In the expression (i, ++i, 1), the comma used is the comma operator the comma operator (represented by the token ,) is a binary operator that evaluates its first operand and discards the result, and then evaluates the second operand and returns this value (and type). Because it discards its first operand, it is generally … Read more
Try this: string.Equals(a, b, StringComparison.CurrentCultureIgnoreCase);
That’s two operators, = and !, not one. It might be an obfuscated way of writing a = !b; if (a) { // whatever } setting a to the logical inverse of b, and testing whether the result is true (or, equivalently, whether b was false). Or it might be a mistyping of a != … Read more
The * is the splat operator. It expands an Array into a list of arguments, in this case a list of arguments to the Hash.[] method. (To be more precise, it expands any object that responds to to_ary/to_a, or to_a in Ruby 1.9.) To illustrate, the following two statements are equal: method arg1, arg2, arg3 … Read more
TL;DR It’s the NULL safe equal operator. Like the regular = operator, two values are compared and the result is either 0 (not equal) or 1 (equal); in other words: ‘a’ <=> ‘b’ yields 0 and ‘a’ <=> ‘a’ yields 1. Unlike the regular = operator, values of NULL don’t have a special meaning and … Read more
What you have is a float literal without the trailing zero, which you then access the __truediv__ method of. It’s not an operator in itself; the first dot is part of the float value, and the second is the dot operator to access the objects properties and methods. You can reach the same point by … Read more
No, it doesn’t, see: R Language Definition: Operators
Use the If() operator with two arguments (Microsoft documentation): ‘ Variable first is a nullable type. Dim first? As Integer = 3 Dim second As Integer = 6 ‘ Variable first <> Nothing, so its value, 3, is returned. Console.WriteLine(If(first, second)) second = Nothing ‘ Variable first <> Nothing, so the value of first is … Read more
It’s c = i && (&i);, with the second part being redundant, since &i will never evaluate to false. For a user-defined type, where you can actually overload unary operator &, it might be different, but it’s still a very bad idea. If you turn on warnings, you’ll get something like: warning: the address of … Read more
The expression: (expression1, expression2) First expression1 is evaluated, then expression2 is evaluated, and the value of expression2 is returned for the whole expression.