integer

问题 I have the following problem. Having a list of integers, I want to split it, into a list of lists, whenever the step between two elements of the original input list is not 1. For example: input = [0, 1, 3, 5, 6, 7], output = [[0, 1], [3], [5, 6, 7]] I wrote the following function, but it's uggly as hell, and I was wondering if anyone of you guys would help me get a nicer solution. I tried to use itertools, but couldn't solve it. Here's my solution: def _get_parts(list_of_indices): lv = list

问题 There is something that baffles me with integer arithmetic in tutorials. To be precise, integer division. The seemingly preferred method is by casting the divisor into a float, then rounding the float to the nearest whole number, then cast that back into integer: #include <cmath> int round_divide_by_float_casting(int a, int b){ return (int) std::roundf( a / (float) b); } Yet this seems like scratching your left ear with your right hand. I use: int round_divide (int a, int b){ return a / b + a

问题 This question already has answers here : What's the first double that deviates from its corresponding long by delta? (4 answers) Which is the first integer that an IEEE 754 float is incapable of representing exactly? (2 answers) Closed 6 years ago . What is the exact range of (contiguous) integers that can be expressed as a double (resp. float?) The reason I ask is because I am curious for questions such as this one when a loss of accuracy will occur. That is What is the least positive

问题 package practicejava; public class Query { public static void main(String[] args) { char ch = 66; System.out.println("character= " + ch); ch++; System.out.println("character = " + ch); } } Technically ch++; and ch=ch+1; are the same but why do I get an error when I write ch=ch+1; instead of ch++; ? 回答1: You need to provide a cast in order to do that : ch = (char) (ch + 1); This is because the expression ch + 1 is is promoted ( upcast ) to an int . In order for you to reassign this expression

问题 In Java the int type is signed, but it has a method that compares two ints as if they were unsigned: public static int compareUnsigned(int x, int y) { return compare(x + MIN_VALUE, y + MIN_VALUE); } It adds Integer.MIN_VALUE to each argument, then calls the normal signed comparison method, which is: public static int compare(int x, int y) { return (x < y) ? -1 : ((x == y) ? 0 : 1); } How does adding MIN_VALUE to each argument magically make the comparison unsigned? 回答1: This technique works

问题 I am wondering how to take an EditText area that the user can enter a number in, then make it an integer so it can be used in the program for adding, subtracting, dividing, etc. Basicaly I need the test the enter to be able to be used in a calculator which will be within the code and then it needs to be put into a TextView or string so the final answer can be seen by the user. More info: I have three EditText areas and the user has to fill in all three then press "equals" and it will tell

问题 I have come across with the following two codes. Why does it not throw an exception for floating point where as in other case it will throw a runtime exception. class FloatingPoint { public static void main(String [] args) { float a=1000f; float b=a/0; System.out.println("b=" +b); } } OUTPUT:b=Infinity . If I try with int values then it will throw a runtime exception. Why is it like this? 回答1: The short answer Integral types (JLS 4.2.1) are categorically different from floating point types

问题 I often have to convert a retreived value (usually as a string) - and then convert it to an int. But in C# (.Net) you have to choose either int16, int32 or int64 - how do you know which one to choose when you don't know how big your retrieved number will be? 回答1: Everyone here who has mentioned that declaring an Int16 saves ram should get a downvote. The answer to your question is to use the keyword "int" (or if you feel like it, use "Int32"). That gives you a range of up to 2.4 billion

问题 I have a small problem. I have numbers like 5421, -1 and 1. I need to print them in four bytes, like: 5421 -> 0x0000152D -1 -> 0xFFFFFFFF 1 -> 0x00000001 Also, I have floating point numbers like 1.2, 58.654: 8.25f -> 0x41040000 8.26 -> 0x410428f6 0.7 -> 0x3f333333 I need convert both types of numbers into their hexadecimal version, but they must be exactly four bytes long (four pairs of hexadecimal digits). Does anybody know how is this possible in Java? Please help. 回答1: Here are two

问题 Okay, so I'm admittedly a newbie to programming, but I can't determine how to get python v3.2 to generate a random positive integer between parameters I've given it. Just so you can understand the context, I'm trying to create a guessing-game where the user inputs parameters (say 1 to 50), and the computer generates a random number between the given numbers. The user would then have to guess the value that the computer has chosen. I've searched long and hard, but all of the solutions I can