comparison

问题 I've faced a weird case with a double variable after adding some values to it. The problem occurs when adding (0.2) to a double variable more than one time -I think it only happens with (0.2)- for example: consider this code: Dim i As Double = 2 i = i + 0.2 MsgBox(i) '2.2 MsgBox(i > 2.2) 'False >> No problem But if I add (0.2) more than one time: Dim i As Double = 2 i = i + 0.2 i = i + 0.2 MsgBox(i) '2.4 Msgbox(i > 2.4) 'True >> !!!! Also Dim i As Double = 2 For x As Integer = 1 to 5 i = i +

问题 I have two objects. key1 is of type *rsa.PublicKey . key2 is of type *ssh.PublicKey which is an interface hiding a *ssh.rsaPublicKey object. ssh.rsaPublicKey is defined as: type ssh.rsaPublicKey rsa.PublicKey And it has a few extra methods. However, I can't cast either key to an ssh.rsaPublicKey since that class is "not exported", I can't cast key2 to an rsa.PublicKey since that doesn't implement ssh.PublicKey , and I can't access the N or e from key2 because I am not supposed to know I have

问题 I tried doing some floating point comparison and here is what I found: 130 === 130.000000000000014210854715 // true 130 === 130.000000000000014210854716 // false 9 === 9.0000000000000008881784197001 // true 9 === 9.0000000000000008881784197002 // false 0.1 === 0.100000000000000012490009027033 // true 0.1 === 0.100000000000000012490009027034 // false I tried running those on Firefox and Chrome with the same results. Okay, I KNOW that floating point comparison is a bad practice and has

问题 If I want to compare two values for equality there are a number of options, such as: eq for symbols = for numbers char-equal for characters string-equal for strings eql for symbols, numbers and strings equal for everything but symbols … (And I hope I got this right so far.) Now, as a Lisp beginner, my question is: Why is that? Is this just for "historical reasons" or is there a real benefit of having all these possibilities? I know that why -questions are always hard to answer and may be

问题 Consider this: set A: 1 2 3 4 set B: 3 4 5 6 set C: 4 5 6 7 set D: 1 I want to compare D with the rest and get as a result a set of numbers as most relevant. The result should be in this order: 4 (as D has a common number with A and 4 is in A and also in B and C), 3 (as D has a common number with A and 3 is in A and B), 2 (as D has a common number with A and 2 is also in A), then 5, 6, 7. Is there some algorithm to do this in an efficient way in PHP/MySQL? I don't want to reinvent the wheel,

问题 Update: Unfortunately, the answer provided here did not give me what I wanted - My mistake for accepting too quickly without really doing a deep enough check. I have reported the question here. Suppose I have the following two tables in my SQL Server (2012) DB: Tbl1: ID: Col1: Col2: Col3: 1 Val1 Val2 Val3 2 <NULL> Val2 <NULL> 3 Val1 <NULL> Val3 4 Val1 <NULL> Val3 Tbl2: ID: Col1: Col2: Col3: 1 Val1 Val2 Val3 2 <NULL> Val2 <NULL> 3 <NULL> <NULL> Val3 5 <NULL> <NULL> Val3 And, at this point, all

问题 I am new to iPhone developer, i want to compare two Dates, first date will come from Database let say, newDate and second is todays date, if today is greater then newDate then i want to display alert. Here is my code snippet, NSString *newDate = [formatter stringFromDate:st]; NSDate *today=[[NSDate alloc]init]; if ([today compare:newDate] == NSOrderedDescending) { NSLog(@"today is later than newDate"); } else if ([today compare:newDate] == NSOrderedAscending) { NSLog(@"today is earlier than

问题 If I have 20 pairs of coordinates, whose x and y values are say : x y 27 182 180 81 154 52 183 24 124 168 146 11 16 90 184 153 138 133 122 79 192 183 39 25 194 63 129 107 115 161 33 14 47 65 65 2 1 124 93 79 Now if I randomly generate 15 pairs of coordinates (x,y) and want to compare with these 20 pairs of coordinates given above, how can I do that most efficiently without nested loops? 回答1: If you're trying to see if any of your 15 randomly generated coordinate pairs are equal to any of your

问题 This question already has answers here : Compare two dates with JavaScript (36 answers) Closed 4 years ago . Here is my javascript code: var prevDate = new Date('1/25/2011'); // the string contains a date which // comes from a server-side script // may/may not be the same as current date var currDate = new Date(); // this variable contains current date currDate.setHours(0, 0, 0, 0); // the time portion is zeroed-out console.log(prevDate); // Tue Jan 25 2011 00:00:00 GMT+0500 (West Asia

问题 I have to compare two strings such as INTU and IXTE and check if two or more of the characters are the same. With the previous two strings, I'd want to return true , since the I and the T are the same. Order of letters in the string ends up being irrelevant as each character can not appear in different positions in the string. It seems like there should be an easy way to do this. 回答1: look at similar_text() . The following code is untested but i think it would work as you want. $a = "INTU";