Linux CMAKE 中的操作系统特定说明：如何？
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OS specific instructions in CMAKE: How to?
提问by Prasad
I am a beginner to CMAKE. Below is a simple cmake file which works well in mingw environment windows. The problem is clearly with target_link_libraries()function of CMAKE where I am linking libwsock32.a. In windows this works and I get the results.
我是 CMAKE 的初学者。下面是一个简单的 cmake 文件，它在 mingw 环境窗口中运行良好。问题显然target_link_libraries()出在我链接 libwsock32.a 的 CMAKE 功能上。在 Windows 中，这有效，我得到了结果。
However, as expected, in Linux, the /usr/bin/ldwill look for -lwsock32which is NOT there on the Linux OS.
然而，正如预期的那样，在 Linux 中，/usr/bin/ld将查找-lwsock32Linux 操作系统上不存在的内容。
My Problem is: How do I instruct CMAKE to avoid linking wsock32 library in Linux OS???
我的问题是：如何指示 CMAKE 避免在 Linux 操作系统中链接 wsock32 库？？？
Any help will be greatly appreciated.
任何帮助将不胜感激。
My Simple CMake file:
我的简单 CMake 文件：
PROJECT(biourl)
set (${PROJECT_NAME}_headers ./BioSocketAddress.h ./BioSocketBase.h ./BioSocketBuffer.h ./BioSocketCommon.h ./BioSocketListener.h ./BioSocketPrivate.h ./BioSocketStream.h ./BioUrl.h BioDatabase.h )
set (${PROJECT_NAME}_sources BioSocketAddress.C BioSocketBase.C BioSocketCommon.C BioSocketStream.C BioUrl.C BioDatabase.C )
add_library(${PROJECT_NAME} STATIC ${${PROJECT_NAME}_headers} ${${PROJECT_NAME}_sources} )
# linkers
#find_library(ws NAMES wsock32 PATHS ${PROJECT_SOURCE_DIR} NO_SYSTEM_ENVIRONMENT_PATH NO_DEFAULT_PATH)
target_link_libraries(${PROJECT_NAME} bioutils wsock32)
install (TARGETS ${PROJECT_NAME}
RUNTIME DESTINATION bin
LIBRARY DESTINATION lib
ARCHIVE DESTINATION lib/archive )
采纳答案by relaxxx
Use
用
if (WIN32)
#do something
endif (WIN32)
or
或者
if (UNIX)
#do something
endif (UNIX)
or
或者
if (MSVC)
#do something
endif (MSVC)
or similar
或类似
回答by tibur
回答by Barun Parichha
Use some preprocessor macro to check if it's in windows or linux. For example
使用一些预处理器宏来检查它是在 windows 还是 linux 中。例如
#ifdef WIN32
LIB=
#elif __GNUC__
LIB=wsock32
#endif
include -l$(LIB) in you build command.
在构建命令中包含 -l$(LIB)。
You can also specify some command line argument to differentiate both.
您还可以指定一些命令行参数来区分两者。
回答by Bruno Soares
You have some special words from CMAKE, take a look:
你有一些来自 CMAKE 的特殊词，看看：
if(${CMAKE_SYSTEM_NAME} STREQUAL "Linux")
// do something for Linux
else
// do something for other OS
回答by mlvljr
Given this is such a common issue, geronto-posting:
鉴于这是一个如此普遍的问题，geronto-posting：
if(UNIX AND NOT APPLE)
set(LINUX TRUE)
endif()
# if(NOT LINUX) should work, too, if you need that
if(LINUX)
message(STATUS ">>> Linux")
# linux stuff here
else()
message(STATUS ">>> Not Linux")
# stuff that should happen not on Linux
endif()
回答by Afr
In General
一般来说
You can detect and specify variables for several operating systems like that:
您可以为多个操作系统检测和指定变量，如下所示：
Detect Microsoft Windows
检测 Microsoft Windows
if(WIN32)
# for Windows operating system in general
endif()
Or:
或者：
if(MSVC OR MSYS OR MINGW)
# for detecting Windows compilers
endif()
Detect Apple MacOS
检测苹果 MacOS
if(APPLE)
# for MacOS X or iOS, watchOS, tvOS (since 3.10.3)
endif()
Detect Unix and Linux
检测 Unix 和 Linux
if(UNIX AND NOT APPLE)
# for Linux, BSD, Solaris, Minix
endif()
Your specific linker issue
您的特定链接器问题
To solve your issue with the Windows-specific wsock32library, just remove it from other systems, like that:
要解决 Windows 特定wsock32库的问题，只需将其从其他系统中删除，如下所示：
if(WIN32)
target_link_libraries(${PROJECT_NAME} bioutils wsock32)
else
target_link_libraries(${PROJECT_NAME} bioutils)
endif()
回答by Cascades
Generator expressions are also possible:
生成器表达式也是可能的：
target_link_libraries(
target_name
PUBLIC
libA
$<$<PLATFORM_ID:Windows>:wsock32>
PRIVATE
$<$<PLATFORM_ID:Linux>:libB>
libC
)
This will link libA, wsock32 & libC on Windows and link libA, libB & libC on Linux
这将在 Windows 上链接 libA、wsock32 和 libC，在 Linux 上链接 libA、libB 和 libC
回答by Nico Heidtke
I want to leave this here because I struggled with this when compiling for Android in Windows with the Android SDK.
我想把它留在这里，因为我在使用 Android SDK 在 Windows 中编译 Android 时遇到了这个问题。
CMake distinguishes between TARGET and HOST platform.
CMake 区分了 TARGET 和 HOST 平台。
My TARGET was Android so the variables like CMAKE_SYSTEM_NAME had the value "Android" and the variable WIN32 from the other answer here was not defined. But I wanted to know if my HOST system was Windows because I needed to do a few things differently when compiling on either Windows or Linux or IOs. To do that I used CMAKE_HOST_SYSTEM_NAME which I found is barely known or mentioned anywhere because for most people TARGEt and HOST are the same or they don't care.
我的目标是 Android，所以像 CMAKE_SYSTEM_NAME 这样的变量的值是“Android”，而另一个答案中的变量 WIN32 没有定义。但是我想知道我的 HOST 系统是否是 Windows，因为在 Windows、Linux 或 IO 上编译时我需要做一些不同的事情。为此，我使用了 CMAKE_HOST_SYSTEM_NAME，我发现它在任何地方都鲜为人知或被提及，因为对于大多数人来说，TARGEt 和 HOST 是相同的，或者他们不在乎。
Hope this helps someone somewhere...
希望这可以帮助某人某处...
