"jena" <[EMAIL PROTECTED]> wrote: > when i create list of lambdas: > l=[lambda:x.upper() for x in ['a','b','c']] > then l[0]() returns 'C', i think, it should be 'A'
the "x" variable contains "c" when you leave the loop: >>> l=[lambda:x.upper() for x in ['a','b','c']] >>> x 'c' so x.upper() will of course return 'C' for all three lambdas: >>> l[0]() 'C' > it is OK or it is bug? it's not a bug. free variables bind to names, not objects. > can i do it correctly simplier than with helper X class? bind to the object instead of the name: >>> l=[lambda x=x:x.upper() for x in ['a','b','c']] >>> x 'c' >>> l[0]() 'A' >>> l[1]() 'B' >>> l[2]() 'C' </F> -- http://mail.python.org/mailman/listinfo/python-list
