Python 3 exec assignment error

Question

I have this Python 3 code which is causing me issues:

def setStartingVariable(inputType, text, code = None, customErrorMessage = "Error: No error message"):
    while True:
        try:
            variable = inputType(input(text)) # Test if the inputted variable is of the right type. Keep in mind inputType is a variable, not a function.
        except BaseException: 
            print("Error! Try again!")
            continue
        err = 0 # Reset
        if code != None:
            exec(code)
        if err == 0:
            print("No error")
            return variable
        elif err == 1:
            print(customErrorMessage)
        else:
            print("Error: var err != 1 and != 0")


def inputStartingVariables():
    global wallet
    wallet = setStartingVariable(float, "Just type a number in... ", "err = 1", "-!- error message -!-")

inputStartingVariables()

This should cause the prompt...

Just type a number in...

...And it does. And if I type in something other than a float, it gives the error...

Error! Try again!

...And re-prompts me. So far, so good. However, if I put in a normal number, it displays...

No Error

...When it should display variable customErrorMessage, in this case being...

-!- error message -!-

I originally thought that the issue was that in the exec() function, err wasn't being treated as a global variable, but using global err; err = 1 instead of just err = 1 doesn't fix it.

Please provide a minimal example. In particular, either the input or its evaluation or the loop shouldn't be necessary. See also https://stackoverflow.com/help/how-to-ask. — Ulrich Eckhardt, Jul 31 '15 at 05:35
This _is_ the stripped-down version... It shows the basic idea of what I'm trying to do while eliminating the unecessary code. — Quelklef, Jul 31 '15 at 05:39
The problem with your code is that when you are testing for input `variable = inputType(input(text))` , it is actually converting what ever you input to the inputType. So when you type a integer, it converts the number to type float. — Digvijayad, Jul 31 '15 at 06:07
You want an error when the user inputs a integer, but your input command converts the input to float, hence no error. try printing the variable after this. `variable = inputType(input(text))` `print(variable)`. You will see what i mean — Digvijayad, Jul 31 '15 at 06:20
I want an error when the user inputs something that is not an integer... — Quelklef, Jul 31 '15 at 06:20
'> However, if I put in a normal number, it displays... No Error' that's what you put in your question. So from my understanding, if I input `10` you want your custom error. — Digvijayad, Jul 31 '15 at 06:22
Like, I said, I want an error the the user inputs something that is _not_ an integer. That part is working fine. — Quelklef, Jul 31 '15 at 06:26

score 1 · Accepted Answer · edited May 23 '17 at 11:51

To be straight forward:

your err value is never changed .It is always 0

using exec() doesn't change it

This changes will answer your question:

def setStartingVariable(inputType, text, code = None, customErrorMessage = "Error: No error message"):
    while True:
        try:
            variable = inputType(input(text)) # Test if the inputted variable is of the right type. Keep in mind inputType is a variable, not a function.
        except BaseException: 
            print("Error! Try again!")
            continue
        exec_scope = {"err" :0}
         # Reset
        if code != None:
            print ("here")
            exec(code,exec_scope)

        print (code)
        if exec_scope["err"] == 0:
            print("No error")
            return variable
        elif exec_scope["err"] == 1:
            print(customErrorMessage)
        else:
            print("Error: var err != 1 and != 0")


def inputStartingVariables():
    global wallet
    wallet = setStartingVariable(float, "Just type a number in... ", "err = 1", "-!- error message -!-")
inputStartingVariables()

cause of problem:

1.exec is a function in python3 so if you do any variable assignment in it it will not change the variable content it will only be available for the exec function

i.e)

def a():
    exec("s=1")
    print (s)
a()

Traceback (most recent call last):
File "python", line 4, in <module>
File "python", line 3, in a
NameError: name 's' is not defined

For more on variable assignment you can see this both question so by martin and blaknight

edit:

def setStartingVariable(inputType, text, code = None, customErrorMessage = "Error: No error message"):
    global err
    while True:
        try:
            variable = inputType(input(text)) # Test if the inputted variable is of the right type. Keep in mind inputType is a variable, not a function.
        except BaseException: 
            print("Error! Try again!")
            continue

        err = 0 # Reset
        if code != None:
            exec("global err;"+code)
            print (err)
        if err == 0:
            print("No error")
            return variable
        elif err == 1:
            print(customErrorMessage)
        else:
            print("Error: var err != 1 and != 0")


def inputStartingVariables():
    global wallet
    wallet = setStartingVariable(float, "Just type a number in... ", "err = 1", "-!- error message -!-")

inputStartingVariables()

I thought I addressed this issue under my "**I originally thought**" section? — Quelklef, Jul 31 '15 at 06:16
Did you see my section? It should work, as proven by the fact that [this code](http://pastebin.com/Fm5Uk7t7) works — Quelklef, Jul 31 '15 at 06:23
Also: I'm a bit of a Python noob, so I'm having some troubles understanding your links. — Quelklef, Jul 31 '15 at 06:25
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/84765/discussion-between-vignesh-kalai-and-quelklef). — The6thSense, Jul 31 '15 at 06:31
OK, it works, kinda odd to me that you have to `global` it in the function, too, and not just the `exec()` — Quelklef, Jul 31 '15 at 06:32

Python 3 exec assignment error

1 Answers1