Another approach to convert an unsigned integer to a string and write it: section .text global _start _start: mov eax, 1234567890 mov ebx, 5 add eax, ebx ; Convert EAX to ASCII and store it onto the stack sub esp, 16 ; reserve space on the stack mov ecx, 10 mov ebx, 16 .L1: xor … Read more
X-Code 4.2 will have iOS 5 simulator and library only. If you want lower version simulator and library with X-Code just goto X-Code->Prefrences-> Downloads Tab. In downloads tab you’ll have two tabs: 1) Components – Here you will have option to download iOS 4.3 simulator(~600 Mb), iOS 4.0-4.1(~670 MB) Device debugging support, iOS 3.0-3.2.2(686.3 MB) … Read more
It’s not a JS problem but a more general computer one. Floating number can’t store properly all decimal numbers, because they store stuff in binary For example: 0.5 is store as b0.1 but 0.1 = 1/10 so it’s 1/16 + (1/10-1/16) = 1/16 + 0.0375 0.0375 = 1/32 + (0.0375-1/32) = 1/32 + 00625 … … Read more
Let’s begin by looking the multiplication code. The idea is actually pretty clever. Suppose that you have n1 and n2 written in binary. Then you can think of n1 as a sum of powers of two: n2 = c30 230 + c29 229 + … + c1 21 + c0 20, where each ci is … Read more
+ and . have the same operator precedence, but are left associative. Means after the first concatenation: ‘(‘ . (int)$data[‘event_id’] The string got added with your key, e.g. “($data[‘event_id’]” + $key So the string gets converted into an integer in that numerical context and disappears. To solve this use parentheses () around your addition.
Use Integer.parseInt(String, int radix). public static String addBinary(){ // The two input Strings, containing the binary representation of the two values: String input0 = “1010”; String input1 = “10”; // Use as radix 2 because it’s binary int number0 = Integer.parseInt(input0, 2); int number1 = Integer.parseInt(input1, 2); int sum = number0 + number1; return Integer.toBinaryString(sum); … Read more
In C, with bitwise operators: #include<stdio.h> int add(int x, int y) { int a, b; do { a = x & y; b = x ^ y; x = a << 1; y = b; } while (a); return b; } int main( void ){ printf( “2 + 3 = %d”, add(2,3)); return 0; } … Read more
Your code concatenates three strings, then converts the result to a number. You need to convert each variable to a number by calling parseFloat() around each one. total = parseFloat(myInt1) + parseFloat(myInt2) + parseFloat(myInt3);
UPDATE: TensorFlow 1.0 includes a tf.scatter_nd() operator, which can be used to create delta below without creating a tf.SparseTensor. This is actually surprisingly tricky with the existing ops! Perhaps somebody can suggest a nicer way to wrap up the following, but here’s one way to do it. Let’s say you have a tf.constant() tensor: c … Read more
PHP has no operator overloading* so + with objects makes PHP trying it to convert them to string first, but DateInterval does not support that: interval 1: 03:05 interval 2: 05:00 Total interval : 08:05 Instead you need to create a new DateTime object, then use the add function to add the intervals and finally … Read more