Hello, I am having trouble with the following code that causes the program to crash. I tried debugging and I think the problem is that the stack is empty somewhere and the operations for pop/ top are performed, however I don't know where. I am using visual studios. Please help me find the problem so that the code works and evaluates the postfix expression, Thank you for your help in advance

This is displayed:

The code is below:

For main:

#include

#include

#include

#include

#include

#include

#include "EvalBool.h"

using namespace std;

//Function to return precedence of operators

int prec(char);

/*

this function takes in an Infix string expression and turns it into a postfix

expression, it returns the postfix string expression

*/

string InfixToPostfix(string);

// Generates a random boolean expression of length int

string rand_gen_logExp(int);

// This function should receive the string passed from the Postfix function and evaluate the expression, returning true or false

char EvaluateBooleanPostFixExpression(string);

// Operands are T or F, they will be evaluated

bool IsOperand(char);

// The Operators perform the evaluation

bool IsOperator(char);

int main()

{

string expression = "T+!(T+T)*(T)*(!F)*T*T+!T+!T+F+T*!!F*!F*(F+!F*F)*F+F";

string postfix = InfixToPostfix(expression);

cout

cout

cout

return 0;

}

For the header file:

#include

#include

#include

#include

using namespace std;

int prec(char);

string InfixToPostfix(string);

string rand_gen_logExp(int);

char EvaluateBooleanPostFixExpression(string);

bool IsOperand(char);

bool IsOperator(char);

//Function to return precedence of operators

int prec(char c)

{

if (c == '!')

return 3;

else if (c == '*')

return 2;

else if (c == '+')

return 1;

else

return -1;

}

/*

this function takes in an Infix string expression and turns it into a postfix

expression, it returns the postfix string expression

*/

string InfixToPostfix(string Expression)

{

// initialize stack S

stackS;

// this will serve as the first element of the stack so it wont be empty

S.push('L');

// initialize output string Tokens to be empty

string Tokens;

int l = Expression.length();

// for i = 0 to n do the following cases (there is already an element in the first stack so it is for i =1 to jump it )

for (int i = 0; i

{

// process the array of Expression[1..n]

// If the scanned character is an operand, add it to output string, Tokens.

// all operands need to be copied to Tokens first

if ((Expression[i] == 'T') || (Expression[i] == 'F'))

Tokens += Expression[i];

// If the scanned character is an (, push it to the stack.*** it wont be popped until the closing parenthese

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

S.push('(');

// If the scanned character is an ), pop both ()... eventually, but first all the operators need to be cppied into the output string

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

{

//push all operators of the string to to the stack until an ( is encountered.

while (S.top() != 'L' && S.top() != '(')

{

// is it a (? No? then pop the operator

char c = S.top();

S.pop();

//and put it in Tokens

Tokens += c;

}

// when the parentheses close then pop them from the stack

if (S.top() == '(')

{

char c = S.top();

S.pop();

}

}

else {

// pop out the operators from the stack according to their precedence, greatest to least

// this is case II in the pseudocode from the powerpoint slide 32

while (S.top() != 'L' && prec(Expression[i])

{

char c = S.top();

S.pop();

// The operators popped are according to their precedence, add the operators popped into the ouput string

Tokens += c;

}

S.push(Expression[i]);

}

}

// at the end of the input, Pop all the remaining symbols from the stack to the end of the output

while (S.top() != 'L')

{

char c = S.top();

S.pop();

Tokens += c;

}

cout

return Tokens;

}

// This function should receive the string passed from the Postfix function and evaluate the expression, returning true or false

char EvaluateBooleanPostFixExpression(string x)

{

// length of expression

int n = x.length();

char Op1, Op2, symbol;

// initialize a stack to be empty

stackY;

// for i = 0 to length n of the Expression do

for (int i = 0; i

/// if character is an operand(T or F) then push it into the stack as it is scanned

if (IsOperand(x[i]))

{

// push character in the stack

Y.push(x[i]);

}

/*else if the character is an operator, pop two items n1 and n2 from the

stack, calculate the op of n1 and n2, and then push the result into the stack*/

else if (IsOperator(x[i]))

{

switch (x[i])

{

// case when character is NOT then pop one element from stack and push it's negation (i.e if !T, then push F into the stack)

case '!':

// stacks are last in first out, so Operator1 is inserted in the end of the stack

// top then pushes the operator into the last element of the stack

Op1 = Y.top();

// pop the item

Y.pop();

// evaluate the NOT operation,k

if (Op1 == 'T')

Y.push('F');

// and then push the result into the stack

else

Y.push('T');

break;

// case for AND when * is encountered

case '*':

// pop n1 and n2 from the stack

Op1 = Y.top();

Y.pop();

Op2 = Y.top();

Y.pop();

// evaluate using the Truth operation for AND, and then push the result into the stack

if (Op1 == 'T' && Op2 == 'T')

Y.push('T');

else

Y.push('F');

break;

// Case for OR when + is encountered

case '+':

// pop n1 and n2 from the stack

Op1 = Y.top();

Y.pop();

Op2 = Y.top();

Y.pop();

// evaluate the OR Truth operation on the popped elements, then push the result back into the stack

if (Op1 == 'F' && Op2 == 'F')

Y.push('F');

else

Y.push('T');

break;

// if there are no sufficient operands report error and exit

default: exit(-1); break; // incase of error

}

}

}

// output the only remaining item in the stack, the answer to the expression

return Y.top();

}

// Operands are T or F, they will be evaluated

bool IsOperand(char Operand)

{

if(Operand == 'T' && Operand == 'F')

return true;

else

return false;

}

// The Operators perform the evaluation

bool IsOperator(char Operator)

{

if(Operator == '!' && Operator == '*' && Operator =='+')

return true;

else

return false;

}

string rand_gen_logExp(int n) {

if (n

return "";

//possible symbols in the expression

string symbols = "FT()!*+";

//possible symbols after each current symbol

string succ[] = { "*+)","*+)", "!FT", "*+)", "FT(!", "FT(!", "FT(!" };

//Ouput string initialized;

string logical_exp = "";

/umber of left apparantheses generated so far

int num_of_lp = 0;

//initializz first symbol of the output

char first_symbol = ' ';

//randomly generate first symbol in the output

//seed for rand

srand(time(0));

switch (rand() % 4)

{

case 0: first_symbol = 'F'; break;

case 1: first_symbol = 'T'; break;

case 2: first_symbol = '!'; break;

case 3: first_symbol = '('; num_of_lp++;

break;

}

logical_exp += first_symbol;

// The last symbol generated

char last_symbol = first_symbol;

// The main loop that generates the rest of the expression

for (int i = 1; i

{

//string s = succ[symbols.indexOf(last_symbol)];

string s = succ[symbols.find(last_symbol)];

char next_symbol;

bool again;

// genrating the next symbol in the expression and

// make sure that it conforms to format of loogical expressions

do

{

again = false;

next_symbol = s.at(rand() % (s.length()));

switch (next_symbol)

{

case ')': // All parantheses seen so far matched

// ) is not allowed for next symbol

if (num_of_lp == 0)

{

again = true;

}

else

num_of_lp--;

break;

case '(': num_of_lp++;

break;

}

} while (again);

logical_exp += next_symbol;

last_symbol = next_symbol;

}

// check if the last symbol is an operator and hence need an operand

if ((last_symbol == '(' || last_symbol == '*' || last_symbol == '+' || last_symbol == '!'))

logical_exp += rand() % 2 == 0 ? 'F' : 'T';

// adding necessary closing )'s

for (int i = 0; i

logical_exp += ')';

return logical_exp;

}

C:AWINDOWS1system32\cmd.exe Input T+!(T+T)*(T) (F)*TT+F+T*F*FF)*F+F Microsoft Visual C+Runtime Library Debug Assertion Failed! Program: CAWINDOWSLSYSTEM32 MSVCP140D.dll File: cprogram files (x86) microsoft visual studio 2017 communityvctools\msvc 14.14.26428 includelde que Line: 321 Expression: deque iterator not dereferencable For information on how your program can cause an assertion failure, see the Visual C++ documentation on asserts. (Press Retry to debug the application) Abort Ignore

<>