For Eric Hogue (see question comments): origDiv := Number getSlot(“https://stackoverflow.com/”) 10 origDiv(5) println # => 2 10 origDiv(0) println # => inf Number / := method (i, if (i != 0, self origDiv(i), 0) ) (10 / 5) println # => 2 (10 / 0) println # => 0
It’s the same operator. The difference is merely one of style. Consult your project documentation, or ask your boss, or your wife, or flip a coin.
The manpage for bash v3.2.48 says: […] The format for arithmetic expansion is: $((expression)) The old format $[expression] is deprecated and will be removed in upcoming versions of bash. So $[…] is old syntax that should not be used anymore.
C# compiler converts overloaded operator to functions with name op_XXXX where XXXX is the operation. For example, operator + is compiled as op_Addition. Here is the full list of overloadable operators and their respective method names: ┌──────────────────────────┬───────────────────────┬──────────────────────────┐ │ Operator │ Method Name │ Description │ ├──────────────────────────┼───────────────────────┼──────────────────────────┤ │ operator + │ op_UnaryPlus │ Unary │ │ … Read more
Try this: [model1, model2].map(&:valid?).all? It’ll return true if both are valid, and create the errors on both instances.
Of course you can use IfThen(SomeBooleanExpression, IfTrueReturnValue, IfFalseReturnValue) where the return values are numeric (uses Math) or string (uses StrUtils). But notice that this will evaluate both arguments in all cases — there is no lazy evaluation, so it is not as efficient as the ?: operator in C#, where only the right operand is … Read more
That is a bitwise or with assignment. It is equivalent to object.attribute = object.attribute | variable Read more here.
Another solution if You don’t want to return from function just yet: x?.let{ y = it } Which checks if x is non-null then passes it as the only parameter to the lambda block. This is also a safe call in case your x is a var.
From the MSDN library for +=: Using this operator is almost the same as specifying result = result + expression, except that result is only evaluated once. So they are not identical and that is why x += 1 will be more efficient. Update: I just noticed that my MSDN Library link was to the … Read more
I can’t give you the rationale, but I can tell why the compiler has that behavior from the stand point of the rules the compiler has to follow (which might not really be what you’re interesting in knowing). From an old copy of the C# spec (I should probably download a newer version), emphasis added: … Read more