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以编程方式为显卡注册应用程序

[英]Register Application programmatically for Graphic card

I am trying to write a Setup for an application with Inno Setup, which is pretty neat by the way ;) , but now I am stuck with a problem regarding the graphiccard.我正在尝试使用 Inno Setup 为应用程序编写一个安装程序,顺便说一下,这非常简洁;),但现在我遇到了有关图形卡的问题。

I am installing a 3D application and want to configure the program so that it uses a certain graphic card as default.我正在安装 3D 应用程序并希望配置该程序,使其默认使用某个图形卡。 It sounded logical to me that somewhere in the registry must be a something to configure that.对我来说,注册表中的某个地方必须是配置它的东西,这对我来说是合乎逻辑的。 I tried to change the default graphiccard manually und compare the changes in the registry but couldn't find any entrys relative to the changes我尝试手动更改默认图形卡并比较注册表中的更改,但找不到与更改相关的任何条目

I know that programs like Adobe Photoshop are using the high performance graphiccards by default so there must be some kind of solution to my problem.我知道像 Adob​​e Photoshop 这样的程序默认使用高性能图形卡,因此必须有某种解决方案来解决我的问题。

I am using inno script to write my setup but I can call external programs to do some tasks so I could write a C# program to execute this task.我正在使用 inno 脚本来编写我的设置,但我可以调用外部程序来执行一些任务,因此我可以编写一个 C# 程序来执行此任务。

This option is mostly relevant for Windows 7 and Windows 8 because I only saw it there.此选项主要与 Windows 7 和 Windows 8 相关,因为我只在那里看到过。

Does anyone of you know how to change the default graphiccard of an application programmatically?你们中有人知道如何以编程方式更改应用程序的默认图形卡吗?

Your approach, sorry, sucks.对不起,你的方法很糟糕。 This is not something you should do during setup.这不是您在设置过程中应该做的事情。 What if the card changes?卡改了怎么办?

Check it during application start.在应用程序启动期间检查它。 Let the user choose if you do not find anything.如果您没有找到任何内容,请让用户选择。 Let him update.让他更新。 Not everyone has only one graphics card.不是每个人都只有一张显卡。 People do update the graphics card.人们确实更新了显卡。 Why should I rerun setup just to get this done?为什么我要重新运行安装程序才能完成这项工作?

And it is quite easy to identify graphics cards, performance is another story.而且很容易识别显卡,性能又是另外一回事了。 Which one is "high performance" in your eyes?你眼中的“高性能”是哪一种?

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