Case 1: String str = “Good”; str = str + ” Morning”; In the above code you create 3 String Objects. “Good” it goes into the String Pool. ” Morning” it goes into the String Pool as well. “Good Morning” created by concatenating “Good” and ” Morning”. This guy goes on the Heap. Note: Strings … Read more
It allows the differentiation of bitwise const and logical const. Logical const is when an object doesn’t change in a way that is visible through the public interface, like your locking example. Another example would be a class that computes a value the first time it is requested, and caches the result. Since c++11 mutable … Read more
You may want to re-read The Rust Programming Language, specifically the sections on: mutability mutable references vectors We can also iterate over mutable references to each element in a mutable vector in order to make changes to all the elements. The for loop in Listing 8-9 will add 50 to each element. let mut v … Read more
Yes, it is a bad idea. The reason is that if in-place and non-in-place operations have apparently identical output, then programmers will frequently mix up in-place operations and non-in-place operations (List.sort() vs. sorted()) and that results in hard-to-detect errors. In-place operations returning themselves can allow you to perform “method chaining”, however, this is bad practice … Read more
Making integers mutable would be very counter-intuitive to the way we are used to working with them. Consider this code fragment: a = 1 # assign 1 to a b = a+2 # assign 3 to b, leave a at 1 After these assignments are executed we expect a to have the value 1 and … Read more
I can only support what gradbot said – when I need mutation, I prefer let mutable. Regarding the implementation and differences between the two – ref cells are essentially implemented by a very simple record that contains a mutable record field. You could write them easily yourself: type ref<‘T> = // ‘ { mutable value … Read more
You created a new list object and bound it to the same name, x. You never mutated the existing list object bound to x at the start. Names in Python are just references. Assignment is binding a name to an object. When you assign to x again, you are pointing that reference to a different … Read more
struct are value types, thus in the for loop you are dealing with a copy. Just as a test you might try this: Swift 3: struct Options { var backgroundColor = UIColor.black } var arrayOfMyStruct = [Options]() for (index, _) in arrayOfMyStruct.enumerated() { arrayOfMyStruct[index].backgroundColor = UIColor.red } Swift 2: struct Options { var backgroundColor = … Read more
The first way: def change(array): array.append(4) change(array) is the most idiomatic way to do it. Generally, in python, we expect a function to either mutate the arguments, or return something1. The reason for this is because if a function doesn’t return anything, then it makes it abundantly clear that the function must have had some … Read more
This is called the ‘mutable defaults trap’. See: http://www.ferg.org/projects/python_gotchas.html#contents_item_6 Basically, a_list is initialized when the program is first interpreted, not each time you call the function (as you might expect from other languages). So you’re not getting a new list each time you call the function, but you’re reusing the same one. I guess the … Read more