Given “where T : new()”, does “new T()” use Activator.CreateInstance internally?

问题

If I have a type parameter constraint new():

void Foo<T>() where T : new()
{
    var t = new T();
}

Is it true that new T() will internally use the Activator.CreateInstance method (i.e. reflection)?

回答1:

Yes, this is true. Edit 2: Here's a good explanation of the how and why.

http://www.simple-talk.com/community/blogs/simonc/archive/2010/11/17/95700.aspx

For verification I compiled the following method:

public static T Create<T>() where T: new() {
    return new T();
}

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 
}

Edit: 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 
}

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.

回答2:

Yes. It does for reference types.

Using ILSpy on the following release-compiled code:

    public static void DoWork<T>() where T: new()
    {
        T t = new T();
        Console.WriteLine(t.ToString());
    }

Yielded

.method public hidebysig 
    instance void DoWork<.ctor T> () cil managed 
{
    // Method begins at RVA 0x2064
    // Code size 52 (0x34)
    .maxstack 2
    .locals init (
        [0] !!T t,
        [1] !!T CS$0$0000,
        [2] !!T CS$0$0001
    )

    IL_0000: ldloca.s CS$0$0000
    IL_0002: initobj !!T
    IL_0008: ldloc.1
    IL_0009: box !!T
    IL_000e: brfalse.s IL_001b

    IL_0010: ldloca.s CS$0$0001
    IL_0012: initobj !!T
    IL_0018: ldloc.2
    IL_0019: br.s IL_0020

    IL_001b: call !!0 [mscorlib]System.Activator::CreateInstance<!!T>()

    IL_0020: stloc.0
    IL_0021: ldloca.s t
    IL_0023: constrained. !!T
    IL_0029: callvirt instance string [mscorlib]System.Object::ToString()
    IL_002e: call void [mscorlib]System.Console::WriteLine(string)
    IL_0033: ret
} // end of method Program::DoWork

Or in C#:

public void DoWork<T>() where T : new()
{
    T t = (default(T) == null) ? Activator.CreateInstance<T>() : default(T);
    Console.WriteLine(t.ToString());
}

JIT will create different compiled instructions for each different value type parameter passed in, but will use the same instructions for reference types -- hence the Activator.CreateInstance()

来源：https://stackoverflow.com/questions/6708459/given-where-t-new-does-new-t-use-activator-createinstance-internally

标签

generics

reflection

activator

type-constraints