short-circuiting

Why doesn't bool? support lifted && and || ? They could have lifted the true and false operators which would have indirectly added lifted && and || . The operators | and & are already lifted and implement the correct Three-valued logic . But of course they are not short circuiting like || and && . The question is why they decided not to lift those operators when creating the specification. So "It's like this because the spec says so" is no answer to the "why?". When lifting true and false so that null is neither true nor false : public static bool operator true(bool? x) { return x.HasValue &&

In the PHP code if(a() && b()) when the first operand evaluates to false , b() will not be evaluated. Similarly, in if (a() || b()) when the first operand evaluates to true , b() will not be evaluated.. Is this true for all languages, like Java, C#, etc? This is the test code we used. <?php function a(){ echo 'a'; return false; } function b(){ echo 'b'; return true; } if(a() && b()){ echo 'c'; } ?> This is called short-circuit evaluation . It is generally true for languages derived from C (C, C++, Java, C#) but not true for all languages. For example, VB6 does not do this, nor was it done in

This question already has an answer here: Is the shortcircuit behaviour of Python's any/all explicit? 4 answers I wish to use the Python all() function to help me compute something, but this something could take substantially longer if the all() does not evaluate as soon as it hits a False . I'm thinking it probably is short-circuit evaluated, but I just wanted to make sure. Also, is there a way to tell in Python how the function gets evaluated? TerryA Yes, it short-circuits: >>> def test(): ... yield True ... print('one') ... yield False ... print('two') ... yield True ... print('three') ...

I know what is short circuit evaluation in C. a && b (operand b is not checked if a = 0) a || b (operand b is not checked if a = non zero) But I am stuck at this question int x = 0; if (5 || 2 && ++x) printf("%d", x); This outputs 0 . My first thinking goes as follows: According to precedence table , precedence is ++ , && , || (descending order) ++x : evaluated.x becomes 1. 2 && ++x evaluated. Both operands are evaluated. || is evaluated. But according to this, 1 should be printed, not 0 . My second thinking goes as this: 5 || anything anything is not evaluated because of short circuit

Why would a language NOT use Short-circuit evaluation ? Are there any benefits of not using it? I see that it could lead to some performances issues... is that true? Why? Related question : Benefits of using short-circuit evaluation RBarryYoung Reasons NOT to use short-circuit evaluation: Because it will behave differently and produce different results if your functions, property Gets or operator methods have side-effects. And this may conflict with: A) Language Standards, B) previous versions of your language, or C) the default assumptions of your languages typical users. These are the

In the Wikipedia page describing short-circuit evaluation , & and | are listed as eager operators in Python. What does this mean and when are they used in the language? The wikipedia page is wrong, I've corrected it. | and & are not boolean operators, even though they are eager operators, which just means that they are not short circuit operators. As you probably know, here's how the python and and or operators work: >>> def talk(x): ... print "Evaluating: ", bool(x) ... return x ... >>> talk(1 == 1) or talk(2 == 1) # 2 == 1 is not evaluated Evaluating: True True >>> talk(1 == 1) and talk(2 ==

I'm new to c++ and am curious how the compiler handles lazy evaluation of booleans. For example, if(A == 1 || B == 2){...} If A does equal 1, is the B==2 part ever evaluated? James McNellis No, the B==2 part is not evaluated. This is called short-circuit evaluation . Edit: As Robert C. Cartaino rightly points out , if the logical operator is overloaded, short-circuit evaluation does not take place (that having been said, why someone would overload a logical operator is beyond me). Unless the || operator is overloaded , the second expression will not be evaluated. This is called "short-circuit