[version] Display version when booting operator

[surajssd]

Operator shows it's version when it starts. This version information
is embedded inside the binary while building it. It is a good
practice to show the operator version when it is booting up.

Fixes #348

[surajssd]

@krnowak I did not understand the part where you are saying

And our release steps will need to be extended with a check for a proper version of the operator.

[krnowak]

I did not understand the part where you are saying

@surajssd: Basically, when doing a release of the operator, we want to make sure that the version it prints is something like v0.7.2, not v0.7.2-31-g9eae1e5d nor v0.7.2-31-g9eae1e5d-dirty nor v0.7.2-dirty.

[surajssd]

@krnowak

And our release steps will need to be extended with a check for a proper version of the operator.

this is done :-)

[krnowak]

What I actually meant is to see if the operator, after being built without overriding the VERSION Make variable, prints a version like 0.7.2 without commit checksum or "dirty" string. The reason is to guard against building the operator from the dirty tree or from untagged commit by accident.

But this kind of check might be hard/annoying to execute… So I just got an idea that you could add a rule to Makefile like:

print-version:
	@echo $(VERSION)

and then, in this script, right after the readonly VERSION="v$(cat VERSION)" line, you could compare the output of make print-version with the VERSION variable. If they are different, bail out.

What do you think?

[surajssd]

@krnowak is this check needed? Because this will be only called from hack/publish-release.sh which makes sure that the directory is setup properly and then binary build happens?

[krnowak]

One nit, otherwise LFAD.

[krnowak]

LFAD if tests pass.