使用预处理器指令定义输出路径

发布于 2024-09-01 08:20:51 字数 199 浏览 5 评论 0原文

使用以下伪代码：

#define BUILD_PATH "C:/MyBuild/"
#define BUILD_NAME "mydll.dll"
// Set build path here

表示我希望如何将当前项目（dll）构建到 C:/MyBuild/mydll.dll 中，如何仅使用预处理器指令来完成此任务？

原文

Using the following pseudo-code:

#define BUILD_PATH "C:/MyBuild/"
#define BUILD_NAME "mydll.dll"
// Set build path here

representing how I would like to build the current project (a dll) into C:/MyBuild/mydll.dll, how would I accomplish this by only using preprocessor directives?

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瘫痪情歌 2024-09-08 08:20:51

我可能会误解，但我真的无法理解你为什么要这样做，但它是可行的：

#pragma comment( linker, "/out:c:\mydll.dll" )

我无法准确地重申你有多少不想这样做......

如果你想获取输出那么，恐怕……你不能。该信息来自预处理器之后的几个步骤，因此预处理器无法获取该信息。

I may be misunderstanding, but I really cannot understand WHY you want to do this but it is doable:

#pragma comment( linker, "/out:c:\mydll.dll" )

I cannot re-iterate enough exactly how much you don't want to be doing this though ...

If you want to GET the output path via pre-processor info then, I'm afraid ... you can't. That info comes from several steps after the pre-processor so there is no way the pre-processor could get that info.

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