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Michael Cook
Full Stack JavaScript Techdegree Graduate 28,975 PointsDo I understand this?
If I understand it correctly, the someCount variable does not reset because when the outer function is called and set to a variable, that variable itself contains the state - the someCount variable. Then the returned inner function can be called repeatedly, modifying the state contained in the variable. Is that right?
1 Answer
Brady Snuggs
9,603 PointsYou understand it correctly.
I would add that the variable that's initialized with the outer-function serves as a memory address, like that of a class, to that new instance of the outer-function. Every new variable name, that has the outer-function called, is a new memory address that contains its own 'counter' variable - as well as every inner function that's modifying its respective counter variable.
So, another way of putting it would be, each memory address created with a new variable has its own 'counter' state that is being modified each time the inner-function is executed like you described.