I wrote the stack in Python

What is a stack in the first place? Yes, store and retrieve data. That's all. However, I want to be able to retrieve the last stored data first. Σ(oдΟ;)

For example, as shown below, while stacking 6 data vertically Suppose you have a box that you can store.

First, empty the box, store A, and then remove it. After storing and retrieving A, it returned to its original empty state. It's easy.

Now, after naming the data storage as Push and the retrieval as Pop, I tried to figure out the movement of Push / Pop of the data a little more. I pushed A and B, and finally popped. The last thing left in the box is A, isn't it? In this way, you can now imagine the behavior of the stack that pops the last Pushed data first.

I wish I didn't have to have a box depth of 6 !? I don't care about the details (laughs). Now let's write in Python.

Before that, if you don't understand the description here, We recommend studying with Progate !! https://prog-8.com/

Let's first prepare a box to realize the stack. The image looks like this. I prepared a box named str and set the capacity as how many data can be stored. ptr represents the position from 1 to 6. I added the movement of ptr to the previous figure. Like this, ptr (pointer) shows the position of the box that can be pushed. Now that you have an image of ptr, it's finally Python. I made a box.

stack.py

    def __init__(self,capacity:int = 10):
        self.str  = [None] * capacity
        self.capa = capacity
        self.ptr  = 0

"def __ init __" Determine the initial value in the prepared program! It is a cliché. The story doesn't start unless you load it first when the program runs. Next is Push.

stack.py

    def push(self,value):
        if self.ptr >= self.capa:
            raise top.full
        self.str[self.ptr] = value 
        print(self.str[self.ptr])
        self.ptr += 1

There is a print in between, but you can ignore it. The important thing is to first make sure that the ptr position is not Full. Full if ptr == capacity If ptr> capacity, the program is out of control !! You can see that we are checking with if self.ptr> = self.capa :. If it is not Full, set self.str [self.ptr] = value and Push is completed successfully. Increment ptr by one in preparation for the next Push. That's all we're doing with Push.

Next is Pop.

stack.py

    def pop(self):
        if self.ptr <= 0:
            raise top.empty
        self.ptr -= 1
        return self.str[self.ptr]

If the pointer ptr was negative, Because you will penetrate the bottom and screw the data into the ground For the time being, check with the if statement. I'm sure some of you may have wondered here. As a side note, in the Python world, storage doesn't start at 1. It starts from 0. I'm sorry the first figure wasn't good. So if ptr = 0, it means empty. It's Empty. If it is not empty, decrement ptr by 1 as shown below. Specify the data to retrieve. After that, you can retrieve the stored data by doing return self.str [self.ptr].

No, I was able to return the data with return, The specified area is not empty! ?? Yes! That's right !! Bakon! (----) / ☆ (* _ *)

Actually, the stack manages the data at the position of ptr. Even if you haven't emptied it, the next Push will automatically rewrite it, right? That's why you can stack by just looking at the movement of ptr. That's why I will put the whole picture of the stack prepared in a hurry.

stack.py

class top:
    class full(Exception):
        pass
    class empty(Exception):
        pass    
    
    def __init__(self,capacity:int = 10):
        self.str  = [None] * capacity
        self.capa = capacity
        self.ptr  = 0
        
    def push(self,value):
        if self.ptr >= self.capa:
            raise top.full
        self.str[self.ptr] = value 
        print(self.str[self.ptr])
        self.ptr += 1
        
    def pop(self):
        if self.ptr <= 0:
            raise top.empty
        self.ptr -= 1
        return self.str[self.ptr]

test = top()

while True:
    num = int(input("select 1.push , 2.pop : "))
    
    if num == 1 :
        s = int(input("enter data: "))
        try:
            test.push(s)
        except test.full:
            print("full!")
    elif num == 2:
        try:
            x = test.pop()
            print(x)
        except test.empty:
            print("empty!")
    else:
        break

Maybe "__ init __", push, pop is the key, so if you know that, I hope you understand.

This is my first post, so if there is something missing or difficult to understand Please kindly point out !! m (_ _) m