Thread: Equation Verifier! Almost complete!

Code:

void checkSymbols(string infix){
	string acceptedSymbols = "0123456789+-*/^()", 
	       acceptedNumbers = "0123456789", 
		   acceptedOperators = "+-*/^";
	int numberCount = 0, opCount = 0, leftParCount = 0, rightParCount = 0;

	for(unsigned int i = 0; i < infix.length(); i++){
	
		if (acceptedSymbols.find(infix[i], 0) == string::npos)
			outputPostfixError(1);
			
		if (acceptedNumbers.find(infix[i], 0) != string::npos)
			numberCount++;
			
		if (acceptedOperators.find(infix[i], 0) != string::npos)
			opCount++;
			
		if (infix[i] == '(')
			leftParCount++;
			
		if (infix[i] == ')')
			rightParCount++;
	}

	if ((numberCount - 1) != opCount) 
		outputPostfixError(2);
		
	if (leftParCount != rightParCount)
		outputPostfixError(3);
	
} //end void checkSymbols

This is a code to verify if a given equation is correct. Like if I enter "3 + 4 - (6 * 5) / 2" it is correct, but if I do "a + 5" or "a ++ 3)" or "((1 + 2)" they will all give errors, for it must be a solvable equation with no unknown variables.

However, I am pretty sure all these errors have been fixed, but I do not know how to check for incorrect parenthesis direction. Like if the user enters ")1 + 2( - 2" or something to that extent. Parenthesis can be nested so this makes it a bit harder, like this would be accepted: ((1 + 5) / 3)

But of course, I don't want ")(1 + 5( / 3)"

Any way I can solve this?

I had an assignment similar to this!

You can use stack to solve this problem:

PHP Code:


for (every character in the infix)
{

case '(':
push to the stack

case ')':
  if (stack is not empty)
     pop the stack

  else
   return 0;

default:
 // do nothing

}
if (stack is not empty)
 return 0; 


Lazy Student: Thank you! That method worked beautifully!