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Does CMAKE_BUILD_TYPE=Release implicitly imply -DNDEBUG?
If not: isn't it reasonable to expect that this implication takes place?
I want to know if following CMake code is redundant in my CMakeLists.txt:
if (NOT CMAKE_BUILD_TYPE MATCHES Debug) add_definitions(-DNDEBUG) endif() 
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Specifies the build type on single-configuration generators. This statically specifies what build type (configuration) will be built in this build tree. Possible values are empty, Debug , Release , RelWithDebInfo and MinSizeRel .
exerpt: "RelwithDebInfo is quite similar to Release mode. It produces fully optimized code, but also builds the program database, and inserts debug line information to give a debugger a good chance at guessing where in the code you are at any time."
Simply delete the CMakeFiles/ directory inside your build directory. rm -rf CMakeFiles/ cmake --build . This causes CMake to rerun, and build system files are regenerated. Your build will also start from scratch.
In this answer, it says Debug is the default cmake build configuration.
Yes, it is set by CMake. Grepping through the CMake code reveals, that for a host of compilers it is set. Probably they set it only for these compilers, which accepts this flag. Here one of the lines concerning GCC:
Modules/Compiler/GNU.cmake: set(CMAKE_${lang}_FLAGS_RELEASE_INIT "-O3 -DNDEBUG") But be aware that many projects overwrite release/debug flags without preserving the initial setting and also overwriting user's definitions.
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