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Asked: 2026-05-22T22:45:23+00:00

I have the following bits in my makefile : GLFW_FLAG := -m32 -O2 -Iglfw/include

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I have the following bits in my makefile:

GLFW_FLAG := -m32 -O2 -Iglfw/include -Iglfw/lib -Iglfw/lib/cocoa $(CFLAGS)
...
$(BUILD_DIR)/%.o : %.c
    $(CC) -c $(GLFW_FLAG) $< -o $@
$(BUILD_DIR)/%.o : %.m
    $(CC) -c $(GLFW_FLAG) $< -o $@

The -m32 instructs GCC to generate 32bit code. It’s there because on some configurations GHC is set to build 32bit code but GCC’s default is sometimes 64bit. I would like to generalize this so that it autodetects whether GHC is building 32bit or 64bit code and then passes the correct flag to GCC.

Question: How can I ask GHC what type of code (32bit vs. 64bit) it will build?

PS: My cabal file calls this makefile during the build to workaround limitations in cabal. I do wish I could just list these as c-sources in my cabal file.

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1 Answer

  1. Editorial Team

    Thanks to Ed’ka I know the correct answer now.

    The Makefile now has a rule like this:

    GCCFLAGS  := $(shell ghc --info | ghc -e "fmap read getContents >>= putStrLn . unwords . read . Data.Maybe.fromJust . lookup \"Gcc Linker flags\"")

    It’s a bit long, but all it does is extract the “Gcc Linker flags” from ghc’s output. Note: This is the output of ghc --info and not ghc +RTS --info.

    This is better than the other suggested ways as it gives me all the flags that need to be specified and not just the -m flag. It’s also empty when no flags are needed.

    Thank you everyone!

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