Sample pseudocode for the Task 2: for (each character ch in the infix expression) { switch (ch) // Append operand to end of postfix expression-step 1 case operand: postfixExp = postfixExp • ch break...

Answer fast otherwise I’ll downvote/thumbs down it

Extracted text: Sample pseudocode for the Task 2: for (each character ch in the infix expression) { switch (ch) // Append operand to end of postfix expression-step 1 case operand: postfixExp = postfixExp • ch break // Save '(' on stack-step 2 case '(': operatorStack.push(ch) break case operator: // Process stack operators of greater precedence-step 3 while (!operatorStack.isEmpty() and operatorStack.peek() is not a '(' and precedence(ch) <- precedence(operatorstack.peek()))="" append="" operatorstack.peek()to="" the="" end="" of="" postfixexp="" operatorstack.popo="" operatorstack.push(ch)="" save="" the="" operator="" break="" pop="" stack="" until="" matching="" '('–step="" 4="" case="" '):="" while="" (operatorstack.peek()="" is="" not="" a="" '().="" {="" append="" operatorstack.peek()="" to="" the="" end="" of="" postfixexp="" operatorstack.pop()="" operatorstack.popo="" remove="" the="" open="" parenthesis="" break="" append="" to="" postfixexp="" the="" operators="" remaining="" in="" the="" stack-step="" 5="" while="" (!operatorstack.isempty())="" {="" append="" operatorstack.peek()="" to="" the="" end="" of="" postfixexp="" operatorstack.popo="">