Struct Unknown

pub struct Unknown<T = ()>(/* private fields */);

Available on crate feature serde only.

Easily convert a non-Specta type into a Specta compatible type. This will be typed as unknown in Typescript.

Examples

This can be used as a type override.

use serde::Serialize;
use specta::{Type, Unknown};

#[derive(Serialize, Type)]
pub struct Demo {
    #[specta(type = Unknown)]
    pub field: String,
}

Or it can be used as a wrapper type.

use serde::Serialize;
use specta::{Type, Unknown};

#[derive(Serialize, Type)]
pub struct Demo {
    pub field: Unknown<String>,
}

Trait Implementations

impl<T: Clone> Clone for Unknown<T>

fn clone(&self) -> Self

Returns a copy of the value.

fn clone_from(&mut self, source: &Self)

Performs copy-assignment from source.

impl<T: Debug> Debug for Unknown<T>

fn fmt(&self, f: &mut Formatter<'_>) -> Result

Formats the value using the given formatter.

impl<T: Default> Default for Unknown<T>

fn default() -> Self

Returns the “default value” for a type.

impl<T: Serialize> Serialize for Unknown<T>

fn serialize<S>(&self, serializer: S) -> Result<S::Ok, S::Error>

where S: Serializer,

Serialize this value into the given Serde serializer.

impl<T> Type for Unknown<T>

fn inline(_: &mut TypeCollection, _: Generics<'_>) -> DataType

Returns the definition of a type using the provided generics.

fn reference(type_map: &mut TypeCollection, generics: &[DataType]) -> Reference

Generates a datatype corresponding to a reference to this type, as determined by its category. Getting a reference to a type implies that it should belong in the type map (since it has to be referenced from somewhere), so the output of definition will be put into the type map.