boxing

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

不羁岁月 提交于 2019-11-27 06:39:22
问题 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); 回答1: 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

Boxing Occurrence in C#

☆樱花仙子☆ 提交于 2019-11-27 06:03:40
I'm trying to collect all of the situations in which boxing occurs in C#: Converting value type to System.Object type: struct S { } object box = new S(); Converting value type to System.ValueType type: struct S { } System.ValueType box = new S(); Converting value of enumeration type to System.Enum type: enum E { A } System.Enum box = E.A; Converting value type into interface reference: interface I { } struct S : I { } I box = new S(); Using value types in C# string concatenation: char c = F(); string s1 = "char value will box" + c; note: constants of char type are concatenated at compile time

Does autoboxing call valueOf()?

别说谁变了你拦得住时间么 提交于 2019-11-27 05:46:07
问题 I'm trying to determine whether the following statements are guaranteed to be true: ((Boolean)true) == Boolean.TRUE ((Boolean)true) == Boolean.valueOf(true) ((Integer)1) == Integer.valueOf(1) I've always assumed that autoboxing was equivalent to calling valueOf() on the corresponding type. Every discussion that I've seen on the topic seems to support my assumption. But all I could find in the JLS was the following (§5.1.7): If the value p being boxed is an integer literal of type int between

Does int.class equal Integer.class or Integer.TYPE in Java?

只谈情不闲聊 提交于 2019-11-27 04:24:24
问题 Let's imagine one retrieves the declaring type of a Field using reflection. Which of the following tests will correctly indicate whether one is dealing with an int or an Integer ? Field f = ... Class<?> c = f.getDeclaringClass(); boolean isInteger; isInteger = c.equals(Integer.class); isInteger = c.equals(Integer.TYPE); isInteger = c.equals(int.class); isInteger = ( c == Integer.class); isInteger = ( c == Integer.TYPE); isInteger = ( c == int.class); 回答1: Based on Field.getType() (instead of

Does var keyword in C# cause boxing?

自作多情 提交于 2019-11-27 03:43:07
问题 My boss forbids me to use var as it would cause boxing and slowing down the app. Is that true? 回答1: An approach that might work is to write these two methods: public static void WithInt() { int x = 5; Console.WriteLine(x); } public static void WithVar() { var x = 5; Console.WriteLine(x); } Compile, and use ildasm to examine the produced CIL. Show your boss. edit @ck has done all but the last step for you :) 回答2: Following on from Aakash's answer, here is the IL: (thanks LINQPad) WithInt: IL

What is the difference between boxing/unboxing and type casting?

£可爱£侵袭症+ 提交于 2019-11-27 03:02:41
What is the difference between boxing/unboxing and type casting? Often, the terms seem to be used interchangeably. Boxing refers to a conversion of a non-nullable-value type into a reference type or the conversion of a value type to some interface that it implements (say int to IComparable<int> ). Further, the conversion of an underlying value type to a nullable type is also a boxing conversion. (Caveat: Most discussions of this subject will ignore the latter two types of conversions.) For example, int i = 5; object o = i; converts i to an instance of type object . Unboxing refers to an

How is the boxing/unboxing behavior of Nullable<T> possible?

风流意气都作罢 提交于 2019-11-26 22:43:36
Something just occurred to me earlier today that has got me scratching my head. Any variable of type Nullable<T> can be assigned to null . For instance: int? i = null; At first I couldn't see how this would be possible without somehow defining an implicit conversion from object to Nullable<T> : public static implicit operator Nullable<T>(object box); But the above operator clearly does not exist, as if it did then the following would also have to be legal, at least at compile-time (which it isn't): int? i = new object(); Then I realized that perhaps the Nullable<T> type could define an

Boxing Occurrence in C#

Deadly 提交于 2019-11-26 22:15:32
问题 I'm trying to collect all of the situations in which boxing occurs in C#: Converting value type to System.Object type: struct S { } object box = new S(); Converting value type to System.ValueType type: struct S { } System.ValueType box = new S(); Converting value of enumeration type to System.Enum type: enum E { A } System.Enum box = E.A; Converting value type into interface reference: interface I { } struct S : I { } I box = new S(); Using value types in C# string concatenation: char c = F()

Does calling a method on a value type result in boxing in .NET?

て烟熏妆下的殇ゞ 提交于 2019-11-26 19:51:27
问题 I was just participating in Stack Overflow question Is everything in .NET an object? . And one poster (in comments of accepted answer) seemed to think that performing a method call on a value type resulted in boxing. He pointed me to Boxing and Unboxing (C# Programming Guide) which doesn't exactly specify the use case we're describing. I'm not one to trust a single source, so I just wanted to get further feedback on the question. My intuition is that there is no boxing but my intuition does

C# - Issues with boxing / unboxing / typecasting ints. I don't understand

*爱你&永不变心* 提交于 2019-11-26 17:50:32
问题 I'm having a hard time understanding this. Consider the following example: protected void Page_Load(object sender, EventArgs e) { // No surprise that this works Int16 firstTest = Convert.ToInt16(0); int firstTest2 = (int)firstTest; // This also works object secondTest = 0; int secondTest2 = (int)secondTest; // But this fails! object thirdTest = Convert.ToInt16(0); int thirdtest2 = (int)thirdTest; // It blows up on this line. } The specific error that I get at runtime is Specified cast is not