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locals() and globals()

Hi

this exaple:

def lcl():
n=1
x=locals()
x["n"]=100
print "n in lcl() is:" +str(n)
#This will say Name error
#x["new"]=1
#print new
n=1
x=globals()
x["n"]=100
print "gobal n is:" +str(n)
x["new"]=1
print "new is:" +str(new)
lcl()

produces

gobal n is:100
new is:1
n in lcl() is:1

shouldn't be n in lcl() 100 too?

why accessing the names dictionary globals() and locals() gives
different results?
This example was made using
Python 2.4.3 (#69, Mar 29 2006, 17:35:34) [MSC v.1310 32 bit (Intel)] on win32

PAolo
Oct 14 '06 #1
1 1448
Paolo Pantaleo wrote:
Hi

this exaple:

def lcl():
n=1
x=locals()
x["n"]=100
print "n in lcl() is:" +str(n)
#This will say Name error
#x["new"]=1
#print new
n=1
x=globals()
x["n"]=100
print "gobal n is:" +str(n)
x["new"]=1
print "new is:" +str(new)
lcl()

produces

gobal n is:100
new is:1
n in lcl() is:1

shouldn't be n in lcl() 100 too?

why accessing the names dictionary globals() and locals() gives
different results?
Python is statically scoped. The name bindings are fixed at compile
time. Hence you can't achieve dynamic scoping by using the locals()
dictionary and add new names to the locals() which might eventually be
used as local variables in the body of the function. The function still
controls its own state and does not permit delegating state changes via
locals(). That's what objects are for.

Oct 14 '06 #2

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