here is the algorithm and a code below in converting infix notation

1) if you encounter "(" push on it on stack

Code:

if(x[i]=='('){stack.push(x[i]);}

2) if you encounter operators (+,-,*,/) then compare the operators with the top of the stack.As long as the top of the stack has operators with higher or equal priority to the scanned operators you have to keep popping the stack and store it in result

* * If the token is an operator, o1, then:

Code:

else if((x[i]=='+')||(x[i]=='-')||(x[i]=='*')||(x[i]=='/'))

* while there is an operator, o2, at the top of the stack, and either

o1 is associative or left-associative and its precedence is less than (lower precedence) or equal to that of o2, or

o1 is right-associative and its precedence is less than (lower precedence) that of o2, (i got confised with the algorithm)

pop o2 off the stack, onto the output queue;

Code:

q=q+stack.pop(); //where q=is a string and stack.pop()=is an o2;

* push o1 onto the stack.

Code:

stack.push(x[i]); //where x[i] = o1

Lastly you have to push the scanned operator on the stack

Code:

stack.push(x[i]); //where x[i] = o1

3)If you encounter any digits or alphabets then store in in result

Code:

else if(isdigit(x[i])){q=q+x[i];}

4)if you encounter ")" keep popping the stack unless you encounter "(" on the stack and store it in result

Code:

else if(x[i]==')'){q=q+stack.pop();}

Lastly decrement top by 1 as "(" is of no importance to us

as i see, that I confused on the scanning of precedence of operators, can you show me clear understanding about the precedence of the operators?