CODEWITHSUNDEEP
pythonexceptiontry-catch
garen
garen

Reputation: 358

Python Try-Except inside of Function

I've got a pretty good understanding of python's try-except clause, but I'm encountering problems when trying to put it inside of a function.

>>> def tryAppend(child, parent):
...     try:
...             parent.append(child)
...     except NameError:
...             print "WRONG NAME"
>>> var1 = []
>>> var2 = 'test2'
>>> tryAppend(var2, var1)  #works, no error
>>> tryAppend(foo, var1)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
NameError: name 'foo' is not defined

it is almost like python doesn't see the try: statement. Any help is appreciated.

Upvotes: 6

Views: 53193

Answers (8)

Temu&#231;in Yılmaz Suveren
Temu&#231;in Yılmaz Suveren

Reputation: 1

def tryAppend(child, parent):
    try:
        global temp_list
        temp_list = [child]
        script = "global " + str(parent) +  '\n' + str(parent) + " = " + str(parent) + " + temp_list" + '\n' + "print('List already exists, only append')"
        return exec(script)
    except NameError:
        print("List does not exist, creating list")
        string_e_1 = "global " + str(parent) +  '\n' + str(parent) + " = temp_list"
        return exec(string_e_1)

Upvotes: 0

jam7co
jam7co

Reputation: 43

(Duplicates of this question also here and here)

In Python 3, you can avoid this issue by passing a lambda function into the function containing the try-except. For example, this first block will throw an error:

def trying_fn1(foo):
    try:
        print(foo)
    except ZeroDivisionError:
        print('Cannot divide a number by zero.')

trying_fn1(10/0)

Traceback (most recent call last):
    File "<stdin>", line 1, in <module> trying_fn1(10/0)
ZeroDivisionError: division by zero

But this next function accepts a lambda function which it executes inside the try-except:

def trying_fn2(lambda_foo):
    try:
        print(lambda_foo())
    except ZeroDivisionError:
        print('Cannot divide a number by zero.')
    
lambda_foo = lambda: 10/0
trying_fn2(lambda_foo)
Cannot divide a number by zero.

Hope this helps! If there's a new or better way to do this, happy to take suggestions!

Upvotes: 1

user395760
user395760

Reputation:

tryAppend(foo, var1) is evaluated (roughly) in this order:

  1. Fetch the object tryAppend references
  2. Fetch the object foo references
  3. Fetch the object var1 references
  4. Call the first with the second and third as arguments (=do whatever the function tryAppend does, uncluding the try-except)

The error occurs at #2, long before the function and the try block is entered. In fact, the try block cannot to throw a NameError, as the only names used are parent and child, both being arguments and thus always available (if .append does not exist, that's an AttributeError).

In the same way, the following code will not print "caught it" because the exception is raised before the try block is executed:

raise Exception("Catch me if you can")
try:
    pass # do nothing
except:
    print "caught it"

Upvotes: 11

Kliment Merzlyakov
Kliment Merzlyakov

Reputation: 1083

For someone who is looking for how to use try except construction inside of the function. I am not sure whether it is a good programming style, but it works.

You can put string arguments to the function. It will be evaluated correctly and then you can use exec inside of the function:

def tryAppend(child, parent):
    try:
        script = parent + '.append(' + child + ')'
        exec script
        return parent
    except NameError:
        print "WRONG NAME"
var1 = []
var2 = 'test2'
tryAppend('var2', 'var1')
tryAppend('foo', 'var1')

Upvotes: 1

Srikar Appalaraju
Srikar Appalaraju

Reputation: 73588

foo exception happens even before you enter the function tryAppend() i.e. outside of the function.

Upvotes: 1

Russell Borogove
Russell Borogove

Reputation: 19037

The NameError is being thrown when the name 'foo' is evaluated, which is before entering the function. Therefore the try/except within the function isn't relevant.

Upvotes: 2

Jonathan Sternberg
Jonathan Sternberg

Reputation: 6647

The name error is happening before it ever gets into tryAppend. It evaluates the value of foo when trying to pass it to the function. This works:

def tryAppend(child, parent):
    parent.append(child)

var1 = []
try:
    tryAppend(foo, var1)
except NameError:
    print 'WRONG NAME'

Upvotes: 5

cababunga
cababunga

Reputation: 3114

This has nothing to do with your exception handler. The error you are seeing is because "foo" is not defined anywhere.

Upvotes: 5

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