非托管代码 [英] Unmanaged code

问题描述

你好，

我遇到了以下问题:

我有一个变量

Hello,

I''ved came accross the following problem :

I have a variable

IntPtr pointer

，我知道我已经为其分配了一个点结构(x和y为浮点数).
我需要获取x值.
如果尝试编写x = pointer-> x，我会得到:"*和->运算符必须应用于指针".
我怎么能得到指针-> x的值?

在此先谢谢您！

to which I know that I have assigned a point structure ( x and y as floats ) .
I need to get the x value.
If I try to write x=pointer->x I get : "The * and -> operators must be applied to a pointer".
How could I get the value of pointer->x ?

Thanks in advance !

推荐答案

您首先需要创建一个指向point结构的指针并将其指针对象投射到该结构，然后您可以通过以下方式访问该结构的内容点参考.像这样的东西:

You first need to create a pointer to a point structure and cast your pointer object to it, then you can access the structure contents through the point reference. Something like:

IntPtr pointer = ...// your initialization here
Point* pt = (Point*)pointer;
float fx = pt->x;

我在这里可能会觉得很陌生，但是您的主题行和标记很少提供有关您使用哪种语言(例如C ++，C ++/CLI，C#)的线索.

I may be wide of the mark here but your subject line and tag gives very few clues as to what language you are using (i.e C++, C++/CLI, C#).

它"自从我做过C ++以来已经有一段时间了，但是尝试pointer.x

It''s been a while since I did any C++, but try pointer.x

我相信您可能必须使用这样的语法.和John一样，自从我从事C ++工作以来已经很长时间了，所以我可能只提供了无用的信息，但是在我可能没错的情况下……

I believe you may have to use syntax like this. Like John, it has been a very long time since I''ve done work in C++, so I might just be providing useless info, but on the off chance that I''m right...

*pointer.x

或

((point)*pointer).x

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