How to link against to specific packageid in Conan - conan

Question

I have a package with 2 binaries. The binaries only differ by a single option.
The package is a library. How can now link this package to a specific package that I require?

Answer 1

There are few options:
You can use cmake_paths and discover the library name:
First, you add a conanfile.txt, declaring package name and cmake_paths generator
[requires]
my_package/0.1.0#user/channel
[generators]
cmake_paths
Second, you search the desired library from the package, by its name in your cmake file:
cmake_minimum_required(VERSION 3.0)
project(myapp)
find_library(MY_LIBRARY foo REQUIRED) # the library name foo in this example
add_executable(myapp app.cpp)
target_link_libraries (myapp ${MY_LIBRARY})
And finally, you pass the cmake_paths to cmake, so it will find your library
mkdir build && cd build
conan install ..
cmake .. -DCMAKE_TOOLCHAIN_FILE=conan_paths.cmake
cmake --build .
./myapp
It works, but it's a bit fragile, as the consumer need to now the library name and there is no warning from CMake when the library is found, you have to add a condition to check it.
The second possible option is using Components feature, but requires a recipe modification, you can provide a different target for each library:
First, you need to update your conanfile.py, adding the components:
from conans import ConanFile, CMake
class MyPackage(ConanFile):
name = "my_package"
version = "0.1.0"
settings = "os", "compiler", "build_type", "arch"
options = {"shared": [True, False]}
default_options = {"shared": False}
generators = "cmake"
exports_sources = "src/*"
def build(self):
cmake = CMake(self)
cmake.configure(source_folder="src")
cmake.build()
def package(self):
self.copy("*.h", dst="include", src="src")
self.copy("*.lib", dst="lib", keep_path=False)
self.copy("*.dll", dst="bin", keep_path=False)
self.copy("*.dylib*", dst="lib", keep_path=False)
self.copy("*.so", dst="lib", keep_path=False)
self.copy("*.a", dst="lib", keep_path=False)
def package_info(self):
self.cpp_info.names["cmake_find_package"] = "MyPackage"
self.cpp_info.names["cmake_find_package_multi"] = "MyPackage"
self.cpp_info.components["libfoo"].names["cmake_find_package"] = "foo"
self.cpp_info.components["libfoo"].names["cmake_find_package_multi"] = "foo"
self.cpp_info.components["libfoo"].libs = ["foo"]
self.cpp_info.components["libbar"].names["cmake_find_package"] = "bar"
self.cpp_info.components["libbar"].names["cmake_find_package_multi"] = "bar"
self.cpp_info.components["libbar"].libs = ["bar"]
As you can see, the recipe builds a package with 2 libraries, foo and bar, and use different components for each one. As CamelCase is preferred for CMake targets, the name MyPackage will be used for the file name and target namespace. The result will be MyPackage::foo and MyPackage::bar.
As consumer, you have to update your project too:
Now we will use cmake_find_package generator, to avoid CMAKE_TOOLCHAIN_FILE in the command line:
[requires]
my_package/0.1.0#user/channel
[generators]
cmake_find_package
The CMake file now requires the package name, but it's checked by CMake:
cmake_minimum_required(VERSION 3.0)
project(myapp)
find_package(MyPackage REQUIRED)
add_executable(myapp app.cpp)
target_link_libraries (myapp MyPackage::foo) # We only need libfoo here
And finally, but simpler now, the command line:
mkdir build && cd build
conan install ..
cmake ..
cmake --build .
./myapp
Conan will generate FindMyPackage.cmake in build/ which will be loaded by your CMakeLists.txt.
Both demonstrations achieve what you asked, but I prefer the second, because is safer, as you can create a specific target, and avoid any mistake from the customer side.
NOTE: The feature Components requires Conan >=1.27.