I don’t quite understand the question you are asking. I mean, you have already given the answer yourself, before even asking the question: && binds tighter than = while and binds less tightly than =. So, in the first case, the expression is evaluated as follows: ( a=f(2) ) and ( b=f(4) ) ( a= … Read more
It is impossible for both (?=foo) and (?=baz) to match at the same time. It would require the next character to be both f and b simultaneously which is impossible. Perhaps you want this instead: (?=.*foo)(?=.*baz) This says that foo must appear anywhere and baz must appear anywhere, not necessarily in that order and possibly … Read more
Yes, the && operator in C++ uses short-circuit evaluation so that if bool1 evaluates to false it doesn’t bother evaluating bool2. “Short-circuit evaluation” is the fancy term that you want to Google and look for in indexes. The same happens with the || operator, if bool1 evaluates to true then the whole expression will evaluate … Read more
Both AND and OR operators can shortcut. So && only tries the second expression if the first is true (truth-like, more specifically). The fact that the second operation does stuff (whatever the contents of foo() does) doesn’t matter because it’s not executed unless that first expression evaluates to something truthy. If it is truthy, it … Read more
Basically, the Logical AND operator (&&), will return the value of the second operand if the first is truthy, and it will return the value of the first operand if it is by itself falsy, for example: true && “foo”; // “foo” NaN && “anything”; // NaN 0 && “anything”; // 0 Note that falsy … Read more
The shorter ones are vectorized, meaning they can return a vector, like this: ((-2:2) >= 0) & ((-2:2) <= 0) # [1] FALSE FALSE TRUE FALSE FALSE The longer form evaluates left to right examining only the first element of each vector, so the above gives ((-2:2) >= 0) && ((-2:2) <= 0) # [1] … Read more