test: use void to indicate "useless" expressions

[Trott]

Instead of disabling the no-unused-expressions ESLint rule, mark
intentional "useless" expressions with void.

Refs: https://eslint.org/docs/rules/no-unused-expressions#options

[nodejs-github-bot]

CI: https://ci.nodejs.org/job/node-test-pull-request/42474/

[targos]

The .out files of the message tests have to be updated.

[benjamingr]

(I don't think this improves code quality and I think it introduces a pretty rarely used language feature to help with an eslint rule)

[Trott]

(I don't think this improves code quality and I think it introduces a pretty rarely used language feature to help with an eslint rule)

I think you're right. I want to believe there's something more semantic about void foo.bar; instead of foo.bar; but explicit is better than implicit, and explicit would be a comment and not void.