;
else
return 0;
}

void evaluation()
{
int i,x,y;
char postfix[MAX];
printf("Enter postfix: ");
scanf("%s",&postfix);
for(i=0;i<strlen(postfix);)
{
if(checkoperand(postfix[i]))
push(postfix[i++]-'0');
else if(operator(postfix[i]))
{
x=peek();
pop();
y=peek();
pop();
push(operation(x,y,postfix[i]));
i++;
}
}
printf("%d",s.str[s.top]);
}

int main()
{
s.top=-1;
evaluation();
return 0;
}

@Ishikanagar07
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Ishikanagar07 commented May 26, 2021

I made minor changes in this code
It works fine now
Segementation error removed

/*
Evaluation Of postfix Expression in C++
Input Postfix expression must be in a desired format.
Operands must be single-digit integers and there should be space in between two operands.
Only '+' , '-' , '*' and '/' operators are expected.
*/

#include
#include
#include

using namespace std;

// Function to evaluate Postfix expression and return output
int EvaluatePostfix(string expression);

// Function to perform an operation and return output.
int PerformOperation(char operation, int operand1, int operand2);

// Function to verify whether a character is operator symbol or not.
bool IsOperator(char C);

// Function to verify whether a character is numeric digit.
bool IsNumericDigit(char C);

int main()
{
string expression;
cout<<"Enter Postfix Expression \n";
getline(cin,expression);
int result = EvaluatePostfix(expression);
cout<<"Output = "<<result<<"\n";
}

// Function to evaluate Postfix expression and return output
int EvaluatePostfix(string expression)
{
// Declaring a Stack from Standard template library in C++.
stack S;

for(int i = 0;i< expression.length();i++) {

	// Scanning each character from left. 
	// If character is a delimitter, move on. 
	if(expression[i] == ' ' || expression[i] == ',') continue; 

	// If character is operator, pop two elements from stack, perform operation and push the result back. 
	else if(IsOperator(expression[i])) {
		// Pop two operands. 
		int operand2 = S.top(); S.pop();
		int operand1 = S.top(); S.pop();
		// Perform operation
		int result = PerformOperation(expression[i], operand1, operand2);
		//Push back result of operation on stack. 
		S.push(result);
	}


           else if(IsNumericDigit(expression[i])) {		    
                    int operand;		    
                    operand = expression[i] - '0';	
			
                   // Push operand on stack. 			
                   S.push(operand);		
             }

}
// If expression is in correct format, Stack will finally have one element. This will be the output. 
return S.top();

}

// Function to verify whether a character is numeric digit.
bool IsNumericDigit(char C)
{
if(C >= '0' && C <= '9') return true;
return false;
}

// Function to verify whether a character is operator symbol or not.
bool IsOperator(char C)
{
if(C == '+' || C == '-' || C == '*' || C == '/')
return true;

return false;

}

// Function to perform an operation and return output.
int PerformOperation(char operation, int operand1, int operand2)
{
if(operation == '+') return operand1 +operand2;
else if(operation == '-') return operand1 - operand2;
else if(operation == '*') return operand1 * operand2;
else if(operation == '/') return operand1 / operand2;

else cout<<"Unexpected Error \n";
return -1; 

}

// here we are only considering single-digit integers, because if we start considering integers > 9 then 2 single-digit integer might get interpreted as 1 two-digit integer and we will get segmentaion error (because PerformOperation function is expecting two integers in stack whereas only 1 is found)

@vk7024
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vk7024 commented Jun 4, 2021

Why -'0' in line 64
为了把字符转换成数字

@Ishikanagar07
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char in C++ evaluates to the decimal equivalent.
'0' is 48
'1' is 49
'2' is 50
.
.
.
if expression[i] = '2' which evaluates to 50
so (expression[i] - '0') in this case will be (50 - 48) = 2

@nofrie
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nofrie commented Aug 26, 2022

The reason segmentation fault happens is that you don't put space between each operand.
For example, if you type: "11+", it's buggy. It must be "1 1+" or "1 1 +".
By applying this logic, it can count up multi digits numbers.

@saku123hh
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Here is my implemention, I have fixed the while loop error:
#include
using namespace std;
#include
#include

bool IsOperator(char x) {
if (x == '+' || x == '-' || x == '*') {
return true;
}
return false;
}

bool IsOperand(char x) {
// cout<<"x"<<x<<endl;
if (x <= '9' && x >= '0') return true;
return false;

}

int PerformOperation(char c, int x, int y) {
if (c == '+') return(x + y);
else if (c == '-') return(x - y);
else if (c == '*') return(x * y);
else cout << "unexpected error" << endl;
return -1;
}

void Print(stack S) {
while (!S.empty()) {
cout << S.top() << endl;
S.pop();
}
cout << "" << endl;
}

int EvaluatePostfix(string expression) {
stack S;

for (int i = 0; i < expression.length(); i++) {
    // cout<<IsOperator(expression[i])<<endl;
   //  cout<<IsOperand(expression[i])<<endl;
    if (expression[i] == ' ' || expression[i] == ',') continue;
    else if (IsOperator(expression[i])) {
        int operand2 = S.top(); S.pop();
        if (S.empty()) { cout << "Error: check your postfix" << endl; return -1; }
        int operand1 = S.top(); S.pop();
        int result = PerformOperation(expression[i], operand1, operand2);
        S.push(result);
        cout << "进入operatorS is:" << endl;
        Print(S);
    }

    else if (IsOperand(expression[i])) {
        int operand = 0;
        while (IsOperand(expression[i]) && i < expression.length()) {
            operand = operand * 10 + (expression[i] - '0');
            i++;
            cout << i << endl;
        }
        i--;  //比如35 6+ 在遇到5后i++=2 在这次程序结束后还要自动+1 会跳过对i=2的判断  所以需要-1
        S.push(operand);
        cout << "进入operand S is:" << endl;
        Print(S);
    }


    //  cout << i << endl;
}
return S.top();

}

int main()
{
cout << "input postfix" << endl;
string expression;
// cin>>expression;
getline(cin, expression);
cout << "expression:" << expression << endl;
int result = EvaluatePostfix(expression);
cout << "result is " << result << endl;

}

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