Calculator final project in Python
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$begingroup$
In my final project for my Python course, I am required to create a calculator program. I really need someone to look over my code and tell me if I have met all the requirements for this project.
Here are the requirements:
For the final project, you are to write a calculator program in Python
with the following characteristics:
- It must be at least 50 lines of code excluding comments
- It must have the following functions:+,-,*,/, SQRT, and work with decimal numbers.
- It must be all text, no high resolution graphics
- It must automatically copy the result to the operating system clipboard
- It must work as a functional calculator (calculations can be entered over and over)
Please submit the source code only (.py), no object code (I will
compile on receipt)
NOTE:
This assignment is graded on a pass/fail basis; there is no partial credit or resubmissions. Be sure you understand ALL of the
requirements.
Solutions that require the user install any software will NOT be accepted (in case you are not sure, this means NO Pyperclip.)
WARNING: All submissions will be checked for plagiarism using
SafeAssign or other plagiarism checker.
I completed the project, but because this is a pass/fail project I want to be 200% sure I have met every requirement. I am also open to any other feedback that would sharpen my Python skills.
Menu
def main():
#Menu
print("=======================")
print("Welcome to Calculator")
print("By: Tyler Harris")
print("=======================")
print("Menu: ")
print("[1] Calculator")
print("[2] Instructions")
print("[3] Exit")
print("=======================")
choice = input("Please select an option: ")
if choice == '1':
calculator()
elif choice == '2':
instructions()
main()
elif choice == '3':
print("Thank you for using Calculator.")
else:
print("Not an option, try again:")
main()
Calculator
def calculator():
result=0 #Resets calulator.
try:
while True:
print("Last Result: "+str(result))
#Number 1 input.
if result == 0:
num1= int(input("First Number: "))
if result != 0:
num1=result
#Operator input.
#Checks for input of special operators 'c', 'sqrt', and 'esc'.
operator=input("Opertaor: ").lower()
if operator == "c":
print("Calculator cleared")
calculator()
if operator=="sqrt":
result=(sqrt(num1))
print("Result: "+str(result))
continue
if operator=="esc":
main()
if operator.isdigit(): #Throw error if operator is a number.
raise Exception()
#Number 2 input.
num2= int(input("Second Number: "))
#Operator calls.
if operator=="+":
result=(add(num1,num2))
elif operator=="-":
result=(sub(num1,num2))
elif operator=="*":
result=(mult(num1,num2))
elif operator=="/":
result=(div(num1,num2))
#Copy result to System's clipboard and display the result.
copy2clip(str(result))
print("=======================")
print("Result: "+str(result))
print("copied to clipboard")
print("=======================")
#Catch any errors and reset calculator
except Exception:
print("=======================")
print("Error Please try again.")
calculator()
else:
calculator()
Operators
def add(num1,num2):
return num1 + num2
def sub(num1,num2):
return num1 - num2
def mult(num1,num2):
return num1 * num2
def div(num1,num2):
return num1 / num2
def sqrt(num1):
return (num1**(1/2.0))
Instructions
def instructions():
print("=======================")
print(format('Calculator','^25s'))
print("=======================")
print("Available operators")
print("+: Addition")
print("-: Subtraction")
print("*: Multiplication")
print("/: Division")
print("sqrt: Square Root")
print("c: Clear Calculator")
print("esc: Exit Calculator")
print("=======================")
def copy2clip(txt):
cmd='echo '+txt.strip()+'|clip'
return subprocess.check_call(cmd, shell=True)
if __name__=='__main__':
main()
python python-3.x homework calculator
$endgroup$
migrated from stackoverflow.com Mar 31 at 0:50
This question came from our site for professional and enthusiast programmers.
add a comment |
$begingroup$
In my final project for my Python course, I am required to create a calculator program. I really need someone to look over my code and tell me if I have met all the requirements for this project.
Here are the requirements:
For the final project, you are to write a calculator program in Python
with the following characteristics:
- It must be at least 50 lines of code excluding comments
- It must have the following functions:+,-,*,/, SQRT, and work with decimal numbers.
- It must be all text, no high resolution graphics
- It must automatically copy the result to the operating system clipboard
- It must work as a functional calculator (calculations can be entered over and over)
Please submit the source code only (.py), no object code (I will
compile on receipt)
NOTE:
This assignment is graded on a pass/fail basis; there is no partial credit or resubmissions. Be sure you understand ALL of the
requirements.
Solutions that require the user install any software will NOT be accepted (in case you are not sure, this means NO Pyperclip.)
WARNING: All submissions will be checked for plagiarism using
SafeAssign or other plagiarism checker.
I completed the project, but because this is a pass/fail project I want to be 200% sure I have met every requirement. I am also open to any other feedback that would sharpen my Python skills.
Menu
def main():
#Menu
print("=======================")
print("Welcome to Calculator")
print("By: Tyler Harris")
print("=======================")
print("Menu: ")
print("[1] Calculator")
print("[2] Instructions")
print("[3] Exit")
print("=======================")
choice = input("Please select an option: ")
if choice == '1':
calculator()
elif choice == '2':
instructions()
main()
elif choice == '3':
print("Thank you for using Calculator.")
else:
print("Not an option, try again:")
main()
Calculator
def calculator():
result=0 #Resets calulator.
try:
while True:
print("Last Result: "+str(result))
#Number 1 input.
if result == 0:
num1= int(input("First Number: "))
if result != 0:
num1=result
#Operator input.
#Checks for input of special operators 'c', 'sqrt', and 'esc'.
operator=input("Opertaor: ").lower()
if operator == "c":
print("Calculator cleared")
calculator()
if operator=="sqrt":
result=(sqrt(num1))
print("Result: "+str(result))
continue
if operator=="esc":
main()
if operator.isdigit(): #Throw error if operator is a number.
raise Exception()
#Number 2 input.
num2= int(input("Second Number: "))
#Operator calls.
if operator=="+":
result=(add(num1,num2))
elif operator=="-":
result=(sub(num1,num2))
elif operator=="*":
result=(mult(num1,num2))
elif operator=="/":
result=(div(num1,num2))
#Copy result to System's clipboard and display the result.
copy2clip(str(result))
print("=======================")
print("Result: "+str(result))
print("copied to clipboard")
print("=======================")
#Catch any errors and reset calculator
except Exception:
print("=======================")
print("Error Please try again.")
calculator()
else:
calculator()
Operators
def add(num1,num2):
return num1 + num2
def sub(num1,num2):
return num1 - num2
def mult(num1,num2):
return num1 * num2
def div(num1,num2):
return num1 / num2
def sqrt(num1):
return (num1**(1/2.0))
Instructions
def instructions():
print("=======================")
print(format('Calculator','^25s'))
print("=======================")
print("Available operators")
print("+: Addition")
print("-: Subtraction")
print("*: Multiplication")
print("/: Division")
print("sqrt: Square Root")
print("c: Clear Calculator")
print("esc: Exit Calculator")
print("=======================")
def copy2clip(txt):
cmd='echo '+txt.strip()+'|clip'
return subprocess.check_call(cmd, shell=True)
if __name__=='__main__':
main()
python python-3.x homework calculator
$endgroup$
migrated from stackoverflow.com Mar 31 at 0:50
This question came from our site for professional and enthusiast programmers.
1
$begingroup$
I would suggest you don't useraise Exception
, instead provide a bit more info i.e.raise ValueError('Operator must be a digit')
.
$endgroup$
– David Waterworth
Mar 31 at 8:32
$begingroup$
I'm afraid I don't have time for a full review, but one thing I did pick up on and wants checking is the requirement to "work with decimal numbers." It may be worth clarifying whether it's fine to use floating point numbers, or whether you're specifically expected to use the decimal library: docs.python.org/3.7/library/decimal.html
$endgroup$
– Josiah
Mar 31 at 17:13
add a comment |
$begingroup$
In my final project for my Python course, I am required to create a calculator program. I really need someone to look over my code and tell me if I have met all the requirements for this project.
Here are the requirements:
For the final project, you are to write a calculator program in Python
with the following characteristics:
- It must be at least 50 lines of code excluding comments
- It must have the following functions:+,-,*,/, SQRT, and work with decimal numbers.
- It must be all text, no high resolution graphics
- It must automatically copy the result to the operating system clipboard
- It must work as a functional calculator (calculations can be entered over and over)
Please submit the source code only (.py), no object code (I will
compile on receipt)
NOTE:
This assignment is graded on a pass/fail basis; there is no partial credit or resubmissions. Be sure you understand ALL of the
requirements.
Solutions that require the user install any software will NOT be accepted (in case you are not sure, this means NO Pyperclip.)
WARNING: All submissions will be checked for plagiarism using
SafeAssign or other plagiarism checker.
I completed the project, but because this is a pass/fail project I want to be 200% sure I have met every requirement. I am also open to any other feedback that would sharpen my Python skills.
Menu
def main():
#Menu
print("=======================")
print("Welcome to Calculator")
print("By: Tyler Harris")
print("=======================")
print("Menu: ")
print("[1] Calculator")
print("[2] Instructions")
print("[3] Exit")
print("=======================")
choice = input("Please select an option: ")
if choice == '1':
calculator()
elif choice == '2':
instructions()
main()
elif choice == '3':
print("Thank you for using Calculator.")
else:
print("Not an option, try again:")
main()
Calculator
def calculator():
result=0 #Resets calulator.
try:
while True:
print("Last Result: "+str(result))
#Number 1 input.
if result == 0:
num1= int(input("First Number: "))
if result != 0:
num1=result
#Operator input.
#Checks for input of special operators 'c', 'sqrt', and 'esc'.
operator=input("Opertaor: ").lower()
if operator == "c":
print("Calculator cleared")
calculator()
if operator=="sqrt":
result=(sqrt(num1))
print("Result: "+str(result))
continue
if operator=="esc":
main()
if operator.isdigit(): #Throw error if operator is a number.
raise Exception()
#Number 2 input.
num2= int(input("Second Number: "))
#Operator calls.
if operator=="+":
result=(add(num1,num2))
elif operator=="-":
result=(sub(num1,num2))
elif operator=="*":
result=(mult(num1,num2))
elif operator=="/":
result=(div(num1,num2))
#Copy result to System's clipboard and display the result.
copy2clip(str(result))
print("=======================")
print("Result: "+str(result))
print("copied to clipboard")
print("=======================")
#Catch any errors and reset calculator
except Exception:
print("=======================")
print("Error Please try again.")
calculator()
else:
calculator()
Operators
def add(num1,num2):
return num1 + num2
def sub(num1,num2):
return num1 - num2
def mult(num1,num2):
return num1 * num2
def div(num1,num2):
return num1 / num2
def sqrt(num1):
return (num1**(1/2.0))
Instructions
def instructions():
print("=======================")
print(format('Calculator','^25s'))
print("=======================")
print("Available operators")
print("+: Addition")
print("-: Subtraction")
print("*: Multiplication")
print("/: Division")
print("sqrt: Square Root")
print("c: Clear Calculator")
print("esc: Exit Calculator")
print("=======================")
def copy2clip(txt):
cmd='echo '+txt.strip()+'|clip'
return subprocess.check_call(cmd, shell=True)
if __name__=='__main__':
main()
python python-3.x homework calculator
$endgroup$
In my final project for my Python course, I am required to create a calculator program. I really need someone to look over my code and tell me if I have met all the requirements for this project.
Here are the requirements:
For the final project, you are to write a calculator program in Python
with the following characteristics:
- It must be at least 50 lines of code excluding comments
- It must have the following functions:+,-,*,/, SQRT, and work with decimal numbers.
- It must be all text, no high resolution graphics
- It must automatically copy the result to the operating system clipboard
- It must work as a functional calculator (calculations can be entered over and over)
Please submit the source code only (.py), no object code (I will
compile on receipt)
NOTE:
This assignment is graded on a pass/fail basis; there is no partial credit or resubmissions. Be sure you understand ALL of the
requirements.
Solutions that require the user install any software will NOT be accepted (in case you are not sure, this means NO Pyperclip.)
WARNING: All submissions will be checked for plagiarism using
SafeAssign or other plagiarism checker.
I completed the project, but because this is a pass/fail project I want to be 200% sure I have met every requirement. I am also open to any other feedback that would sharpen my Python skills.
Menu
def main():
#Menu
print("=======================")
print("Welcome to Calculator")
print("By: Tyler Harris")
print("=======================")
print("Menu: ")
print("[1] Calculator")
print("[2] Instructions")
print("[3] Exit")
print("=======================")
choice = input("Please select an option: ")
if choice == '1':
calculator()
elif choice == '2':
instructions()
main()
elif choice == '3':
print("Thank you for using Calculator.")
else:
print("Not an option, try again:")
main()
Calculator
def calculator():
result=0 #Resets calulator.
try:
while True:
print("Last Result: "+str(result))
#Number 1 input.
if result == 0:
num1= int(input("First Number: "))
if result != 0:
num1=result
#Operator input.
#Checks for input of special operators 'c', 'sqrt', and 'esc'.
operator=input("Opertaor: ").lower()
if operator == "c":
print("Calculator cleared")
calculator()
if operator=="sqrt":
result=(sqrt(num1))
print("Result: "+str(result))
continue
if operator=="esc":
main()
if operator.isdigit(): #Throw error if operator is a number.
raise Exception()
#Number 2 input.
num2= int(input("Second Number: "))
#Operator calls.
if operator=="+":
result=(add(num1,num2))
elif operator=="-":
result=(sub(num1,num2))
elif operator=="*":
result=(mult(num1,num2))
elif operator=="/":
result=(div(num1,num2))
#Copy result to System's clipboard and display the result.
copy2clip(str(result))
print("=======================")
print("Result: "+str(result))
print("copied to clipboard")
print("=======================")
#Catch any errors and reset calculator
except Exception:
print("=======================")
print("Error Please try again.")
calculator()
else:
calculator()
Operators
def add(num1,num2):
return num1 + num2
def sub(num1,num2):
return num1 - num2
def mult(num1,num2):
return num1 * num2
def div(num1,num2):
return num1 / num2
def sqrt(num1):
return (num1**(1/2.0))
Instructions
def instructions():
print("=======================")
print(format('Calculator','^25s'))
print("=======================")
print("Available operators")
print("+: Addition")
print("-: Subtraction")
print("*: Multiplication")
print("/: Division")
print("sqrt: Square Root")
print("c: Clear Calculator")
print("esc: Exit Calculator")
print("=======================")
def copy2clip(txt):
cmd='echo '+txt.strip()+'|clip'
return subprocess.check_call(cmd, shell=True)
if __name__=='__main__':
main()
python python-3.x homework calculator
python python-3.x homework calculator
edited Mar 31 at 0:56
200_success
131k17157422
131k17157422
asked Mar 30 at 23:31
user3590710
migrated from stackoverflow.com Mar 31 at 0:50
This question came from our site for professional and enthusiast programmers.
migrated from stackoverflow.com Mar 31 at 0:50
This question came from our site for professional and enthusiast programmers.
1
$begingroup$
I would suggest you don't useraise Exception
, instead provide a bit more info i.e.raise ValueError('Operator must be a digit')
.
$endgroup$
– David Waterworth
Mar 31 at 8:32
$begingroup$
I'm afraid I don't have time for a full review, but one thing I did pick up on and wants checking is the requirement to "work with decimal numbers." It may be worth clarifying whether it's fine to use floating point numbers, or whether you're specifically expected to use the decimal library: docs.python.org/3.7/library/decimal.html
$endgroup$
– Josiah
Mar 31 at 17:13
add a comment |
1
$begingroup$
I would suggest you don't useraise Exception
, instead provide a bit more info i.e.raise ValueError('Operator must be a digit')
.
$endgroup$
– David Waterworth
Mar 31 at 8:32
$begingroup$
I'm afraid I don't have time for a full review, but one thing I did pick up on and wants checking is the requirement to "work with decimal numbers." It may be worth clarifying whether it's fine to use floating point numbers, or whether you're specifically expected to use the decimal library: docs.python.org/3.7/library/decimal.html
$endgroup$
– Josiah
Mar 31 at 17:13
1
1
$begingroup$
I would suggest you don't use
raise Exception
, instead provide a bit more info i.e. raise ValueError('Operator must be a digit')
.$endgroup$
– David Waterworth
Mar 31 at 8:32
$begingroup$
I would suggest you don't use
raise Exception
, instead provide a bit more info i.e. raise ValueError('Operator must be a digit')
.$endgroup$
– David Waterworth
Mar 31 at 8:32
$begingroup$
I'm afraid I don't have time for a full review, but one thing I did pick up on and wants checking is the requirement to "work with decimal numbers." It may be worth clarifying whether it's fine to use floating point numbers, or whether you're specifically expected to use the decimal library: docs.python.org/3.7/library/decimal.html
$endgroup$
– Josiah
Mar 31 at 17:13
$begingroup$
I'm afraid I don't have time for a full review, but one thing I did pick up on and wants checking is the requirement to "work with decimal numbers." It may be worth clarifying whether it's fine to use floating point numbers, or whether you're specifically expected to use the decimal library: docs.python.org/3.7/library/decimal.html
$endgroup$
– Josiah
Mar 31 at 17:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You are using function calls as if they were goto labels. That is a huge sin: it makes your program spaghetti code. If you want a loop, then write a loop. For example, main()
should look like:
def main():
while True:
#Menu
print("=======================")
print("Welcome to Calculator")
print("By: Tyler Harris")
print("=======================")
print("Menu: ")
print("[1] Calculator")
print("[2] Instructions")
print("[3] Exit")
print("=======================")
choice = input("Please select an option: ")
if choice == '1':
calculator()
elif choice == '2':
instructions()
elif choice == '3':
print("Thank you for using Calculator.")
break
else:
print("Not an option, try again:")
$endgroup$
add a comment |
$begingroup$
elif choice == '2':
instructions()
main()
Ummm, that's probably not exactly what you want.
Python lacks a goto
statement, so the usual idiom would be to wrap the whole thing in some sort of while
loop, perhaps while True
.
Consider an input sequence of "2 2 2 2 2", or "z z z z z".
You'll wind up pushing main
onto the call stack again, and again, and again....
Your locals never go out of scope,
so their ref count never decrements and they won't be deallocated and tidied up.
tl;dr: leaking stack frames is Bad, don't do it.
print("Last Result: "+str(result))
Usual idiom would be print('Last Result:', result)
, but no harm done.
num1 = int(input("First Number: "))
The requirements seemed pretty clear about "...must work with decimal numbers."
You chose to invoke int()
rather than float()
.
if result != 0:
num1 = result
I can't imagine how that corresponds to The Right Thing.
If you want a sentinel, None
would be much better than 0
,
just consider a sequence like 2 + 1 = - 3 =
(which yields zero).
Perhaps there's a reason for assigning the accumulator to num1
, but I'm not seeing it.
It appears you have discarded the user input.
operator = input("Opertaor: ").lower()
Typo.
calculator()
Same criticism as above. Python lacks goto
, and you're leaking stack frames here.
if operator == "sqrt":
result = (sqrt(num1))
print("Result: "+str(result))
continue
It is apparent that I'm not understanding why you are using the result
accumulator versus the num1
user input.
On most calculators, repeated clicking of the √ key would yield repeated assignment of result = sqrt(result)
.
I'm a little concerned about num1
versus result
confusion.
if operator=="esc":
main()
Same remark about leaking stack frames.
A return
statement would be usual, here.
if operator.isdigit():
This seems a bit restrictive.
Better to unconditionally raise an "unrecognized operator" error.
num2 = int(input("Second Number: "))
Same criticism as above, the "decimal" instructions are pretty clear about calling float()
.
result = (add(num1,num2))
All four of these assignments seem pretty weird,
as for a typical calculator we'd be accumulating result = add(result, num2)
.
This is a consequence of the num1
assignment above.
Also, please discard the (
extraneous parentheses)
.
def sqrt(num1):
return (num1**(1/2.0))
This works fine, but do consider the usual idiom of simply calling math.sqrt()
.
$endgroup$
1
$begingroup$
Just to elaborate on the first point of why it is bad to keep calling main() - basically each time you call a method, you are 'entering' that method. But you never exit the method - you just keep going deeper into the wormhole with each main() call. As JH is saying, each time you call a method you are putting more data onto the stack, but since the methods never are allowed to exit, the data is never freed, which eventually means you run out of space and the application will stop working.
$endgroup$
– fabspro
Mar 31 at 12:17
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You are using function calls as if they were goto labels. That is a huge sin: it makes your program spaghetti code. If you want a loop, then write a loop. For example, main()
should look like:
def main():
while True:
#Menu
print("=======================")
print("Welcome to Calculator")
print("By: Tyler Harris")
print("=======================")
print("Menu: ")
print("[1] Calculator")
print("[2] Instructions")
print("[3] Exit")
print("=======================")
choice = input("Please select an option: ")
if choice == '1':
calculator()
elif choice == '2':
instructions()
elif choice == '3':
print("Thank you for using Calculator.")
break
else:
print("Not an option, try again:")
$endgroup$
add a comment |
$begingroup$
You are using function calls as if they were goto labels. That is a huge sin: it makes your program spaghetti code. If you want a loop, then write a loop. For example, main()
should look like:
def main():
while True:
#Menu
print("=======================")
print("Welcome to Calculator")
print("By: Tyler Harris")
print("=======================")
print("Menu: ")
print("[1] Calculator")
print("[2] Instructions")
print("[3] Exit")
print("=======================")
choice = input("Please select an option: ")
if choice == '1':
calculator()
elif choice == '2':
instructions()
elif choice == '3':
print("Thank you for using Calculator.")
break
else:
print("Not an option, try again:")
$endgroup$
add a comment |
$begingroup$
You are using function calls as if they were goto labels. That is a huge sin: it makes your program spaghetti code. If you want a loop, then write a loop. For example, main()
should look like:
def main():
while True:
#Menu
print("=======================")
print("Welcome to Calculator")
print("By: Tyler Harris")
print("=======================")
print("Menu: ")
print("[1] Calculator")
print("[2] Instructions")
print("[3] Exit")
print("=======================")
choice = input("Please select an option: ")
if choice == '1':
calculator()
elif choice == '2':
instructions()
elif choice == '3':
print("Thank you for using Calculator.")
break
else:
print("Not an option, try again:")
$endgroup$
You are using function calls as if they were goto labels. That is a huge sin: it makes your program spaghetti code. If you want a loop, then write a loop. For example, main()
should look like:
def main():
while True:
#Menu
print("=======================")
print("Welcome to Calculator")
print("By: Tyler Harris")
print("=======================")
print("Menu: ")
print("[1] Calculator")
print("[2] Instructions")
print("[3] Exit")
print("=======================")
choice = input("Please select an option: ")
if choice == '1':
calculator()
elif choice == '2':
instructions()
elif choice == '3':
print("Thank you for using Calculator.")
break
else:
print("Not an option, try again:")
answered Mar 31 at 1:00
200_success200_success
131k17157422
131k17157422
add a comment |
add a comment |
$begingroup$
elif choice == '2':
instructions()
main()
Ummm, that's probably not exactly what you want.
Python lacks a goto
statement, so the usual idiom would be to wrap the whole thing in some sort of while
loop, perhaps while True
.
Consider an input sequence of "2 2 2 2 2", or "z z z z z".
You'll wind up pushing main
onto the call stack again, and again, and again....
Your locals never go out of scope,
so their ref count never decrements and they won't be deallocated and tidied up.
tl;dr: leaking stack frames is Bad, don't do it.
print("Last Result: "+str(result))
Usual idiom would be print('Last Result:', result)
, but no harm done.
num1 = int(input("First Number: "))
The requirements seemed pretty clear about "...must work with decimal numbers."
You chose to invoke int()
rather than float()
.
if result != 0:
num1 = result
I can't imagine how that corresponds to The Right Thing.
If you want a sentinel, None
would be much better than 0
,
just consider a sequence like 2 + 1 = - 3 =
(which yields zero).
Perhaps there's a reason for assigning the accumulator to num1
, but I'm not seeing it.
It appears you have discarded the user input.
operator = input("Opertaor: ").lower()
Typo.
calculator()
Same criticism as above. Python lacks goto
, and you're leaking stack frames here.
if operator == "sqrt":
result = (sqrt(num1))
print("Result: "+str(result))
continue
It is apparent that I'm not understanding why you are using the result
accumulator versus the num1
user input.
On most calculators, repeated clicking of the √ key would yield repeated assignment of result = sqrt(result)
.
I'm a little concerned about num1
versus result
confusion.
if operator=="esc":
main()
Same remark about leaking stack frames.
A return
statement would be usual, here.
if operator.isdigit():
This seems a bit restrictive.
Better to unconditionally raise an "unrecognized operator" error.
num2 = int(input("Second Number: "))
Same criticism as above, the "decimal" instructions are pretty clear about calling float()
.
result = (add(num1,num2))
All four of these assignments seem pretty weird,
as for a typical calculator we'd be accumulating result = add(result, num2)
.
This is a consequence of the num1
assignment above.
Also, please discard the (
extraneous parentheses)
.
def sqrt(num1):
return (num1**(1/2.0))
This works fine, but do consider the usual idiom of simply calling math.sqrt()
.
$endgroup$
1
$begingroup$
Just to elaborate on the first point of why it is bad to keep calling main() - basically each time you call a method, you are 'entering' that method. But you never exit the method - you just keep going deeper into the wormhole with each main() call. As JH is saying, each time you call a method you are putting more data onto the stack, but since the methods never are allowed to exit, the data is never freed, which eventually means you run out of space and the application will stop working.
$endgroup$
– fabspro
Mar 31 at 12:17
add a comment |
$begingroup$
elif choice == '2':
instructions()
main()
Ummm, that's probably not exactly what you want.
Python lacks a goto
statement, so the usual idiom would be to wrap the whole thing in some sort of while
loop, perhaps while True
.
Consider an input sequence of "2 2 2 2 2", or "z z z z z".
You'll wind up pushing main
onto the call stack again, and again, and again....
Your locals never go out of scope,
so their ref count never decrements and they won't be deallocated and tidied up.
tl;dr: leaking stack frames is Bad, don't do it.
print("Last Result: "+str(result))
Usual idiom would be print('Last Result:', result)
, but no harm done.
num1 = int(input("First Number: "))
The requirements seemed pretty clear about "...must work with decimal numbers."
You chose to invoke int()
rather than float()
.
if result != 0:
num1 = result
I can't imagine how that corresponds to The Right Thing.
If you want a sentinel, None
would be much better than 0
,
just consider a sequence like 2 + 1 = - 3 =
(which yields zero).
Perhaps there's a reason for assigning the accumulator to num1
, but I'm not seeing it.
It appears you have discarded the user input.
operator = input("Opertaor: ").lower()
Typo.
calculator()
Same criticism as above. Python lacks goto
, and you're leaking stack frames here.
if operator == "sqrt":
result = (sqrt(num1))
print("Result: "+str(result))
continue
It is apparent that I'm not understanding why you are using the result
accumulator versus the num1
user input.
On most calculators, repeated clicking of the √ key would yield repeated assignment of result = sqrt(result)
.
I'm a little concerned about num1
versus result
confusion.
if operator=="esc":
main()
Same remark about leaking stack frames.
A return
statement would be usual, here.
if operator.isdigit():
This seems a bit restrictive.
Better to unconditionally raise an "unrecognized operator" error.
num2 = int(input("Second Number: "))
Same criticism as above, the "decimal" instructions are pretty clear about calling float()
.
result = (add(num1,num2))
All four of these assignments seem pretty weird,
as for a typical calculator we'd be accumulating result = add(result, num2)
.
This is a consequence of the num1
assignment above.
Also, please discard the (
extraneous parentheses)
.
def sqrt(num1):
return (num1**(1/2.0))
This works fine, but do consider the usual idiom of simply calling math.sqrt()
.
$endgroup$
1
$begingroup$
Just to elaborate on the first point of why it is bad to keep calling main() - basically each time you call a method, you are 'entering' that method. But you never exit the method - you just keep going deeper into the wormhole with each main() call. As JH is saying, each time you call a method you are putting more data onto the stack, but since the methods never are allowed to exit, the data is never freed, which eventually means you run out of space and the application will stop working.
$endgroup$
– fabspro
Mar 31 at 12:17
add a comment |
$begingroup$
elif choice == '2':
instructions()
main()
Ummm, that's probably not exactly what you want.
Python lacks a goto
statement, so the usual idiom would be to wrap the whole thing in some sort of while
loop, perhaps while True
.
Consider an input sequence of "2 2 2 2 2", or "z z z z z".
You'll wind up pushing main
onto the call stack again, and again, and again....
Your locals never go out of scope,
so their ref count never decrements and they won't be deallocated and tidied up.
tl;dr: leaking stack frames is Bad, don't do it.
print("Last Result: "+str(result))
Usual idiom would be print('Last Result:', result)
, but no harm done.
num1 = int(input("First Number: "))
The requirements seemed pretty clear about "...must work with decimal numbers."
You chose to invoke int()
rather than float()
.
if result != 0:
num1 = result
I can't imagine how that corresponds to The Right Thing.
If you want a sentinel, None
would be much better than 0
,
just consider a sequence like 2 + 1 = - 3 =
(which yields zero).
Perhaps there's a reason for assigning the accumulator to num1
, but I'm not seeing it.
It appears you have discarded the user input.
operator = input("Opertaor: ").lower()
Typo.
calculator()
Same criticism as above. Python lacks goto
, and you're leaking stack frames here.
if operator == "sqrt":
result = (sqrt(num1))
print("Result: "+str(result))
continue
It is apparent that I'm not understanding why you are using the result
accumulator versus the num1
user input.
On most calculators, repeated clicking of the √ key would yield repeated assignment of result = sqrt(result)
.
I'm a little concerned about num1
versus result
confusion.
if operator=="esc":
main()
Same remark about leaking stack frames.
A return
statement would be usual, here.
if operator.isdigit():
This seems a bit restrictive.
Better to unconditionally raise an "unrecognized operator" error.
num2 = int(input("Second Number: "))
Same criticism as above, the "decimal" instructions are pretty clear about calling float()
.
result = (add(num1,num2))
All four of these assignments seem pretty weird,
as for a typical calculator we'd be accumulating result = add(result, num2)
.
This is a consequence of the num1
assignment above.
Also, please discard the (
extraneous parentheses)
.
def sqrt(num1):
return (num1**(1/2.0))
This works fine, but do consider the usual idiom of simply calling math.sqrt()
.
$endgroup$
elif choice == '2':
instructions()
main()
Ummm, that's probably not exactly what you want.
Python lacks a goto
statement, so the usual idiom would be to wrap the whole thing in some sort of while
loop, perhaps while True
.
Consider an input sequence of "2 2 2 2 2", or "z z z z z".
You'll wind up pushing main
onto the call stack again, and again, and again....
Your locals never go out of scope,
so their ref count never decrements and they won't be deallocated and tidied up.
tl;dr: leaking stack frames is Bad, don't do it.
print("Last Result: "+str(result))
Usual idiom would be print('Last Result:', result)
, but no harm done.
num1 = int(input("First Number: "))
The requirements seemed pretty clear about "...must work with decimal numbers."
You chose to invoke int()
rather than float()
.
if result != 0:
num1 = result
I can't imagine how that corresponds to The Right Thing.
If you want a sentinel, None
would be much better than 0
,
just consider a sequence like 2 + 1 = - 3 =
(which yields zero).
Perhaps there's a reason for assigning the accumulator to num1
, but I'm not seeing it.
It appears you have discarded the user input.
operator = input("Opertaor: ").lower()
Typo.
calculator()
Same criticism as above. Python lacks goto
, and you're leaking stack frames here.
if operator == "sqrt":
result = (sqrt(num1))
print("Result: "+str(result))
continue
It is apparent that I'm not understanding why you are using the result
accumulator versus the num1
user input.
On most calculators, repeated clicking of the √ key would yield repeated assignment of result = sqrt(result)
.
I'm a little concerned about num1
versus result
confusion.
if operator=="esc":
main()
Same remark about leaking stack frames.
A return
statement would be usual, here.
if operator.isdigit():
This seems a bit restrictive.
Better to unconditionally raise an "unrecognized operator" error.
num2 = int(input("Second Number: "))
Same criticism as above, the "decimal" instructions are pretty clear about calling float()
.
result = (add(num1,num2))
All four of these assignments seem pretty weird,
as for a typical calculator we'd be accumulating result = add(result, num2)
.
This is a consequence of the num1
assignment above.
Also, please discard the (
extraneous parentheses)
.
def sqrt(num1):
return (num1**(1/2.0))
This works fine, but do consider the usual idiom of simply calling math.sqrt()
.
edited Mar 31 at 17:22
answered Mar 31 at 1:01
J_HJ_H
4,707137
4,707137
1
$begingroup$
Just to elaborate on the first point of why it is bad to keep calling main() - basically each time you call a method, you are 'entering' that method. But you never exit the method - you just keep going deeper into the wormhole with each main() call. As JH is saying, each time you call a method you are putting more data onto the stack, but since the methods never are allowed to exit, the data is never freed, which eventually means you run out of space and the application will stop working.
$endgroup$
– fabspro
Mar 31 at 12:17
add a comment |
1
$begingroup$
Just to elaborate on the first point of why it is bad to keep calling main() - basically each time you call a method, you are 'entering' that method. But you never exit the method - you just keep going deeper into the wormhole with each main() call. As JH is saying, each time you call a method you are putting more data onto the stack, but since the methods never are allowed to exit, the data is never freed, which eventually means you run out of space and the application will stop working.
$endgroup$
– fabspro
Mar 31 at 12:17
1
1
$begingroup$
Just to elaborate on the first point of why it is bad to keep calling main() - basically each time you call a method, you are 'entering' that method. But you never exit the method - you just keep going deeper into the wormhole with each main() call. As JH is saying, each time you call a method you are putting more data onto the stack, but since the methods never are allowed to exit, the data is never freed, which eventually means you run out of space and the application will stop working.
$endgroup$
– fabspro
Mar 31 at 12:17
$begingroup$
Just to elaborate on the first point of why it is bad to keep calling main() - basically each time you call a method, you are 'entering' that method. But you never exit the method - you just keep going deeper into the wormhole with each main() call. As JH is saying, each time you call a method you are putting more data onto the stack, but since the methods never are allowed to exit, the data is never freed, which eventually means you run out of space and the application will stop working.
$endgroup$
– fabspro
Mar 31 at 12:17
add a comment |
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$begingroup$
I would suggest you don't use
raise Exception
, instead provide a bit more info i.e.raise ValueError('Operator must be a digit')
.$endgroup$
– David Waterworth
Mar 31 at 8:32
$begingroup$
I'm afraid I don't have time for a full review, but one thing I did pick up on and wants checking is the requirement to "work with decimal numbers." It may be worth clarifying whether it's fine to use floating point numbers, or whether you're specifically expected to use the decimal library: docs.python.org/3.7/library/decimal.html
$endgroup$
– Josiah
Mar 31 at 17:13