On Wed, Jan 19, 2022 at 2:56 AM Paul Smith <[email protected]> wrote: > > On Tue, 2022-01-18 at 10:14 +0800, Hongyi Zhao wrote: > > But it seems that the `$?' used above can't obtain the exit status of > > the code block in @(...) when used from outside the @() structure > > [2]. So, I want to know if I can reliably obtain the exit status of > > the code block in @(...) from outside the @() structure. Any hints > > will be greatly appreciated. > > First, of course you must use $$? not $?, because $? is a shell > variable not a make variable so you must escape it from make. > > Second, there's nothing magical about ().
This reminds me of the following similar problems when using GNU Makefile variable assignment as discussed here [1]: VAL = foo VARIABLE = $(VAL) In the above usage, can I also change to the following? VAL = foo VARIABLE = $$VAL [1] https://programmer.help/blogs/gnu-makefile-variable-assignment-what-is-the-difference-between.html Regards, HZ
