问题描述
如果我有一个类型参数约束新()
:
If I have a type parameter constraint new()
:
void Foo<T>() where T : new()
{
var t = new T();
}
这是真的,新T()
将在内部使用的 Activator.CreateInstance
法(即反射)?
Is it true that new T()
will internally use the Activator.CreateInstance
method (i.e. reflection)?
推荐答案
是的,这是事实。 修改2:这里是怎样的一个很好的解释,为什么
Yes, this is true. Edit 2: Here's a good explanation of the how and why.
的
有关验证我整理了以下方法:
For verification I compiled the following method:
public static T Create<T>() where T: new() {
return new T();
}
这是生成的IL时与C#编译器在.NET 3.5编译SP1:
And this is the generated IL when compiled with the C# compiler in .NET 3.5 SP1:
.method public hidebysig static !!T Create<.ctor T>() cil managed
{
.maxstack 2
.locals init (
[0] !!T local,
[1] !!T local2)
L_0000: ldloca.s local
L_0002: initobj !!T
L_0008: ldloc.0
L_0009: box !!T
L_000e: brfalse.s L_001a
L_0010: ldloca.s local2
L_0012: initobj !!T
L_0018: ldloc.1
L_0019: ret
L_001a: call !!0 [mscorlib]System.Activator::CreateInstance<!!T>()
L_001f: ret
}
编辑:的C#4编译器将创建略有不同,但大同小异,代码:
The C# 4 compiler creates slightly different, but similar, code:
.method public hidebysig static !!T Create<.ctor T>() cil managed
{
.maxstack 2
.locals init (
[0] !!T CS$1$0000,
[1] !!T CS$0$0001)
L_0000: nop
L_0001: ldloca.s CS$0$0001
L_0003: initobj !!T
L_0009: ldloc.1
L_000a: box !!T
L_000f: brfalse.s L_001c
L_0011: ldloca.s CS$0$0001
L_0013: initobj !!T
L_0019: ldloc.1
L_001a: br.s L_0021
L_001c: call !!0 [mscorlib]System.Activator::CreateInstance<!!T>()
L_0021: stloc.0
L_0022: br.s L_0024
L_0024: ldloc.0
L_0025: ret
}
在的情况下,值类型不使用催化剂,但仅返回默认(T)
值,否则它调用 Activator.CreateInstance
方法。
In the case of a value type it doesn't use the activator but just returns the default(T)
value, otherwise it invokes the Activator.CreateInstance
method.
这篇关于鉴于"其中T:新的()" ;,确实"新T()"使用Activator.CreateInstance内部?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!