boolean-logic

False or None vs. None or False

丶灬走出姿态 提交于 2019-11-28 04:17:25
In [20]: print None or False -------> print(None or False) False In [21]: print False or None -------> print(False or None) None This behaviour confuses me. Could someone explain to me why is this happening like this? I expected them to both behave the same. The expression x or y evaluates to x if x is true, or y if x is false. Note that "true" and "false" in the above sentence are talking about "truthiness", not the fixed values True and False . Something that is "true" makes an if statement succeed; something that's "false" makes it fail. "false" values include False , None , 0 and [] (an

Why is JavaScript bitwise OR behaving strangely?

别等时光非礼了梦想. 提交于 2019-11-28 03:54:29
问题 In JavaScript, it seems: (4294958077 | 0) == -9219 Why is it not 4294958077 ? It suggests that there's some sort of overflow kicking in (although as I understand it a JavaScript Number type's range is +/- 9007199254740992 so that's odd in itself.) Even if it was an overflow, surely (4294958077 | 0) == 4294958077 should evaluate as true - but it doesn't. Help please 回答1: It has nothing to do with floating point type or overflows. It returns -9219 because the standard mandates it, as all binary

Easiest way to flip a boolean value?

China☆狼群 提交于 2019-11-28 03:17:12
I just want to flip a boolean based on what it already is. If it's true - make it false. If it's false - make it true. Here is my code excerpt: switch(wParam) { case VK_F11: if (flipVal == true) { flipVal = false; } else { flipVal = true; } break; case VK_F12: if (otherVal == true) { otherValVal = false; } else { otherVal = true; } break; default: break; } John T You can flip a value like so: myVal = !myVal; so your code would shorten down to: switch(wParam) { case VK_F11: flipVal = !flipVal; break; case VK_F12: otherVal = !otherVal; break; default: break; } Clearly you need a factory pattern!

Any good boolean expression simplifiers out there? [closed]

允我心安 提交于 2019-11-28 02:58:31
I was refactoring old code and encountered several IF conditions that were way too complex and long and I'm certain they can be simplified. My guess is that those conditions grew so much because of later modifications. Anyway, I was wondering if any of you know of a good online simplifier I can use. I'm not interested in any specific language, just a simplifier that would take in for example: ((A OR B) AND (!B AND C) OR C) And give me a simplified version of the expression, if any. I've looked at the other similar questions but none point me to a good simplifier. Thanks. 500 - Internal Server

How to understand De Morgan Laws Boolean Expression

我们两清 提交于 2019-11-28 02:03:23
I got screwed when trying to understand this expression. I've thought several times but I cant get the meaning. ! (p || q) is equivalent to !p && !q For this one, somehow I can comprehend a little bit. My understanding is " Not (p q) = not p and not q" which is understandable ! (p && q) is equivalent to !p || !q For the second, I'm totally got screwed. How come My understanding is " Not (p q) = Not p or Not q " . How come and and or can be equivalent each other? as for the rule in the truth table between && and || is different. That's how I comprehend each expression, perhaps I have the wrong

What is the difference between Verilog ! and ~?

会有一股神秘感。 提交于 2019-11-27 23:44:33
问题 So it ended up that the bug that had kept me on for days, was a section of code that should have evaluated to False evaluating to True. My initial code went something like: if(~x && ~y) begin //do stuff end i.e. If x is NOT ONE and y is NOT ONE then do stuff. Stepping through the debugger, I realized even though x was 1 the expression in the if-statement still resulted into TRUE and the subsequent code was executed. However, when I changed the statement to: if(x == 0 && y == 0) begin //do

Standard SQL boolean operator IS vs. equals (=) operator

那年仲夏 提交于 2019-11-27 23:08:42
问题 On the Wikipedia page for SQL there are some truth tables about boolean logic in SQL. [1] The Wikipedia page seems to source the SQL:2003 standard. The truth table for the equals operator (=) is different from the IS operator from the SQL:2003 draft. Also, the Wikipedia article notes that "IS NULL" (<null predicate>) is a special case. In the SQL:2003 it seems that there is an "IS" opeartor which is a regular operator like AND, NOT and OR. However, the <null predicate> is still there. Why is

Should I always use the AndAlso and OrElse operators?

懵懂的女人 提交于 2019-11-27 14:33:55
Is there ever a circumstance in which I would not want to use the AndAlso operator rather than the And operator? …or in which I would not want to use the OrElse operator rather than the Or operator? From MSDN : Short-Circuiting Trade-Offs Short-circuiting can improve performance by not evaluating an expression that cannot alter the result of the logical operation. However, if that expression performs additional actions, short-circuiting skips those actions. For example, if the expression includes a call to a Function procedure, that procedure is not called if the expression is short-circuited,

why are Javascript negative numbers not always true or false?

别等时光非礼了梦想. 提交于 2019-11-27 13:01:20
-1 == true; //false -1 == false //false -1 ? true : false; //true Can anyone explain the above output? I know I could work round this by comparing to 0 but I'm interested. I'd expect at least one of the sloppy equals statements to be true as they do implicit type conversion, and I certainly didn't expect the ternary to come up with a totally different result. Andy E In the first two cases, the boolean is cast to a number - 1 for true and 0 for false . In the final case, it is a number that is cast to a boolean and any number except for 0 and NaN will cast to true . So your test cases are

Naming of TypeScript's union and intersection types

北战南征 提交于 2019-11-27 11:31:29
问题 I can't understand the logic behind the terms union types and intersection types in TypeScript. Pragmatically, if the properties of different types are sets of properties, if I combine them with the & operator, the resulting type will be the union of the of those sets. Following that logic, I would expect types like this to be called union types . If I combine them with | , I can only use the common properties of them, the intersection of the sets. Wikipedia seems to back that logic: The