Don't understand why UnboundLocalError occurs
What am I doing wrong here?
counter = 0 def increment(): counter += 1 increment()
The above code throws an
Python doesn't have variable declarations, so it has to figure out the scope of variables itself. It does so by a simple rule: If there is an assignment to a variable inside a function, that variable is considered local. Thus, the line
counter += 1
counter local to
increment(). Trying to execute this line, though, will try to read the value of the local variable
counter before it is assigned, resulting in an
Don, Definition of don. (Entry 1 of 4) transitive verb. 1 : to put on (an article of clothing) donned his hat and gloves. 2 : to wrap oneself in : take on sense 3a the donning of new and more tyrannous moralities— Edward Sapir. Don (also dōn) Used as a courtesy title before the name of a man in a Spanish-speaking area. 2.
You need to use the global statement so that you are modifying the global variable counter, instead of a local variable:
counter = 0 def increment(): global counter counter += 1 increment()
If the enclosing scope that
counter is defined in is not the global scope, on Python 3.x you could use the nonlocal statement. In the same situation on Python 2.x you would have no way to reassign to the nonlocal name
counter, so you would need to make
counter mutable and modify it:
counter =  def increment(): counter += 1 increment() print counter # prints '1'
Don, a petty gangster or ruffian. TAKE THE QUIZ TO FIND OUT. Origin of don. 1. 1515–25; < Spanish (initial capital letter) Mr.; Sir: a Spanish title prefixed to a man's given name. (in Spanish-speaking countries) a lord or gentleman. (initial capital letter) an Italian title of address, especially for a priest. a person of great importance.
To answer the question in your subject line,* yes, there are closures in Python, except they only apply inside a function, and also (in Python 2.x) they are read-only; you can't re-bind the name to a different object (though if the object is mutable, you can modify its contents). In Python 3.x, you can use the
nonlocal keyword to modify a closure variable.
def incrementer(): counter = 0 def increment(): nonlocal counter counter += 1 return counter return increment increment = incrementer() increment() # 1 increment() # 2
* The question origially asked about closures in Python.
Don, DCP DeSilva (Boman Irani) sees a way to bring to justice the feared head of a criminal empire by recruiting a man named Vijay, who looks exactly like the crime boss (Shah Rukh Khan). The ruse works too well, and soon Vijay finds his life in danger Directed by Farhan Akhtar. With Shah Rukh Khan, Priyanka Chopra, Arjun Rampal, Isha Koppikar. Vijay is recruited by a police officer to masquerade as his lookalike Don, the leader of an international gang of smugglers. Things go wrong when the officer is killed and Vijay is left to fend for himself.
The reason of why your code throws an
UnboundLocalError is already well explained in other answers.
But it seems to me that you're trying to build something that works like
So why don't you try it out, and see if it suits your case:
>>> from itertools import count >>> counter = count(0) >>> counter count(0) >>> next(counter) 0 >>> counter count(1) >>> next(counter) 1 >>> counter count(2)
don, Head of any mafia crime family. 2. Dom Perignon. Expensive Champagne. Often called 'Don' in the suburbs and hood as a joke that we don't know what it is. 3. Don, and dom, is derived from the Latin Dominus : a master of a household, a title with background from the Roman Republic in classical antiquity.
Python has lexical scoping by default, which means that although an enclosed scope can access values in its enclosing scope, it cannot modify them (unless they're declared global with the
A closure binds values in the enclosing environment to names in the local environment. The local environment can then use the bound value, and even reassign that name to something else, but it can't modify the binding in the enclosing environment.
In your case you are trying to treat
counter as a local variable rather than a bound value. Note that this code, which binds the value of
x assigned in the enclosing environment, works fine:
>>> x = 1 >>> def f(): >>> return x >>> f() 1
don, A contraction of Middle English do on (“put on”), from Old English dōn on. Compare also doff, dup, dout. VerbEdit. don (third-person singular simple present dons, Don, a wanted criminal, dies in a police chase. DSP D'Silva is the only one who knows about his death, and to get hold of the gang he trains Don-lookalike Vijay. But Vijay faces danger from the police and from within the gang.
DON (verb) definition and synonyms, To don means to put on, as in clothing or hats. A hunter will don his camouflage clothes when he goes hunting. What is the opposite of don, an old-fashioned word Edward Don and Company is the world's leading distributor of foodservice equipment and supplies. Online ordering
DON, don meaning: 1. a lecturer (= a college teacher), especially at Oxford or Cambridge University in England 2. to…. Learn more. A don is a guy that everyone wants to be like. He is not only sexy and muscular but amazing in bed aswell. He can rock anyone. Being a don is a talent that not everyone can achieve.
Don dictionary definition, Don is defined as a Spanish title used to refer to a gentleman, or is a term used to describe a leader in an organized-crime family. An example of Don is the title Dôn, in Celtic mythology, leader of one of two warring families of gods; according to one interpretation, the Children of Dôn were the powers of light, constantly in conflict with the Children of Llyr, the powers of darkness. In another view, the conflict was a struggle between indigenous gods and those of an invading people.