logical-operators

Does JavaScript have non-shortcircuiting boolean operators?

笑着哭i 提交于 2019-11-30 00:24:00
问题 In JavaScript (f1() || f2()) won't execute f2 if f1 returns true which is usually a good thing except for when it isn't. Is there a version of || that doesn't short circuit? Something like var or = function(f, g){var a = f(); var b = g(); return a||b;} 回答1: Nope, JavaScript is not like Java and the only logical operators are the short-circuited https://developer.mozilla.org/en/JavaScript/Reference/Operators/Logical_Operators Maybe this could help you: http://cdmckay.org/blog/2010/09/09/eager

Logical vs bitwise

时光总嘲笑我的痴心妄想 提交于 2019-11-29 22:30:57
问题 What the different between logical operators and , or and bitwise analogs & , | in usage? Is there any difference in efficiency in various solutions? 回答1: Logical operators operate on logical values, while bitwise operators operate on integer bits. Stop thinking about performance, and use them for they're meant for. if x and y: # logical operation ... z = z & 0xFF # bitwise operation 回答2: Logical operators are used for booleans, since true equals 1 and false equals 0. If you use (binary)

If statements and && or ||

不羁的心 提交于 2019-11-29 17:05:16
OK so I'm a beginner with C# and I am having trouble understanding the if statement below. For this example INumber is declared as 8, dPrice is 0 and dTAX is 10. if (iNumber != 8 || iNumber != 9) { dPrice = dPrice + dTAX; } Can somebody explain why it is entering the statement and adding the 10 from dTAX to dPrice? I know changing it to && works, but why? As I understand it, it should only enter the If statement, if iNumber does not equal 8 or 9, which here it does, so it should not enter. Here are my outputs after running it through the || if statement. Inumber is: 8 dPrice was: 0 dPrice is

R: A function to tell whether a single char is vowel or not

时光毁灭记忆、已成空白 提交于 2019-11-29 16:55:25
I'm dealing with if in R and I'm struggling with one of the typical examples: checking if it is a vowel= TRUE and FALSE otherwise. if.function <- function(char){ if (char=('a') or ('e') or ('i') or ('o') or ('u') ) { return(TRUE) } else if (char == 0){ return(FALSE) } Could someone give me a hand? I saw other examples with Python and Java, but I barely know how to use just R. char=('a') or ('e') or ('i') or ('o') or ('u') is illegal. Try isVowel <- function(char) char %in% c('a', 'e', 'i', 'o', 'u') Let's try it: isVowel('a') # [1] TRUE isVowel('b') # [1] FALSE Note that I did not use the or

How is it possible that BITWISE AND operation to take more CPU clocks than ARITHMETIC ADDITION operation in a C program?

半城伤御伤魂 提交于 2019-11-29 16:40:53
I wanted to test if bitwise operations really are faster to execute than arithmetic operation. I thought they were. I wrote a small C program to test this hypothesis and to my surprise the addition takes less on average than bitwise AND operation. This is surprising to me and I cannot understand why this is happening. From what I know for addition the carry from the less significant bits should be carried to the next bits because the result depends on the carry too. It does not make sense to me that a logic operator is slower than addition. My cod is below: #include<stdio.h> #include<time.h>

Shortcircuiting of AND in case of increment / decrement operator

孤街浪徒 提交于 2019-11-29 16:03:39
In the code below: #include <stdio.h> int main() { int a = 1; int b = 1; int c = a || --b; int d = a-- && --b; printf("a = %d, b = %d, c = %d, d = %d", a, b, c, d); return 0; } i was expecting the output to be: a=0,b=1,c=1,d=0 because due to short circuiting in the line below, ie a-- returns 0 so the other part wont get executed right? int d = a-- && --b; The output is: a = 0, b = 0, c = 1, d = 0 can anyone please explain? int c = a || --b; In this line, the C standard requires the C implementation to evaluate a first and, if it is not zero, not to evaluate --b . Although -- has higher

What is the type of the logical operators?

瘦欲@ 提交于 2019-11-29 15:41:32
I want to use them as a parameter to a method of my Region struct: private func combineWith(region: RegionProtocol, combine: (Bool, Bool) -> Bool) -> Region { return Region() {point in combine(self.contains(point), region.contains(point)) } } But apparently, (Bool, Bool) -> Bool) is not what && or || are. If you know, please let me know how you found out. If you "cmd-click" on the word "Swift" in the statement import Swift in Xcode and search for || then you'll find that it is declared as func ||<T : BooleanType>(lhs: T, rhs: @autoclosure () -> Bool) -> Bool The reason is the "short-circuiting

Logical short-circuit inside a function handle

你。 提交于 2019-11-29 13:53:19
I have a function handle that operates on 2d arrays of arbitrary size: R2T = @(DL1,DL2) arrayfun(@(DL1,DL2)... 1/(fzero(@(x)fFitObj1(x)./fFitObj2(x)-... DL1./DL2,[minLim maxLim])) ... ,DL1,DL2) - C1; Here's a bottom-up breakdown of what it does: fzero(@(x)fFitObj1(x)./fFitObj2(x)-DL1./DL2,[minLim maxLim]) - This bit looks for a zero of the considered function on the interval [minLim maxLim] , where fFitObj1 and fFitObj2 are function handles available from before, C1 is some known constant and DL1, DL2 are provided. @(DL1,DL2)1/(fzero(...)) - a wrapper for fzero that allows DL1 and DL2 to be

Using logical operators in building a Pandas DataFrame

房东的猫 提交于 2019-11-29 12:15:42
I have two snippets of pandas code which I think should be equivalent, but the second one doesn't do what I expect. # snippet 1 data = all_data[[((np.isfinite(all_data[self.design_metric][i]) and all_data['Source'][i] == 2)) or ((np.isfinite(all_data[self.actual_metric][i]) and all_data['Source'][i] != 2)) for i in range(len(all_data))]] # snippet 2 data = all_data[(all_data['Source'] == 2 & np.isfinite(all_data[self.design_metric])) | (all_data['Source'] != 2 & np.isfinite(all_data[self.actual_metric]))] Each section (e.g. all_data['Source'] == 2 ) does what I expect on its own but it seems

&&= and ||= operators [duplicate]

夙愿已清 提交于 2019-11-29 11:40:08
问题 This question already has answers here : Closed 9 years ago . Possible Duplicates: Why doesn't Java have compound assignment versions of the conditional-and and conditional-or operators? (&&=, ||=) Why does a “&&=” Operator not exist? Today at work I wrote the following LOC (the real identities of b and b1 are confidential :) b &&= b1; // meaning b = b && b1; I stared at it for a couple of seconds and realized that there exists no such operator. Just to be sure, I clicked on compile and it