在优化关闭的内部版本中(通常是调试内部版本),您将获得以下两个IL指令序列:
IL_0000: nop IL_0000: nop
IL_0001: ldnull IL_0001: ldnull
IL_0002: ldftn x IL_0002: ldftn x
IL_0008: newobj Action<int>..ctor IL_0008: newobj Action<int>..ctor
IL_000D: stloc.0 // foo IL_000D: stloc.0 // foo
IL_000E: ldloc.0 // foo IL_000E: ldloc.0 // foo
IL_000F: ldnull IL_000F: brtrue.s IL_0013
IL_0010: cgt.un IL_0011: br.s IL_001C
IL_0012: stloc.1
IL_0013: ldloc.1
IL_0014: brfalse.s IL_001F
IL_0016: ldloc.0 // foo IL_0013: ldloc.0 // foo
IL_0017: ldc.i4.s 0A IL_0014: ldc.i4.s 0A
IL_0019: callvirt Action<int>.Invoke IL_0016: callvirt Action<int>.Invoke
IL_001E: nop IL_001B: nop
IL_001F: ret IL_001C: ret
关于分支指令,这里有一些细微的差别,但是让我们在启用优化的情况下进行构建(通常是发布构建):
IL_0000: ldnull IL_0000: ldnull
IL_0001: ldftn x IL_0001: ldftn x
IL_0007: newobj Action<int>..ctor IL_0007: newobj Action<int>..ctor
IL_000C: stloc.0 // foo IL_000C: dup
IL_000D: ldloc.0 // foo IL_000D: brtrue.s IL_0011
IL_000E: brfalse.s IL_0018 IL_000F: pop
IL_0010: ldloc.0 // foo IL_0010: ret
IL_0011: ldc.i4.s 0A IL_0011: ldc.i4.s 0A
IL_0013: callvirt Action<int>.Invoke IL_0013: callvirt Action<int>.Invoke
IL_0018: ret IL_0018: ret
让我们尝试不同的方法:
public static void Action1(Action<int> foo)
{
if (foo != null)
foo(10);
}
public static void Action2(Action<int> foo)
{
foo?.Invoke(10);
}
这将被编译(同样,在启用优化的情况下)为:
IL_0000: ldarg.0 IL_0000: ldarg.0
IL_0001: brfalse.s IL_000B IL_0001: brfalse.s IL_000B
IL_0003: ldarg.0 IL_0003: ldarg.0
IL_0004: ldc.i4.s 0A IL_0004: ldc.i4.s 0A
IL_0006: callvirt Action<int>.Invoke IL_0006: callvirt Action<int>.Invoke
IL_000B: ret IL_000B: ret
完全相同的代码
要知道这一点,唯一的方法就是实际进行基准测试。
,我会的
非常
如果这是你需要考虑的事情,你会很惊讶。相反,我会根据您认为最容易编写、阅读和理解的内容来选择代码的样式。