Confusing behaviour of 'let _ = '

[steffahn]

Relevant answer on a previous - somewhat related - thread.

In particular the last paragraph:

@steffahn

wouldn’t let _ = EXPR; be the same as EXPR; in all the cases without a must_use warning, and the same as drop(EXPR) in all must_use cases? I suppose we could just add warnings against every let _ = EXPR; and either suggest EXPR; or drop(EXPR) for replacement depending on whether a must_use warning would be generated.