Compiler suggests invalid syntax for derive attribute

[alexxbb]

Code

#[derive(Debug, Deserialize)]
pub struct Build {
    #[serde(deserialize_with = "as_u64")]
    build: u32,
    product: Product,
    release: String,
    status: String,
    version: String,
}

Current output

error[E0308]: `?` operator has incompatible types
   --> downloader-api\src\lib.rs:140:17
    |
140 | #[derive(Debug, Deserialize)]
    |                 ^^^^^^^^^^^ expected `u32`, found `u64`
    |
    = note: `?` operator cannot convert from `u64` to `u32`
    = note: this error originates in the derive macro `Deserialize` (in Nightly builds, run with -Z macro-backtrace for more info)
help: you can convert a `u64` to a `u32` and panic if the converted value doesn't fit
    |
140 | #[derive(Debug, Deserialize.try_into().unwrap())]
    |                            ++++++++++++++++++++

Desired output

Not sure what the correct output should be, but the current one suggests using an invalid syntax: `#[derive(Debug, Deserialize.try_into().unwrap())]`
If no useful "help" message can be provided in this context, the compiler should not suggest an invalid syntax in the "help" section.

Rationale and extra context

No response

Other cases

No response

Anything else?

rustc 1.76.0-nightly (9a66e4471 2023-11-19)

[long-long-float]

@rustbot claim

[long-long-float]

Short code to reproduce the issue.

use serde::{Deserialize, Deserializer};

#[derive(Debug, Deserialize)]
pub struct Build {
    #[serde(deserialize_with = "as_u64")]
    build: u32,
}

fn as_u64<'de, D>(d: D) -> Result<u64, D::Error>
where
    D: Deserializer<'de>,
{
    u64::deserialize(d)
}

fn main() {}