[solved]what i have to do to fix the error of this code?

Question

it gives an UnboundlocalError, i cant understand what is the mistake i have done.

num1 = int(input())
opr = input()
numx = int(input())
oprx = input()
def operator_check() :
if str(opr) in opr_list :
return opr_list[str(opr)]
else :
pass

def add() :
if oprx == "=" :
numx += num1
return numx
elif str(oprx) == "+" :
numx += num1
return operator_check()
else :
pass
def sub() :
if oprx == "=" :
numx -= num1
return numx
elif oprx == "-" :
numx -= num1
return operator_check()
else :
pass
def mult() :
if oprx == "=" :
numx *= num1
return numx
elif oprx == "*" :
numx *= num1
return operator_check()
else :
pass
def div() :
if oprx == "=" :
numx /= num1
return numx
elif oprx == "/" :
numx /= num1
return operator_check()
else :
pass
def factorial(num1) :
if num1 == 1 :
return 1
else :
return num1*factorial(num1-1)
def expo() :
if oprx == "=" :
numx **= num1
elif oprx == "**" :
numx **= num1
return operator_check()
else :
pass
def sqrt() :
if numx == "sqrt" :
num1 = num1 ** (1/2)
return num1

opr_list = {"+": add(),"-": sub(),"*": mult(),"/": div(),"!": factorial(num1),"**": expo(),"sqrt": sqrt()}
operator_check()

Answer 1

You need to add a line
`global numx`
Before reassigning value to the variable <numx> in each function.

def add() :
if oprx == "=" :
global numx # line added
numx += num1
return numx
elif str(oprx) == "+" :
numx += num1
return operator_check()
else :
pass

Similarly you need to add that line before reassigning in each function.

Python does not have variables declaration like in other languages so it has to figure out the scope of a variable itself.
So it follows a simple rule, if there is an assignment to the variable inside a function, that variable is local to that function.

So what is happening that when you write `numx += num1`
It creates <numx> locally and tries to read its value before it is even assigned.
This results in an UnboundLocalError.

Answer 2

Thanks both of you ! 😊 I totally forgot about that local and global thing

Answer 3

Ayash Sameer

What type of error are you getting?

Answer 4

num1 = 5
opr = *
numx = 2
What is oprx?
Ayash Sameer

Answer 5

try to put the result into another variable rather than overwriting the numx variable by making calculations on global variables. Example:
def add():
result = numx + num1
return result

another way is to pass them into a function and then return or print them. Example

def add(x, y):
x += y
return x
print(add(4,5))

Result:
9

This will be a much more clear program. By the way you are not printing anything with this program and it makes it very hard to test it.

And as an answer to your question when you are using the numx += num1 statement the function thinks you are trying to create a new local variable. You can read further from here : https://stackoverflow.com/questions/10851906/JUMP_LINK__&&__python__&&__JUMP_LINK-3-unboundlocalerror-local-variable-referenced-before-assignment

Answer 6

The Unboundlocalerror: local variable referenced before assignment is raised when you try to use a variable before it has been assigned in the local context. Python doesn't have variable declarations , so it has to figure out the scope of variables itself. It does so by a simple rule: If there is an assignment to a variable inside a function, that variable is considered local . To solve this problem, you can explicitly say it's a global by putting global declaration in you function. The global statement does not have to be at the beginning of the function definition, but that is where it is usually placed. Wherever it is placed, the global declaration makes a variable to global variable everywhere in the function.

http://net-informations.com/python/err/local.htm

Answer 7

I will share the link to above code here, so you can get a better understand https://code.sololearn.com/cwgY3Pe8PVy8/#
sample input :
num1 = 3
opr = +
numx = 4
oprx = =

Answer 8

its an unbound local error, it says the variable "numx" is referenced before assignment