C#集合初始化语法是否避免默认初始化开销

Question

使用新的C#集合初始化语法时:

string[] sarray = new[] { "A", "B", "C", "D" };

编译器是否避免将每个数组槽初始化为默认值,或者它是否等效于:

string[] sarray = new string[4];  // all slots initialized to null
sarray[0] = "A";
sarray[1] = "B";
sarray[2] = "C";
sarray[3] = "D";

Answer 1

编译器仍然使用newarrIL指令,因此CLR仍将初始化数组.

集合初始化只是编译器的魔力 - CLR对它一无所知,所以它仍然认为它必须做清理.

但是,这应该非常非常快 - 它只是擦拭内存.我怀疑在许多情况下这是一个重大的开销.

Answer 2

快速测试:

        string[] arr1 =
        {
            "A","B","C","D"
        };
        arr1.GetHashCode();

        string[] arr2 = new string[4];
        arr2[0] = "A";
        arr2[1] = "B";
        arr2[2] = "C";
        arr2[3] = "D";

        arr2.GetHashCode();

结果在这个IL(注意,它们都是相同的)

  IL_0002:  newarr     [mscorlib]System.String
  IL_0007:  stloc.2
  IL_0008:  ldloc.2
  IL_0009:  ldc.i4.0
  IL_000a:  ldstr      "A"
  IL_000f:  stelem.ref
  IL_0010:  ldloc.2
  IL_0011:  ldc.i4.1
  IL_0012:  ldstr      "B"
  IL_0017:  stelem.ref
  IL_0018:  ldloc.2
  IL_0019:  ldc.i4.2
  IL_001a:  ldstr      "C"
  IL_001f:  stelem.ref
  IL_0020:  ldloc.2
  IL_0021:  ldc.i4.3
  IL_0022:  ldstr      "D"
  IL_0027:  stelem.ref
  IL_0028:  ldloc.2
  IL_0029:  stloc.0
  IL_002a:  ldloc.0
  IL_002b:  callvirt   instance int32 [mscorlib]System.Object::GetHashCode()
  IL_0030:  pop
  IL_0031:  ldc.i4.4
  IL_0032:  newarr     [mscorlib]System.String
  IL_0037:  stloc.1
  IL_0038:  ldloc.1
  IL_0039:  ldc.i4.0
  IL_003a:  ldstr      "A"
  IL_003f:  stelem.ref
  IL_0040:  ldloc.1
  IL_0041:  ldc.i4.1
  IL_0042:  ldstr      "B"
  IL_0047:  stelem.ref
  IL_0048:  ldloc.1
  IL_0049:  ldc.i4.2
  IL_004a:  ldstr      "C"
  IL_004f:  stelem.ref
  IL_0050:  ldloc.1
  IL_0051:  ldc.i4.3
  IL_0052:  ldstr      "D"
  IL_0057:  stelem.ref
  IL_0058:  ldloc.1
  IL_0059:  callvirt   instance int32 [mscorlib]System.Object::GetHashCode()