strange unbound local error?

[righes]

hi folks,

suppose this snipplet:

spam = 42

def eggs():
print spam
spam = spam + 1

if __name__=="__ma in__":
eggs()

This thows an UnboundLocalErr or at line 4 (print statement). But if I
comment out line 5 (variable assignment), no error occurs.

Can you explain me this, please?

Regards,
Enrico

[Pablo Ziliani]

ri****@arcor.de wrote:

spam = 42

def eggs():
print spam
spam = spam + 1

if __name__=="__ma in__":
eggs()

This thows an UnboundLocalErr or at line 4 (print statement). But if I
comment out line 5 (variable assignment), no error occurs.

Can you explain me this, please?

Hi Enrico,

You need to say that you will be modifying the variable in the global
scope, like this:

spam = 42

def eggs():
global spam # <--here
print spam
spam = spam + 1

if __name__=="__ma in__":
eggs()
This can help you:
http://www.pasteur.fr/recherche/unit...thon/ch04.html
Pablo

[thebjorn]

On Sep 29, 8:04 pm, [e-mail] wrote:

hi folks,

suppose this snipplet:

spam = 42

def eggs():
print spam
spam = spam + 1

if __name__=="__ma in__":
eggs()

This thows an UnboundLocalErr or at line 4 (print statement). But if I
comment out line 5 (variable assignment), no error occurs.

Can you explain me this, please?

Regards,
Enrico

If you're going to assign to a global variable in a function, you need
to declare it as such:

spam = 42

def eggs():
global spam
print spam
spam = spam + 1

When Python sees an assignment to an identifier in a function it
creates a variable that is local to the function (otherwise you'd be
unable to create local variables with the same name as any of the
global variables).

-- bjorn