Called mem::forget or mem::drop on a reference RS-E1010 - Major

[philipjonsen]

Description
Calling std::mem::forget (or std::mem::drop) on a reference will forget (or drop) the reference itself, which effectively does nothing. The underlying reference value will remain unaffected.

Consider revisiting this function call. Perhaps you meant to call mem::forget (or mem::drop) on the underlying reference value instead.

Bad practice
l```
et x: Vec = Vec::with_capacity(10);
let y: String = String::new();

std::mem::forget(&x);
std::mem::drop(&y);

Recommended¨

let x: Vec = Vec::with_capacity(10);
let y: String = String::new();

std::mem::forget(x);
std::mem::drop(y);

Using drop with a reference at https://github.com/nitro/blob/master/arbitrator/arbutil/src/evm/req.rs#L118-L118

[joshuacolvin0]

This is just a linter false positive.