0

我知道这是我要问的一个愚蠢的问题，但我只想知道以下给定陈述的区别：

Abc object= new ABC();
object.Age=obj1.Age;
object.Place=obj1.Place;
object.Street=obj1.Street;
object.Number=obj1.Number;
object.POBox=obj1.POBox;

和

Abc object= new ABC()
{
    Age=obj1.Age,
    Place=obj1.Place,
    Street=obj1.Street,
    Number=obj1.Number,
    POBox=obj1.POBox
};

上述编写的代码是否有助于提高性能？我只是想知道在创建类的对象并将值分配给该类对象时是否有任何方法可以提高性能？

4

2 回答 2

5

不会。这些语句被编译为相同的 IL，因此没有性能改进。

第一个：

IL_0031: newobj instance void TestApplication.ABC::.ctor()
IL_0036: stloc.1
IL_0037: ldloc.1
IL_0038: ldc.i4.1
IL_0039: callvirt instance void TestApplication.ABC::set_Age(int32)
IL_003e: nop
IL_003f: ldloc.1
IL_0040: ldc.i4.1
IL_0041: callvirt instance void TestApplication.ABC::set_Place(int32)
IL_0046: nop
IL_0047: ldloc.1
IL_0048: ldc.i4.1
IL_0049: callvirt instance void TestApplication.ABC::set_Street(int32)
IL_004e: nop
IL_004f: ldloc.1
IL_0050: ldc.i4.1
IL_0051: callvirt instance void TestApplication.ABC::set_Number(int32)
IL_0056: nop
IL_0057: ldloc.1
IL_0058: ldc.i4.1
IL_0059: callvirt instance void TestApplication.ABC::set_POBox(int32)

第二个：

IL_0001: newobj instance void TestApplication.ABC::.ctor()
IL_0006: stloc.2
IL_0007: ldloc.2
IL_0008: ldc.i4.1
IL_0009: callvirt instance void TestApplication.ABC::set_Age(int32)
IL_000e: nop
IL_000f: ldloc.2
IL_0010: ldc.i4.1
IL_0011: callvirt instance void TestApplication.ABC::set_Place(int32)
IL_0016: nop
IL_0017: ldloc.2
IL_0018: ldc.i4.1
IL_0019: callvirt instance void TestApplication.ABC::set_Street(int32)
IL_001e: nop
IL_001f: ldloc.2
IL_0020: ldc.i4.1
IL_0021: callvirt instance void TestApplication.ABC::set_Number(int32)
IL_0026: nop
IL_0027: ldloc.2
IL_0028: ldc.i4.1
IL_0029: callvirt instance void TestApplication.ABC::set_POBox(int32)
于 2013-07-31T05:40:53.983 回答
1

我认为有一个细微的区别：第二个语句是原子的；您可以使用它来确保在没有适当的构造函数时将对象初始化为正确的状态。

微粒信息：http ://bartdesmet.net/blogs/bart/archive/2007/11/22/c-3-0-object-initializers-revisited.aspx

于 2013-07-31T05:45:48.340 回答