When you update "x" with x = 2000, a fresh integer object is designed plus the dictionary is up to date to stage at The brand new item. The aged just one thousand object is unchanged (and might or might not be alive according to no matter if anything refers to the item).
When you go an object on the functionality, the thing by itself is handed (item in Python is actually what you'd simply call a worth in other programming languages) not the reference to this object. Quite simply, after you contact:
which of course will never change the thing handed into the perform. Should the function looked similar to this:
As a further, unrealted Notice, I might declare that implementing a string comparison function in this type of harmful style (it modifies
So Should the point you should alter can be a mutable item, there is no challenge, as anything is efficiently handed by reference.
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Nonetheless, const or not, the parameter maintains the reference to the object and reference can't be assigned to stage to a unique object throughout the referred to as function.
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Many insights in answers in this article, but I believe an additional place is just not clearly stated right here explicitly. Quoting from python documentation "In Python, variables which are only referenced inside a purpose are implicitly worldwide.
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The blue strong arrow may be the reference price which was handed into the function (by way of the dashed blue arrow path). The unpleasant darkish yellow is The interior dictionary. (It essentially can be drawn also as a green ellipse. The colour and the shape only says it truly is inside.)
You are obtaining the error for the reason that strToLower returns a char. A char is really a kind of integer in C. That you are assigning it into a char[] that is a pointer. That's why "changing integer to pointer".
It's no more pointing to self.variable. The subsequent code snippet shows what would take place in case you modify the info composition pointed to by var and self.variable, in this case a listing:
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