addition

EDIT 1 DISCLAIMER: I know that +++ is not really an operator but the + and ++ operators without a space. I also know that there's no reason to use this; this question is just out of curiosity. So, I'm interested to see if the space between + and ++var is required in Java. Here is my test code: int i = 0; System.out.println(i); i = i +++i; System.out.println(i); This prints out: 0 1 which works as I would expect, just as if there were a space between the first and second + . Then, I tried it with string concatenation: String s1 = "s " + ++i; System.out.println(s1); // String s2 = "s " +++i;

Sort of trying out some quirks in javascript: First I did console.log("5" + 1); This prints 51, this is normal right, both number and string have a + operator, but since string is the first variable it will convert 1 to a string. Now when I did this: console.log(1 + "5") I expected output to be 6, as I thought it would convert string to a number. However, the magic output was 15. Could anyone more experienced in javascript brighten this up for me? Quoting ECMAScript spec The Addition operator ( + ) section : If Type(lprim) is String or Type(rprim) is String, then Return the String that is the

I am working a simple beginner loop program, where i am trying to get an integer input from the user and calculate the sum. I have a simple menu in which the user can select one of for options, the first being inputting a number and second showing the sum of the last two numbers entered. So i need the program to add the previous two entered numbers together. so if the user selects option 1, they can enter a number and then be returned to the menu where they have to select option 1 again to enter another. option 2 should then calculate the sum and return the value.now lets say the user enters a

I need to create a method in C using bitwise operations which checks if x + y will overflow or not. I can only use a maximum of 20 of the following operations; ! ~ & ^ | + << >> Keep in mind I have to test for both negative and positive numbers. I've tried several times to make it work. Is my logic sound? I'm going by: if (x + y) is less than x, then it has overflowed. Based on that logic, I wrote this; int addOK(int x, int y) { int sum = x + y; int nx = ((~x) + 1); int check = (sum + nx)>>31; return !check; } Thank you! This should work, but it doesn't use only bitwise operator, but it work

I want to add the four components of an SSE register to get a single float. This is how I do it now: float a[4]; _mm_storeu_ps(a, foo128); float x = a[0] + a[1] + a[2] + a[3]; Is there an SSE instruction that directly achieves this? You could probably use the HADDPS SSE3 instruction, or its compiler intrinsic _mm_hadd_ps , For example, see http://msdn.microsoft.com/en-us/library/yd9wecaa(v=vs.80).aspx If you have two registers v1 and v2 : v = _mm_hadd_ps(v1, v2); v = _mm_hadd_ps(v, v); Now, v[0] contains the sum of v1's components, and v[1] contains the sum of v2's components. If you want your

Does anyone know the steps for dividing unsigned binary integers using non-restoring division? It's hard to find any good sources online. i.e if A = 101110 and B = 010111 how do we find A divided by B in non-restoring division? What do the registers look like in each step? Thanks! (My answer is a little late-reply. But I hope it will be useful for future visitors) Algorithm for Non-restoring division is given in below image : In this problem, Dividend (A) = 101110, ie 46, and Divisor (B) = 010111, ie 23. Initialization : Set Register A = Dividend = 000000 Set Register Q = Dividend = 101110 (