logical-operators

问题 I have a doubt in the program below. int main() { int i = -3,j = 2, k = 0,m; m = ++i || ++j && ++k; printf("%d %d %d %d\n", i, j, k, m); return 0; } I get the output as -2 2 0 1 . In OR operation if 1st value is true then it won't evaluate the 2nd one so i = -2 and j =2 . Then comes the AND operation . It will check for both the value to be true.So if k = 1 then m = 1 . So the output should be -2 2 1 1 . I run and check and got output as -2 2 0 1 but I could not understand how. 回答1: You used

问题 I had the need to code a statement of the form a = a || expr; where expr should be evaluated and the result be assigned to a iff a is not set. this relies on the logical OR's short-circuiting capabilities. The shorter way to write the above would, of course, be a ||= expr; but (to my surprise) C does not have logical assignment operators. So my question is twofold. First, is there a shorter way to write the first statement in standard C (the ternary operator is even worse - a = a ? a : expr

问题 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

问题 This question already has answers here : Why do “Not a Number” values equal True when cast as boolean in Python/Numpy? (4 answers) Closed 2 years ago . I wonder why does python pandas / numpy not implement 3-valued logic (so-called Łukasiewicz's logic) with true, false and NA (like for instance R does). I've read (https://www.oreilly.com/learning/handling-missing-data) that this is to some extent due to the fact that pandas uses much more many basic data types than R for example. However,

问题 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.

问题 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? 回答1: int c = a || --b; In this line, the C standard requires the C

问题 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

问题 I have this simple line of code: i = " " if i != "" or i != " ": print("Something") This should be simple, if i is not empty "" OR it's not a space " " , but it is, print Something. Now, why I see Something printed if one of those 2 conditions is False ? 回答1: De Morgan's laws, "not (A and B)" is the same as "(not A) or (not B)" also, "not (A or B)" is the same as "(not A) and (not B)". In your case, as per the first statement, you have effectively written if not (i == "" and i == " "): which

问题 Why does the first line return TRUE, and the third line returns 1? I would expect both lines to return 1. What is the exact meaning of those extra two parentheses in the third line? !is.na(5) + !is.na(NA) # TRUE (!is.na(5)) + (!is.na(NA)) # 1 edit: should check these multiple times. The original problem was with !is.na() , thought it replicated for is.na() . But it didn't :) 回答1: ! has a weird, counter-intuitive precedence in R. Your first code is equivalent to !(is.na(5) + !is.na(NA)) That

问题 tl;dr: Is there a non-short circuit logical AND in C++ (similar to &&)? I've got 2 functions that I want to call, and use the return values to figure out the return value of a 3rd composite function. The issue is that I always want both functions to evaluate (as they output log information about the state of the system) IE: bool Func1(int x, int y){ if( x > y){ cout << "ERROR- X > Y" << endl; } } bool Func2(int z, int q){ if( q * 3 < z){ cout << "ERROR- Q < Z/3" << endl; } } bool Func3(int x,