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Answers
It s because when you do some operation on a (which is expected to be global) python treats it like local.
So when you do a+=1 it searches for local variable and it doesn t find s any and hence the error.
If you trying to increment the global a then you need to mention a with a global keyword before that. Else use some other variable and increment it
http://stackoverflow.com/questions/21456739/unboundlocalerror-local-variable-l-referenced-before-assignment-python http://stackoverflow.com/questions/21456739/unboundlocalerror-local-variable-l-referenced-before-assignment-python
Consider this
c = 0 def increment(): print c c += 1 increment()
if there is an assignment to a variable inside a function, that variable is considered local Thus, the line
c += 1
implicitly makes counter local to increment(). So print c will search for local variable c rather than global variable c I hope that s fine now
Source
License : cc by-sa 3.0
http://stackoverflow.com/questions/30522721/unexpected-behavior-of-pythons-variable-scope
