The extend() method appends to the existing array and returns None. In your case, you are creating an array — [4, 5, 6] — on the fly, extending it and then discarding it. The variable b ends up with the return value of None.
The Framework Design Guidelines suggest returning a copy of the Array. That way, consumers can’t change items from the array. // bad code // could still do Path.InvalidPathChars[0] = ‘A’; public sealed class Path { public static readonly char[] InvalidPathChars = { ‘\”‘, ‘<‘, ‘>’, ‘|’ }; } these are better: public static ReadOnlyCollection<char> GetInvalidPathChars(){ … Read more
String#hashCode: private int hash; … public int hashCode() { int h = hash; if (h == 0 && count > 0) { int off = offset; char val[] = value; int len = count; for (int i = 0; i < len; i++) { h = 31*h + val[off++]; } hash = h; } return … Read more
The way you proposed is perfectly fine, except if in your code you need to make assignment of RockSolid variables, like this: RockSolid a(0,1); RockSolid b(0,1); a = b; This would not work as the copy assignment operator would have been deleted by the compiler. So an alternative is to rewrite your struct as a … Read more
Thanks to @cxrodgers’s comment, I think the fastest way to do this is: df.index = df.index.set_levels(df.index.levels[0].str.replace(‘ ‘, ”), level=0) Old, longer answer: I found that the list comprehension suggested by @Shovalt works but felt slow on my machine (using a dataframe with >10,000 rows). Instead, I was able to use .set_levels method, which was quite … Read more
const, in Rust, is short for constant and is related to compile-time evaluation. It shows up: when declaring constants: const FOO: usize = 3; when declaring compile-time evaluable functions: const fn foo() -> &’static str These kinds of values can be used as generic parameters: [u8; FOO]. For now this is limited to array size, … Read more
Update: You could remove a property from an object with a tricky Destructuring assignment: const doSomething = (obj, prop) => { let {[prop]: omit, …res} = obj return res } Though, if property name you want to remove is static, then you could remove it with a simple one-liner: let {lastname, …o2} = o The … Read more
In order to use an effectively immutable object without final fields in a thread safe manner you need to use one of safe publication idioms when making the object available to other threads after initialization, otherwise these threads can see the object in partially initialized state (from Java Concurrency in Practice): Initializing an object reference … Read more
Yes, you need to override __new__ special method: class MySet(frozenset): def __new__(cls, *args): if args and isinstance (args[0], basestring): args = (args[0].split (),) + args[1:] return super (MySet, cls).__new__(cls, *args) print MySet (‘foo bar baz’) And the output is: MySet([‘baz’, ‘foo’, ‘bar’])
An immutable type is a type of which its properties can only be set at initialization. Once an object is created, nothing can be changed anymore. An immutable property is simply a read-only property. In the following example, ImmutableType is an immutable type with one property Test. Test is a read-only property. It can only … Read more