`Option::ok_or` is not a const fn. What's the reason for it?

1

option.rs - source

    /// Transforms the `Option<T>` into a [`Result<T, E>`], mapping [`Some(v)`] to
    /// [`Ok(v)`] and [`None`] to [`Err(err)`].
    ///
    /// Arguments passed to `ok_or` are eagerly evaluated; if you are passing the
    /// result of a function call, it is recommended to use [`ok_or_else`], which is
    /// lazily evaluated.
    ///
    /// [`Ok(v)`]: Ok
    /// [`Err(err)`]: Err
    /// [`Some(v)`]: Some
    /// [`ok_or_else`]: Option::ok_or_else
    ///
    /// # Examples
    ///
    /// ```
    /// let x = Some("foo");
    /// assert_eq!(x.ok_or(0), Ok("foo"));
    ///
    /// let x: Option<&str> = None;
    /// assert_eq!(x.ok_or(0), Err(0));
    /// ```
    #[inline]
    #[stable(feature = "rust1", since = "1.0.0")]
    pub fn ok_or<E>(self, err: E) -> Result<T, E> {
        match self {
            Some(v) => Ok(v),
            None => Err(err),
        }
    }
2

It's not yet allowed to drop a value of a generic type T inside of a const fn. ok_or() would drop err in the case of Ok, so it may not be const fn.

If this is a problem for you, I recommend writing the match in place of ok_or().

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