Ram*_*rma 11 c++ visual-studio visual-c++ libraries visual-studio-2012
我正在构建一个C++库(标头集,导入库和DLL).我想让任何想要使用它的开发人员尽可能地使用这个库.特别是我不希望这个库的使用者不必担心为项目的所有不同配置(Debug | Release和x86/x64/ARM)手动更改头路径,库路径和链接库.我知道我可以使用属性表来做到这一点.我为此创建了6个不同的属性表(每个配置一个).每个工作表如下所示(仅列出x86 | Debug版本,假设正确定义了宏INCLUDEPATH和LIBPATH):
<?xml version="1.0" encoding="utf-8"?>
<Project xmlns="http://schemas.microsoft.com/developer/msbuild/2003">
<PropertyGroup>
<_PropertySheetDisplayName>MyCPPLib, 1.0</_PropertySheetDisplayName>
</PropertyGroup>
<ItemDefinitionGroup>
<ClCompile>
<AdditionalIncludeDirectories>$INCLUDEPATH;%(AdditionalIncludeDirectories)</AdditionalIncludeDirectories>
</ClCompile>
<Link>
<AdditionalLibraryDirectories>$(AdditionalLibraryDirectories);$LIBPATH\x86\Debug</AdditionalLibraryDirectories>
<AdditionalDependencies>MyCPPLib.lib;$(AdditionalDependencies)</AdditionalDependencies>
</Link>
</ItemDefinitionGroup>
</Project>
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我想知道是否有可能只创建一个道具文件,可以根据用户的活动配置来处理所有6种配置?那个文件怎么样?
您可以将库二进制文件安装在以下结构中:
<toplevelsdkdir>
|-> lib
|-> x86
|-> Debug
|-> Release
|-> x64
|-> Debug
|-> Release
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然后只需创建一个单个项目范围内的props文件,如下所示:
<?xml version="1.0" encoding="utf-8"?>
<Project xmlns="http://schemas.microsoft.com/developer/msbuild/2003">
<PropertyGroup>
<_PropertySheetDisplayName>MyCPPLib, 1.0</_PropertySheetDisplayName>
</PropertyGroup>
<ItemDefinitionGroup>
<ClCompile>
<AdditionalIncludeDirectories>$INCLUDEPATH;%(AdditionalIncludeDirectories)</AdditionalIncludeDirectories>
</ClCompile>
<Link>
<AdditionalLibraryDirectories>$(AdditionalLibraryDirectories);$LIBPATH\$(PlatformTarget)\$(Configuration)</AdditionalLibraryDirectories>
<AdditionalDependencies>MyCPPLib.lib;$(AdditionalDependencies)</AdditionalDependencies>
</Link>
</ItemDefinitionGroup>
</Project>
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如果可以,可以用从注册表中读取的信息替换变量INCLUDEPATH和LIBPATH(可以在安装过程中将其放在其中):
<ClCompile>
<AdditionalIncludeDirectories>$([MSBuild]::GetRegistryValue(`HKEY_LOCAL_MACHINE\Software\MyCompany\MySDK\v1`, `InstallDir`))\INCLUDE;%(AdditionalIncludeDirectories)</AdditionalIncludeDirectories>
</ClCompile>
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