HW Assignment - Checking Brackets using Stack

[YoTaMiX]

Hello to you all ,

I am required to write a program which gets a Math Expression (without checking it is good) and put in Stack only the Brackets .
Implementation of Stack is One-Way Linked List Only.

Examples for Good/Bad Expressions :
(a*{b+c}-4/x +[e-5]) - GOOD
(5+} - BAD
(5+{6*)-2} - BAD
(5+z - BAD

I have build up a Mechanism which checks for each entered char (other than Enter Key and also using closing brackets such as ) or ] or } at the beginning of Expression) what type of bracket it is and pushes it into a stack (we must use LIFO technique for check legal Expression).

I thought of 2 cases :

first case : for each open bracket there is a closing one so i check top of stack to bottom of stack and carry on till i reach to middle .
second case : what happens if i have this sequence - ()[]{} - meaning i need to check top against the next , and so on (top of stack to bottom of stack will not help in this case)

I kinda stuck on Algorithm Think Phase :-)

Code Added . Any Ideas ?

Thank you , Yotam , Israel

[dude543]

The logic should be like this :
   while ( input != 13 )
      {
       if input is '(' or '{' or '[' or ')' or '}' or ']' 
          {
          if input is '(' or '{' or '['
              push it into the stack.
          else if input is  ')' or '}' or ']' 
              {
              if stack is empty
                       report error.
              pop cell from stack
              if cell "matches input" 
                       we are ok.
              else if cell dont "matches input" or stack is empty 
                       report error.
              }
          }
      }
   if  the stack is not empty
        report error.

Also I would suggest replacing your input function
with fgets.
A nice touch would be to add to the switch statment
a default case, checking if it is a number of math experssion.
Don't ever write code like this :

free(top);
    top=top->next;

Once you free(p), dont access it !!!!!!!!!!
It will probaly work, but its wrong, and
also your grade will be like it.

[YoTaMiX]

The logic should be like this :
   while ( input != 13 )
      {
       if input is '(' or '{' or '[' or ')' or '}' or ']' 
          {
          if input is '(' or '{' or '['
              push it into the stack.
          else if input is  ')' or '}' or ']' 
              {
              if stack is empty
                       report error.
              pop cell from stack
              if cell "matches input" 
                       we are ok.
              else if cell dont "matches input" or stack is empty 
                       report error.
              }
          }
      }
   if  the stack is not empty
        report error.

Also I would suggest replacing your input function with fgets. A nice touch would be to add to the switch statment a default case, checking if it is a number of math experssion. Don't ever write code like this :

free(top);
    top=top->next;

Once you free(p), dont access it !!!!!!!!!! It will probaly work, but its wrong, and also your grade will be like it.

I see your point :-)
i build something , but for some reason , it will state all Expression are OK and i tried to Debug it , and i cant figure what went wrong... Code Added . The check is though ASCII code :
'(' - ')' = 1
'['- ']' = 2
'{' - '}' = 2

Thanx

[dude543]

I already told you. Dont do this :

free(top);
    top=top->next;

Its very wrong !!!!!
And still you do it again !!!!
:mad:

First thing better change input by getche to fgets.


Do you think that this is nessary ???
if ( (temp[i]==')') || (temp[i]==']') || (temp[i]=='}') )
{ printf("\n Bad Expression! Run it Again... \n");



add to

case ')': { flag=check(temp[i]);  }
       case ']': { flag=check(temp[i]);  }
       case '}': { flag=check(temp[i]);  }

break statemnts!

If the stack is not empty thats wrong

if ( top )
   {
   printf("Items on the stack\n");
   flag = 0;
   }

You are ignoring the case which pop returns zero.


last thing
'(' - ')' == -1

[YoTaMiX]

I already told you. Dont do this :

free(top);
    top=top->next;

Its very wrong !!!!! And still you do it again !!!! :mad:

First thing better change input by getche to fgets. Do you think that this is nessary ??? if ( (temp[i]==')') || (temp[i]==']') || (temp[i]=='}') ) { printf("\n Bad Expression! Run it Again... \n"); add to

case ')': { flag=check(temp[i]);  }
       case ']': { flag=check(temp[i]);  }
       case '}': { flag=check(temp[i]);  }

break statemnts!

If the stack is not empty thats wrong

if ( top )
   {
   printf("Items on the stack\n");
   flag = 0;
   }

You are ignoring the case which pop returns zero. last thing '(' - ')' == -1

A few things :
1. how do you Take off the Top each time if not in that way?
2. Added Break Statements , still causes problems in Expression check ,
each EXP check gets OK ....
3. the If sentence - no Math Exp. starts with backwoards brackets. :-)
4. should i pass into the POP function a Pointer to FLAG ?

Thanx

[hijab_89]

The proramme was good..n i think output will be perfect..but i am in 3rd semester...it is a little bitdifficult for me to understand..is there any simpler code for this bracket checker.....?????
Hello to you all ,

I am required to write a program which gets a Math Expression (without checking it is good) and put in Stack only the Brackets .
Implementation of Stack is One-Way Linked List Only.

Examples for Good/Bad Expressions :
(a*{b+c}-4/x +[e-5]) - GOOD
(5+} - BAD
(5+{6*)-2} - BAD
(5+z - BAD

I have build up a Mechanism which checks for each entered char (other than Enter Key and also using closing brackets such as ) or ] or } at the beginning of Expression) what type of bracket it is and pushes it into a stack (we must use LIFO technique for check legal Expression).

I thought of 2 cases :

first case : for each open bracket there is a closing one so i check top of stack to bottom of stack and carry on till i reach to middle .
second case : what happens if i have this sequence - ()[]{} - meaning i need to check top against the next , and so on (top of stack to bottom of stack will not help in this case)

I kinda stuck on Algorithm Think Phase :-)

Code Added . Any Ideas ?

Thank you , Yotam , Israel

[Berkut]

write a c++ program that reads from the user a mathematical expression like:
(3+(4-9) * (3/4) * (3+2))

1. your program should check if the expression is correct
1.1 An expression is correct if:
the parantheses are correctly added and are balanced
examples of not correct expressions can be:
)9+8(
(((8-1)

etc

Assume that the numbers are only one digit ( you can extend your program to read more than one digit numbers as a bonus)

2. If the expression is correct, your program should evaluate the expression and display the result.

hint ( use stacks ( certain types of linked lists ) to solve the problem.