批处理模式规则 [nmake] 中的多个源路径

Question

我正在为我的 makefile 使用 batch-mode rule。目前我有以下目标：

DIR_SRC = src
DIR_INCLUDE = include
DIR_LIB = lib
DIR_BIN = bin\x64
DIR_BUILD = build\x64    

{$(DIR_SRC)}.cpp{$(DIR_BUILD)}.obj ::
            @echo Compiling...
cl /c /EHsc /Fo$(DIR_BUILD)\ /MD /I$(DIR_INCLUDE) $<
    
$(EXECUTABLE_NAME) : $(DIR_BUILD)\*.obj
   @echo Linking $(EXECUTABLE_NAME)...
   link /out:$(DIR_BIN)$(EXECUTABLE_NAME) $(DIR_BUILD)\*.obj

“bin/x64/*.obj”目标使用“src”文件夹内的所有 cpp 文件。是否可以为其添加另一个源路径？

我想要这样的东西

{$(DIR_SRC) $(ANOTHER_DIR)}.cpp{$(DIR_BUILD)}.obj ::

Answer 1

如果您添加一个 second 构建目录来匹配您的第二个源目录，您可以让它工作。例如，如果我们将您的 makefile 修改为：

DIR_SRC = src
DIR_SRC2 = another
DIR_INCLUDE = include
DIR_BIN = bin\x64
DIR_BUILD = build\x64
DIR_BUILD2 = build_another\x64
EXECUTABLE_NAME = foo.exe    

{$(DIR_SRC)}.cpp{$(DIR_BUILD)}.obj ::
    @echo Compiling...
    cl -nologo /c /EHsc /Fo$(DIR_BUILD)\ /MD /I$(DIR_INCLUDE) $<
    
{$(DIR_SRC2)}.cpp{$(DIR_BUILD2)}.obj ::
    @echo Compiling...
    cl -nologo /c /EHsc /Fo$(DIR_BUILD2)\ /MD /I$(DIR_INCLUDE) $<
    
$(DIR_BIN)$(EXECUTABLE_NAME) : $(DIR_BUILD)\*.obj $(DIR_BUILD2)\*.obj
    @echo Linking $(EXECUTABLE_NAME)...
    link -nologo /out:$@ $**

那么我们将得到：

nmake -nologo

Compiling...
        cl -nologo /c /EHsc /Fobuild\x64\ /MD /Iinclude src\*.cpp
s1.cpp
s2.cpp
Generating Code...
Compiling...
        cl -nologo /c /EHsc /Fobuild_another\x64\ /MD /Iinclude another\*.cpp
a1.cpp
a2.cpp
Generating Code...
Linking foo.exe...
        link -nologo /out:bin\x64\foo.exe build\x64\*.obj build_another\x64\*.obj

其中 src 包含 s1.cpp 和 s2.spp，而 another 包含 a1.cpp 和 a2.cpp。

批处理模式规则 [nmake] 中的多个源路径

Multiple source paths in batch-mode rule [nmake]

c++

nmake

makefile

compilation