Question

我正在尝试为库中的类编写单元测试。构建lib和测试是用CMake完成的。

├── CMakeLists.txt
├── src
│   ├── Bar.cpp
│   ├── CMakeLists.txt
│   └── Foo.cpp
└── test
    ├── BarTest.cpp
    ├── CMakeLists.txt
    └── FooTest.cpp

src/CMakeLists.txt只是add_library(Example Foo.cpp Bar.cpp)。因为我想在不构建库的情况下构建单元测试（即如果我做了一个改变来说Foo我可以在不进行单元测试的情况下进行单元测试与Bar的互动已经开始。所以test/CMakeLists.txt看起来像这样：

add_executable(FooTest FooTest.cpp ${CMAKE_SOURCE_DIR}/src/Foo.cpp)
add_test(bla::FooTest ${CMAKE_CURRENT_BINARY_DIR}/FooTest)

add_executable(BarTest BarTest.cpp ${CMAKE_SOURCE_DIR}/src/Bar.cpp)
add_test(bla::BarTest ${CMAKE_CURRENT_BINARY_DIR}/BarTest)

但是，这会导致每个源文件都构建两次：

[ 11%] Building CXX object src/CMakeFiles/Example.dir/Foo.o
[ 22%] Building CXX object src/CMakeFiles/Example.dir/Bar.o
[ 33%] Linking CXX static library libExample.a
[ 33%] Built target Example
[ 44%] Building CXX object test/CMakeFiles/FooTest.dir/FooTest.o
[ 55%] Building CXX object test/CMakeFiles/FooTest.dir/__/src/Foo.o
[ 66%] Linking CXX executable FooTest
[ 66%] Built target FooTest
[ 77%] Building CXX object test/CMakeFiles/BarTest.dir/BarTest.o
[ 88%] Building CXX object test/CMakeFiles/BarTest.dir/__/src/Bar.o
[100%] Linking CXX executable BarTest
[100%] Built target BarTest

有可能避免这种情况吗？即如果我之前构建了src/Foo.cpp 单元测试，不要再为库建立它？所有标志都是一样的。

Answer 1

如果你不介意其他目标，那就是CMake object libraries的用途：

add_library(Foo OBJECT Foo.cpp)
add_library(Bar OBJECT Bar.cpp)
add_executable(Example $<TARGET_OBJECTS:Foo> $<TARGET_OBJECTS:Bar>)

然后

add_executable(FooTest FooTest.cpp $<TARGET_OBJECTS:Foo>)
add_executable(BarTest BarTest.cpp $<TARGET_OBJECTS:Bar>)

使用CMake进行单元测试，无需重复编译

1 个答案: