Why make does not use jobs when compiling? #58
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I usually run I have noticed the following output, while my CPU uses only a single core: $ make -j $(nproc)
gcc -o nelua-lua \
-DNDEBUG -DLUA_COMPAT_5_3 -DLUA_ROOT='"/usr/local/"' -DLUA_USE_RPMALLOC -DENABLE_GLOBAL_CACHE=0 -DENABLE_UNLIMITED_CACHE=1 -DBUILD_DYNAMIC_LINK -DLUA_USE_LINUX \
-Ilua \
-Wall -O2 -fno-plt -fno-stack-protector -flto \
-s -Wl,-E \
lua/onelua.c lfs.c sys.c hasher.c lpeglabel/*.c rpmalloc/rpmalloc.c \
-lm -ldl Is this the expected behavior? |
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Replies: 2 comments
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This is for compiling the bundled Lua interpreter that comes with Nelua, this is usually done only once when installing Nelua. It is expected because all the C code are concatenated to a big single C source file, so the compiler can generate a more optimized binary, this optimization technique is known as amalgamation, thus this is expected. Also note that generated code by Nelua compiler also does this. |
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Ah you must have been inspired by SQLite's amalgamation technique. Nice, I like it. Well done 👍 |
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This is for compiling the bundled Lua interpreter that comes with Nelua, this is usually done only once when installing Nelua. It is expected because all the C code are concatenated to a big single C source file, so the compiler can generate a more optimized binary, this optimization technique is known as amalgamation, thus this is expected. Also note that generated code by Nelua compiler also does this.