logic

I need to set the test to succeed if one of the two expectations is met: expect(mySpy.mostRecentCall.args[0]).toEqual(jasmine.any(Number)); expect(mySpy.mostRecentCall.args[0]).toEqual(false); I expected it to look like this: expect(mySpy.mostRecentCall.args[0]).toEqual(jasmine.any(Number)).or.toEqual(false); Is there anything I missed in the docs or do I have to write my own matcher? raina77ow Note: This solution contains syntax for versions prior to Jasmine v2.0. For more information on custom matchers now, see: https://jasmine.github.io/2.0/custom_matcher.html Matchers.js works with a

One of my friend was asked this question in an interview - You have given two integer arrays each of size 10. Both contains 9 equal elements (say 1 to 9) Only one element is different. How will you find the different element? What are different approaches you can take? One simple but lengthy approach would be - sort both arrays,go on comparing each element,on false comparison, you'll get your result. So what are different approaches for this? Specify logic as its expected in an interview. Not expecting a particular code in a specific language. Pseudo code will suffice. (Please submit one

I'm building a PHP class with a private member function that returns a string value such as: 'true && true || false' to a public member function. (This string is the result of some regex matching and property lookups.) What I'd like to do is have PHP parse the returned logic and have the aforementioned public function return whether the boolean result of the parsed logic is true or false. I tried eval(), but I get no output at all. I tried typecasting the boolean returns...but there's no way to typecast operators...hehe Any ideas? (Let me know if you need more information.) Just stumbled upon

I am trying to build up subtraction, addition, division, multiplication and other operations using only following ones: incr(x) - Once this function is called it will assign x + 1 to x assign(x, y) - This function will assign the value of y to x (x = y) zero(x) - This function will assign 0 to x (x = 0) loop X { } - operations written within brackets will be executed X times Using following rules it is straight forward to implement addition (add) like this: ADD (x, y) { loop X { y = incr (y) } return y } However, I'm struggling to implement subtraction . I think that all the other needed

Say you are trying to read this property var town = Staff.HomeAddress.Postcode.Town; Somewhere along the chain a null could exist. What would be the best way of reading Town? I have been experimenting with a couple of extension methods... public static T2 IfNotNull<T1, T2>(this T1 t, Func<T1, T2> fn) where T1 : class { return t != null ? fn(t) : default(T2); } var town = staff.HomeAddress.IfNotNull(x => x.Postcode.IfNotNull(y=> y.Town)); or public static T2 TryGet<T1, T2>(this T1 t, Func<T1, T2> fn) where T1 : class { if (t != null) { try { return fn(t); } catch{ } } return default(T2); } var

I'm working on a higher-order theorem prover, of which unification seems to be the most difficult subproblem. If Huet's algorithm is still considered state-of-the-art, does anyone have any links to explanations of it that are written to be understood by a programmer rather than a mathematician? Or even any examples of where it works and the usual first-order algorithm doesn't? Charles Stewart State of the art — yes, so far as I know all algorithms more or less take the same shape as Huet's (I follow theory of logic programming, although my expertise is tangential) provided you need full higher

Line 294 of java.util.Random source says if ((n & -n) == n) // i.e., n is a power of 2 // rest of the code Why is this? The description is not entirely accurate because (0 & -0) == 0 but 0 is not a power of two. A better way to say it is ((n & -n) == n) when n is a power of two, or the negative of a power of two, or zero. If n is a power of two, then n in binary is a single 1 followed by zeros. -n in two's complement is the inverse + 1 so the bits lines up thus n 0000100...000 -n 1111100...000 n & -n 0000100...000 To see why this work, consider two's complement as inverse + 1, -n == ~n + 1 n