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The following is code for a class assignment. I get the Fully Parenthesised Expression in string FPE with no problems. The second section of assignment is to utilize the following algorithm to evaluate the expression.

1. if it is an operand, push on to the stack.

2. if it is an operator, pop two values from the stack and apply operator and push the value.

I don't see how it is possible to apply the operator which is a character, to the 2 operands that I have popped off the stack. Any feedback is greatly appreciated.

#include <iostream.h>

#include <string.h>

#include "A:\class\bstack.h"

#include <ctype.h>

typedef char string[40];

int main()

{

string expression, FPE;

unsigned i = 0, j = 0;

int x = 0, y = 0, z = 0;

char temp;

cout << "Please enter a Fully Parenthesised Expression. \n";

cout << "For example, ((2 + 4 ) * 3). \n\n";

cin.getline(expression, 39);

IntStack stk;

while (i < strlen(expression))

{

if ((expression[i] == '+' ) || (expression[i] == '-' ) || (expression[i] == '/' )||

(expression[i] == '*'))

{

stk.Push(expression[i]);

}

else if (isdigit(expression[i]) != 0 )

{

FPE[j] = expression[i];

++j;

}

else if (expression[i] == ')')

{

expression[i] = stk.Top();

FPE[j] = expression[i];

++j;

stk.Pop();

}

++i;

}

cout << " \nThe Fully Parenthisised Expression is: " << FPE << "\n\n" ;

j = 0;

while (j < strlen(FPE)) // FPE is my Fully Parenthesised Expression

{

if(isdigit(FPE[j]) != 0)

{

stk.Push(FPE[j]); // pushing the operands here

}

else

{

x = char (stk.Top());

cout << x << endl;

stk.Pop();

y = char(stk.Top());

cout << y << endl;

stk.Pop();

temp = (FPE[j]);

z = x temp y;

cout << z << '\n';

cout << temp;

stk.Push(z);

}

++j;

}

return 0;

}

1. if it is an operand, push on to the stack.

2. if it is an operator, pop two values from the stack and apply operator and push the value.

I don't see how it is possible to apply the operator which is a character, to the 2 operands that I have popped off the stack. Any feedback is greatly appreciated.

#include <iostream.h>

#include <string.h>

#include "A:\class\bstack.h"

#include <ctype.h>

typedef char string[40];

int main()

{

string expression, FPE;

unsigned i = 0, j = 0;

int x = 0, y = 0, z = 0;

char temp;

cout << "Please enter a Fully Parenthesised Expression. \n";

cout << "For example, ((2 + 4 ) * 3). \n\n";

cin.getline(expression, 39);

IntStack stk;

while (i < strlen(expression))

{

if ((expression[i] == '+' ) || (expression[i] == '-' ) || (expression[i] == '/' )||

(expression[i] == '*'))

{

stk.Push(expression[i]);

}

else if (isdigit(expression[i]) != 0 )

{

FPE[j] = expression[i];

++j;

}

else if (expression[i] == ')')

{

expression[i] = stk.Top();

FPE[j] = expression[i];

++j;

stk.Pop();

}

++i;

}

cout << " \nThe Fully Parenthisised Expression is: " << FPE << "\n\n" ;

j = 0;

while (j < strlen(FPE)) // FPE is my Fully Parenthesised Expression

{

if(isdigit(FPE[j]) != 0)

{

stk.Push(FPE[j]); // pushing the operands here

}

else

{

x = char (stk.Top());

cout << x << endl;

stk.Pop();

y = char(stk.Top());

cout << y << endl;

stk.Pop();

temp = (FPE[j]);

z = x temp y;

cout << z << '\n';

cout << temp;

stk.Push(z);

}

++j;

}

return 0;

}

{

case '+' :

result = op1 + op2;

break;

case '-' :

result = op1 - op2;

break;

case '*' :

result = op1 * op2;

break;

case '/' :

result = op1 / op2;

break;

}

stk.Push(result)

>> 2. if it is an operator, pop two values from the stack and apply

>> operator and push the value

I beleive this is the algorithm for a post fix expresion. Not an infix expression.

An infix expression would have the operator in between the operands, like

4+5

Your algorithm won't work with that expresion. It works with a postfix one, where the operator comes after the two operands. like

4 5 +

I think you are missing a step in the project. You probably are supposed to convert infix to postfix before trying to calculate. Look over your assignment again. If you have questions, let me know.

(2 + 3 ) //user entered expression

result = 101

which are the ascii conversions, anyway you know to accomplish conversion so that result is the integer 5?

char Ch5 = '5'

int Int5 = Ch5 - '0'; // This integer will be 5.

You could apply that logic to your program. That would work in your current program and in the examples you tested. However your program (and that method) won't work for numbers that are multiple characters. I assume you need to be able to handle expressions like

123 + 56

right? Your current logic won't handle that at all. when you come to a digit in your processing, you probably should use atoi() to convert the entire number (maybe several digits) to an itneger and then work with the integer. If you need to convert back to a string again, you can use

itoa() to convert back.

let me know iof this helps. If you need to send me another co;y make sure to send your bstack files.. Taht way I can compile and test your code.

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>>character, to the 2 operands that I have popped off the stack.

Use a switch statement. Details follow.