CPP/CLI 中的本机和托管之间是否令人困惑？
如果我使用 /clr 模式编译具有如下内容的代码：
int x = 3;
char ch='A';
int arr[]="Hi";
array<int>^ ManArr1={44};
array<int>^ ManArr2= gcnew array<int> {44};
我现在的问题： int 类型会映射到 System::Int32 吗？那么 char ch 呢？它们被视为本机类型还是托管类型？哪里会被处决！是否通过 MSIL！
我们看到 int arr[] 是一个原生数组，这是否意味着它将在 MSIL 之外执行？
最后一个问题，，对于托管数组 ManArr1 & ManArr2 两个初始化有什么区别？
If i use /clr mode to compile a code that has somthing like the following:
int x = 3;
char ch='A';
int arr[]="Hi";
array<int>^ ManArr1={44};
array<int>^ ManArr2= gcnew array<int> {44};
my questions now:
Would the type int be mapped to System::Int32 ?? and what about char ch ? Are they considerd as native or managed type? Where will be executed! through MSIL or not!!
We see that int arr[] is a native array, does that mean it will be executed out of MSIL?
The last question ,, For both the managed array ManArr1 & ManArr2 what is the difference between the two initialization ??
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（2）
使用
/clr编译时，整个程序都会转换为 MSIL，除非您使用#pragma Managed(off)或#pragma unmanagedint相当于System::Int32char相当于System::SByte（不是System::Char！）When compiling with
/clr, your entire program is converted to MSIL unless you use#pragma managed(off)or#pragma unmanagedintis equivalent toSystem::Int32charis equivalent toSystem::SByte(notSystem::Char!)关于
“对于托管数组ManArr1 & ManArr2这两个初始化有什么区别？？”没有功能上的区别，一个是另一个的简写。
Regarding
"For both the managed array ManArr1 & ManArr2 what is the difference between the two initialization ??"There is no functional difference, one is a shorthand for the other.