Windows 在命令行中指定资源的包含目录
我尝试使用以下行在Windows上构建cURL:Windows 在命令行中指定资源的包含目录,windows,curl,libcurl,nmake,rc,Windows,Curl,Libcurl,Nmake,Rc,我尝试使用以下行在Windows上构建cURL: nmake vc-x64 mode=dll VC=10 MACHINE=x64 现在,我在构建之前已经完成了以下设置: set PATH=C:\Program Files (x86)\Microsoft Visual Studio 10.0\VC\bin\amd64;%PATH% set INCLUDE="C:\Program Files (x86)\Microsoft Visual Studio 10.0\VC\atlmfc\include"
nmake vc-x64 mode=dll VC=10 MACHINE=x64
现在,我在构建之前已经完成了以下设置:
set PATH=C:\Program Files (x86)\Microsoft Visual Studio 10.0\VC\bin\amd64;%PATH%
set INCLUDE="C:\Program Files (x86)\Microsoft Visual Studio 10.0\VC\atlmfc\include";"C:\Program Files (x86)\Microsoft Visual Studio 10.0\VC\include"
set LIB="C:\Program Files (x86)\Microsoft Visual Studio 10.0\VC\atlmfc\lib\amd64";"C:\Program Files (x86)\Microsoft Visual Studio 10.0\VC\lib\amd64"
set LIBPATH="C:\Program Files (x86)\Microsoft Visual Studio 10.0\VC\atlmfc\lib\amd64"
set PATH="C:\Program Files (x86)\Microsoft SDKs\Windows\v7.1A\Bin\x64";%PATH%)
set INCLUDE="C:\Program Files (x86)\Microsoft SDKs\Windows\v7.1A\Include";%INCLUDE%
set LIB="C:\Program Files (x86)\Microsoft SDKs\Windows\v7.1A\Lib\x64";%LIB%
我得到以下错误:
rc /dDEBUGBUILD=0 /Fo ..\builds\libcurl-vc10-x64-release-dll-ipv6-sspi-w
inssl-obj-lib\libcurl.res ..\lib\libcurl.rc
Microsoft (R) Windows (R) Resource Compiler Version 6.1.7600.16385
Copyright (C) Microsoft Corporation. All rights reserved.
..\lib\libcurl.rc(22) : fatal error RC1015: cannot open include file 'winver.h'.
NMAKE : fatal error U1077: '"C:\Program Files (x86)\Microsoft SDKs\Windows\v7.1A
\Bin\x64\rc.EXE"' : return code '0x1'
Stop.
NMAKE : fatal error U1077: '"C:\Program Files (x86)\Microsoft Visual Studio 10.0
\VC\bin\amd64\nmake.EXE"' : return code '0x2'
Stop.
NMAKE : fatal error U1077: '"C:\Program Files (x86)\Microsoft Visual Studio 10.0
\VC\bin\amd64\nmake.EXE"' : return code '0x2'
Stop.
它似乎找不到文件“winver.h”,尽管该文件位于目录“C:\Program Files(x86)\Microsoft SDK\Windows\v7.1A\Include”中,并且该目录位于我的Include
变量中。
看起来,rc
没有考虑变量INCLUDE来查找INCLUDE路径。是否有方法为rc
命令添加包含路径?当然,除非这是完全不同的事情