Python: Eval giving NameError in function

Question

I want to use eval to do operations to see if a and b equals c with some operator.

My code:

def Expression(a, b, c):
 operators = ["+", "-", "*", "/"]
 return len([i for i in operators if eval("a () b".replace("()", i)) == c]) >= 1
Expression(1, 2, 3)

For some reason, this results in a NameError. Error log:

return len([i for i in operators if eval("a () b".replace("()", i)) == c]) >= 1
 File "<string>", line 1, in <module>
NameError: name 'a' is not defined

Since the function has a as a parameter I don't believe a should be undefined. What is the problem here?

• Danilo Cabello: Hi, welcome to Stack Overflow. eval is usually associated with risky operations, what you are missing though is likely using string interpolation you want the value of a not just a variable. Try '{} () {}'.format(a, b).replace....`

Answer 1

The problem is in that case eval try to find a and b in global scope, not function scop(it means that a and b is just valid in the function block). so you can pass the function current scope using locals() to eval like this:

def Expression(a, b, c):
 operators = ["+", "-", "*", "/"]
 scope = locals()
 return len([i for i in operators if eval("a () b".replace("()", i), scope) == c]) >= 1

Then your code will work.

for better understanding try to define a and b in the global scope, then you can see it works, like this:

a=1
b=2
def Expression(c):
 operators = ["+", "-", "*", "/"]
 return len([i for i in operators if eval("a () b".replace("()", i)) == c]) >= 1
Expression(3)

Also, you can pass a custom scope to eval by creating a dictionary and pass it to eval:

scope = {'a':a, 'b':b}

So, it was your code problem. but for the better attitude you can use what @Rakesh said before, Using formatted strings, it gets the current a and b and pass it to eval as what they have inside, like this:

eval(f"{a}{i}{b}") # <Python3.6 
eval("{}{}{}".format(a,i,b)) 

Also you can use any() instead of len() >= 1