Hi
I am running gnu make 4.1 on Ubuntu. Make builds my C++ application that
requires glibc 2.22 or higher. Ubuntu 14.04 has only glibc 2.19 so my makefile
checks the host o/s and, if it's Ubuntu, it links to my private copy of glibc
2.22:
GLIBC_FLAGS=
ifeq ($(DISTRO),debian)
echo "Warnin
This is primarily a shell coding question, not a make question. Once you've
worked out how to do it at the shell level you can wrap the result in the
$(shell ...) make function and, assuming proper escaping, it will work the
same. It looks to me like the version is the last word of the first line,
Hi David
Thanks for your answer to my question. I have one outstanding problem. Part of
the makefile is now:
ifeq ($(DISTRO),debian)
verstr := $(ldd --version | head -1 | awk '{print $NF}')
@echo glibc $(verstr) detected
ifneq ($(call version_ge,$(verstr),2.22),TRUE)
@echo "Warning: sy
Makefiles are divided between make syntax and shell (recipe) syntax. Here
you're trying to use a shell command (echo) in a place that's not a recipe.
Replace it with $(warning ...) which is the make-syntax equivalent.
On Tue, Apr 5, 2016 at 9:24 AM, David Aldrich
wrote:
> Hi David
>
> Thanks for
