# If you want to start coding today is not a bad day. Here’s A quick code for begginners.

For a quick fact: Coding might have  a future unless a giant meteor strikes earth down and send us all in heaven or hell maybe and atheists idk man they’ll also end up somewhere but that’s not the point right. So if you want to learn coding what should you do?  Practice you dummy. So for your practice I have got a good code for you. Yes basic calculators are easy and boring to make also if you make something cool you can just flex on your friends. So either read the code try it out or just copy paste it to flex on your friends. You can see the code below:

 #!/usr/local/env python3 # bodmas_calc.py – A rule-based calculator that follows the rules of BODMAS # 3+((4+6)*9/(2-16+8*(3^2+7)))/3*6*4-7 – copy and paste this if you can’t think of a calculation yourself # 7^3+3*(4*5-2^2-28/7+3)+9+((3+5)/(3-1)) – another test equation # ((3+(2*2)-7)^2)+(((2*2)+3-5)^2) def solve(calculation): operator_precedence = {“^”: 4, “/”: 3, “*”: 2, “+”: 1, “-“: 1} # bodmas_index will store the index of all the operators in the equation # The order of the list is the order in which the calculation will be performed bodmas_index = [] for i in range(len(calculation)): if calculation[i] in operator_precedence: if len(bodmas_index) == 0: # starts off the bodmas_index list with a value (this is only executed once) bodmas_index.append(i) else: # Loops through the current indexes of operators and works out the order in which the operators need to be # arranged by comparing the operator_precedence values and seeing if a particular operator needs to be worked on # before or after another one (e.g all * operations will need to be performed before all – operations) for x in range(len(bodmas_index)): if operator_precedence[calculation[i]] < operator_precedence[calculation[bodmas_index[–1]]]: bodmas_index.append(i) break elif operator_precedence[calculation[i]] > operator_precedence[calculation[bodmas_index[x]]]: bodmas_index.insert(x, i) break elif operator_precedence[calculation[i]] == operator_precedence[calculation[bodmas_index[x]]]: if calculation[i] == “+” or calculation[i] == “-“: bodmas_index.append(i) break else: bodmas_index.insert(x + 1, i) break else: continue while len(bodmas_index) != 0: # Loops through the operator indexes from left to right and performs the required operation, storing the result # in calculation_result if calculation[bodmas_index] == ‘^’: calculation_result = calculation[bodmas_index – 1] ** calculation[bodmas_index + 1] elif calculation[bodmas_index] == ‘/’: calculation_result = calculation[bodmas_index – 1] / calculation[bodmas_index + 1] elif calculation[bodmas_index] == ‘*’: calculation_result = calculation[bodmas_index – 1] * calculation[bodmas_index + 1] elif calculation[bodmas_index] == ‘+’: calculation_result = calculation[bodmas_index – 1] + calculation[bodmas_index + 1] else: calculation_result = calculation[bodmas_index – 1] – calculation[bodmas_index + 1] # calculation_result stores the result which is then inserted into the equation in place of the # two values and operator that was used to calculate it. calculation[bodmas_index–1] = calculation_result calculation.pop(bodmas_index+1) calculation.pop(bodmas_index) # Any operator indexes that are higher than the index stored at bodmas_index will need to have their index position shifted by -2 # to accommodate for the shortening calculation. It’s been about 2 years since I wrote this code and I never commented on its behaviour, # so here I am 2 years later trying to remember why I chose to do the loop this way. print(bodmas_index) for i in range(len(bodmas_index)): if bodmas_index[i] > bodmas_index: bodmas_index.insert(i, bodmas_index[i] – 2) bodmas_index.pop(i + 1) bodmas_index.pop(0) return calculation def bracket_pair_finder(calculation): # Pairs brackets together, so that the program knows which calculations to work out first start_bracket_index_array = [] end_bracket_index_array = [] bracket_pairs = {} for i in range(len(calculation)): if calculation[i] == ‘(‘: start_bracket_index_array.append(i) elif calculation[i] == ‘)’: end_bracket_index_array.append(i) # Finds the innermost brackets so that they can be solved first # Only returns one pair of brackets at a time. # If all of them were returned, they could potentially be solved recursively, but I haven’t tried it for i in range(len(start_bracket_index_array) – 1, –1, –1): for x in range(len(end_bracket_index_array)): if end_bracket_index_array[x] < start_bracket_index_array[i] or end_bracket_index_array[x] in bracket_pairs.values(): continue else: bracket_pairs[start_bracket_index_array[i]] = end_bracket_index_array[x] break break if len(bracket_pairs) != 0: return bracket_pairs def calculator(calculation): brackets = bracket_pair_finder(calculation) answer = [] if brackets is None: return float(solve(calculation)) else: start_bracket_index = list(brackets.keys()) end_bracket_index = brackets[start_bracket_index] answer.append(solve(calculation[start_bracket_index + 1:end_bracket_index])) calculation = calculation[:start_bracket_index] + answer + calculation[end_bracket_index + 1:] print(calculation) return calculator(calculation) def calc_input(): # Splits the calculation into an array of integers and BODMAS operators calculation_array = [] calculation = input(‘Enter your calculation: ‘) number = “” for i in range(len(calculation)): # If the current value is an int if calculation[i].isnumeric() or calculation[i] == “.”: # Because the equation is stored as a string, the numbers can be # appended to (e.g. the string ’34’ can have ‘6’ appended to it, so it becomes ‘346’) number = number + calculation[i] # If the current iteration value is the last in the calculation, it can simply be appended, # as there will be nothing left to append to the array after if i == len(calculation)–1: calculation_array.append(float(number)) else: #If the value being stored is not an int, it will be a BODMAS operator # If there is no number currently being held in number, then the operator can be appended to the array if number == “”: calculation_array.append(calculation[i]) # Else, if there is a number currently being held in number, append it to the array and then append the operator after it # Then reset the number variable to be empty, ready for the next lot of numbers to be stored in it. else: calculation_array.append(float(number)) calculation_array.append(calculation[i]) number = “” return calculator(calculation_array) answer = calc_input() print(answer)

Yup that’s the code easy right. that’s it it doesn’t have numbers for lines cus the creator of this site doesn’t have time for adding those and coding to the article topics cus he’s just a dummy but i am sure he’ll soon till then hang on cus more codes will be incoming as I figure them out.  Bye Bye my Ducks

## 4 thoughts on “If you want to start coding today is not a bad day. Here’s A quick code for begginners.”

1. Aditya Gaurav says:

you should have used preformatted instead of table to give proper indentation to your code, you people can get the chronological order easily.

1. Dojo says:

gandu vo sab badme tu phele ye bata mera post image kyu change kiya Yeah Nah vala image accha tha yaar. thats straight cruality mai avaj uthaunga. uongaa boonga

1. Aditya Gaurav says:

You should use appropriate images for posts, images relating or substituted in place of some adultery or shit like you’ve done are no matter what, they will be taken out and replaced appropriately. Also remembers as mentioned in terms of service, frequently reported users are automatically blocked on server so take care when you write and speak, in a way that people won’t report you.

1. Henry Williams says: