遍布Stack Overflow和互联网我发现保持结构不可变是一个很好的设计原则。不幸的是,我从未看到任何实际导致这些结构真正不可变的实现。
假设一个struct里面没有任何引用类型,我该如何实际构造一个struct不可变?也就是说,如何防止任何原始字段的变异(可能是编译时/运行时异常)?
我编写了一个简单的测试,尝试使结构不可变,但甚至没有使用System.ComponentModel.ImmutableObjectAttribute
工作:
class Program
{
static void Main(string[] args)
{
ImmutableStruct immStruct1 = new ImmutableStruct();
Console.WriteLine(immStruct1); //Before mutation.
immStruct1.field1 = 1;
immStruct1.field2 = "Hello";
immStruct1.field3 = new object();
Console.WriteLine(immStruct1); //After 1st mutation.
immStruct1.field1 = 2;
immStruct1.field2 = "World";
immStruct1.field3 = new object();
Console.WriteLine(immStruct1); //After 2nd mutation.
Console.ReadKey();
}
}
[ImmutableObject(true)]
struct ImmutableStruct
{
public int field1;
public string field2;
public object field3;
public override string ToString()
{
string field3String = "null";
if (field3 != null)
{
field3String = field3.GetHashCode().ToString();
}
return String.Format("Field1: {0}, Field2: {1}, Field3: {2}", field1, field2, field3String);
}
}
答案 0 :(得分:16)
创建字段private readonly
并在构造函数中传递初始值
public struct ImmutableStruct
{
private readonly int _field1;
private readonly string _field2;
private readonly object _field3;
public ImmutableStruct(int f1, string f2, object f3)
{
_field1 = f1;
_field2 = f2;
_field3 = f3;
}
public int Field1 { get { return _field1; } }
public string Field2 { get { return _field2; } }
public object Field3 { get { return _field3; } }
}
从C#6.0开始(Visual Studio 2015)您可以使用仅限getter属性
public struct ImmutableStruct
{
public ImmutableStruct(int f1, string f2, object f3)
{
Field1 = f1;
Field2 = f2;
Field3 = f3;
}
public int Field1 { get; }
public string Field2 { get; }
public object Field3 { get; }
}
请注意,readonly fields和getter only属性可以在构造函数中初始化,也可以在类中使用字段或属性初始值设定项public int Field1 { get; } = 7;
进行初始化。
无法保证构造函数在结构上运行。例如。如果您有一个结构数组,则必须为每个数组元素显式调用初始值设定项。对于引用类型数组,所有元素首先被初始化为null
,这使得您必须在每个元素上调用new
。但很容易忘记像结构这样的值类型。
var immutables = new ImmutableStruct[10];
immutables[0] = new ImmutableStruct(5, "hello", new Person());
从C#7.2开始,您可以使用Read-only structs
答案 1 :(得分:3)
将您的不可变数据保密:
struct ImmutableStruct
{
private int field1;
private string field2;
private object field3;
public ImmutableStruct(int f1, string f2, object f3)
{
field1 = f1;
field2 = f2;
field3 = f3;
}
public int Field1 => field1;
public string Field2 => field2;
public object Field3 => field3;
}
或者不那么杂乱:
struct ImmutableStruct
{
public ImmutableStruct(int f1, string f2, object f3)
{
Field1 = f1;
Field2 = f2;
Field3 = f3;
}
public int Field1 { get; }
public string Field2 { get; }
public object Field3 { get; }
}