unboxing

creating custom instance of UArray

旧时模样 提交于 2019-11-29 10:41:12
Suppose I have a simple data type like: data Cell = Open | Blocked and I'd like to use a UArray Int Cell . Is there an easy way to do this? Can I somehow reuse the definition for UArray Int Bool ? crockeea This answer explains why Vectors are better than Arrays, so I'm going to give you the answer for unboxed vectors. I did try deriving an MArray and IArray instance for Cell based on the Bool instances, but the Bool instances are quite complicated; it would be at least as ugly as manually deriving an Unbox instance for vectors. Unlike vectors, you also can't just derive Storable and use

how equal operator works with primitive and object type data

断了今生、忘了曾经 提交于 2019-11-29 06:12:17
I know its a very basic question but I want to be clear about the concept. I want to know how == operator works in case of primitive and object type. For example Integer a = 1; int b = 1; System.out.println(a == b) how a is compared with b , whereas a contain the ref of object that contains value 1. Can somebody clearify it to me how it works internally? In general the equality operator in Java performs a so called shallow comparison. In other words it compares the values that variables contains. Now the variables of primitive data types contains the value itself while the reference types

Difference in behaviour of the ternary operator on JDK8 and JDK10

我们两清 提交于 2019-11-28 16:36:42
问题 Consider the following code public class JDK10Test { public static void main(String[] args) { Double d = false ? 1.0 : new HashMap<String, Double>().get("1"); System.out.println(d); } } When running on JDK8, this code prints null whereas on JDK10 this code results in NullPointerException Exception in thread "main" java.lang.NullPointerException at JDK10Test.main(JDK10Test.java:5) The bytecode produced by the compilers is almost identical apart from two additional instructions produced by the

Does passing a value type in an “out” parameter cause the variable to be boxed?

一曲冷凌霜 提交于 2019-11-28 11:59:18
I'm aware that boxing and unboxing are relatively expensive in terms of performance. What I'm wondering is: Does passing a value type to a method's out parameter cause boxing/unboxing of the variable (and thus a performance hit)? Can the compiler optimize this away? int number; bool result = Int32.TryParse(value, out number); As others have pointed out, there's no boxing here. When you pass a variable as an argument corresponding to an out or ref parameter, what you are doing is making an alias to the variable . You are not doing anything to the value of the variable. You're making two

Which is better in terms of performance, implicit (auto) unboxing or explicit unboxing?

穿精又带淫゛_ 提交于 2019-11-28 11:50:13
To put it in code - which has better performance (if there is a difference at all)? Given this: public class Customer { .... public Boolean isVIP(){...} ... } Which is faster? public void handleCustomer(Customer customer) { if (customer.isVIP()) // Auto Unboxing { handleNow(customer); } else { sayHandlingNowButQueueForTomorrow(customer); } } or this: public void handleCustomer(Customer customer) { if (customer.isVIP().booleanValue()) // Explicit unboxing { handleNow(customer); } else { sayHandlingNowButQueueForTomorrow(customer); } } Mark D No difference between them, you can verify it in the

does valueType.ToString() does a cast on the valueType?

家住魔仙堡 提交于 2019-11-28 08:23:06
问题 lets say, i have the following code in c# int x = 0; x.ToString(); does this internally does a boxing of x? Is there a way to see this happening from visual studio? 回答1: In this specific case, you are using a System.Int32 (an int ). That type redefines ToString , Equals and GetHashCode , so no boxing. If you use a struct that doesn't redefine ToString what you'll have is a constrained callvirt to System.Object.ToString() . The definition of constrained: When a callvirt method instruction has

Boxing and unboxing with generics

|▌冷眼眸甩不掉的悲伤 提交于 2019-11-28 03:32:26
The .NET 1.0 way of creating collection of integers (for example) was: ArrayList list = new ArrayList(); list.Add(i); /* boxing */ int j = (int)list[0]; /* unboxing */ The penalty of using this is the lack of type safety and performance due to boxing and unboxing. The .NET 2.0 way is to use generics: List<int> list = new List<int>(); list.Add(i); int j = list[0]; The price of boxing (to my understanding) is the need to create an object on the heap, copy the stack allocated integer to the new object and vice-versa for unboxing. How does the use of generics overcome this? Does the stack

how equal operator works with primitive and object type data

前提是你 提交于 2019-11-27 23:41:30
问题 I know its a very basic question but I want to be clear about the concept. I want to know how == operator works in case of primitive and object type. For example Integer a = 1; int b = 1; System.out.println(a == b) how a is compared with b , whereas a contain the ref of object that contains value 1. Can somebody clearify it to me how it works internally? 回答1: In general the equality operator in Java performs a so called shallow comparison. In other words it compares the values that variables

Why does autoboxing in Java allow me to have 3 possible values for a boolean?

こ雲淡風輕ζ 提交于 2019-11-27 23:12:01
Reference: http://java.sun.com/j2se/1.5.0/docs/guide/language/autoboxing.html "If your program tries to autounbox null, it will throw a NullPointerException." javac will give you a compile-time error if you try to assign null to a boolean. makes sense. assigning null to a Boolean is a-ok though. also makes sense, i guess. but let's think about the fact that you'll get a NPE when trying to autounbox null. what this means is that you can't safely perform boolean operations on Booleans without null-checking or exception handling. same goes for doing math operations on an Integer. for a long time,

Why can I not modify the result of an unboxing conversion?

*爱你&永不变心* 提交于 2019-11-27 15:20:11
struct Point { public int x; public int y; } void Main() { Point p; p.x = 1; p.y = 1; Object o = p; ((Point) o).x = 4; // error ((Point) o).x = 5; // error ((Point) o).x = 6; // error p = (Point) o // expect 6 } Why doesn't it compile to ldloc.1 // o unbox Point ldc.i4.4 stfld Point.x Where C++ CLI allows it. For those who don't know, unbox is not required to create a copy of value types , instead it pushes a pointer to the value on to the stack. Only assignment would create a copy. Because of how value types work, the boxed Point is a copy of the original, and "unboxing" it by casting back to