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Editorial Team
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Asked: 2026-05-23T18:09:05+00:00

If I have a type parameter constraint new() : void Foo<T>() where T :

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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)?

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1 Answer

  1. Editorial Team

    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.

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