In postfix notation:

An operand $($number$)$ appears before its operator.

Operators act on the preceding operands immediately when

encountered.

There is no need for parentheses, as the order of operations is

determined by the position of the operators relative to the

operands.

For example:

In infix notation $($conventional notation$),$ we might write $3+4.$

In postfix notation, this expression is written as $34+.$

How Postfix Evaluation Works

Postfix evaluation uses a stack to store operands and manage intermediate

results as it processes each part of the expression. Here's a step$-$by$-$

step explanation:

Read the Expression Left to Right:

The postfix expression is processed from left to right, token

by token.

Each token is either an operand $($number$)$ or an operator $($e$.$g$.,$

$+,-,*,/).$

Operands Are Pushed Onto the Stack:

When an operand is encountered, it is pushed onto the stack.

This allows it to be stored temporarily until an operator

needs to act on it$.$

Operators Act on Stack Values:

When an operator is encountered, the calculator pops two

operands from the stack:

The first operand popped is the right operand.

The second operand popped is the left operand.

The operator then performs its action on these two operands.

The result of the operation is then pushed back onto the stack

as an intermediate result.

Repeat Until the End:

Continue reading tokens, pushing operands, and applying

operators until you reach the end of the expression.

Final Result:

At the end of the expression, the stack should contain only

one value: the final result.

This value is popped from the stack to obtain the final

answer.

Example of Postfix Evaluation

Consider the postfix expression: $53+2*$

Let's evaluate it step by step:

Token $5$: It's an operand, so push $5$ onto the stack.

Stack: $[5]$

Token $3$: It's an operand, so push $3$ onto the stack.

Stack: $[5,3]$

Token $+$: It's an operator, so pop the top two values from the stack

$(3$ and $5).$

Operation: $5+3=8$

Push the result $(8)$ back onto the stack.

Stack: $[8]$

Token $2$: It's an operand, so push $2$ onto the stack.

Stack: $[8,2]$

Token $*$: It's an operator, so pop the top two values from the stack

$(2$ and $8).$

Operation: $8*2=16$

Push the result $(16)$ back onto the stack.

Stack: $[16]$

End of Expression: The stack contains a single value, $16,$ which is

the final result.

Advantages of Postfix Notation

No Parentheses Needed: Postfix notation does not require

parentheses, making expressions simpler to write and evaluate.

Simplified Computation with a Stack: The LIFO structure of a stack

naturally handles the order of operations without needing to

consider precedence rules or parentheses.

Q$1)$ Write an assembly code using floating point unit to compute the value

expression

Y$=6\mathrm{.}0**(2\mathrm{.}0+3\mathrm{.}0)-4\mathrm{.}0$

You must convert the above expression to postfix notation using pencil

and paper and then use that expression, store the same in a data structure

and use FPU and write an assembly program. You don't have to write a

tokenizer.

$10$ Points

Q$2)$ Trace this program and showing the contents of FPU stack after

execution of each Instruction of your code.

$10$ Points

please help, I$\text{'}$m having trouble with this, if you could answer the questions that would be extremely helpful, masm is preferred please