Partial with

[jdforsythe]

TypeScript Version: 2.9.2

Search Terms: partial, spread, optional, required

Code

TS Playground

interface Foo {
  bar: string;
  name: string;
}

const common: Partial<Foo> = { bar: 'foobar' };
const foos: Foo[] = [
  { ...common, name: 'one' },
  { ...common, name: 'two' },
]; // error

const foosTwo: Foo[] = [
  { ...<Foo>common, name: 'one' },
  { ...<Foo>common, name: 'two' },
]; // type coercion works 

Expected behavior:

Typescript knows all the properties are provided and there are no errors

Actual behavior:

[ts]
Type '{ name: string; bar?: string; }[]' is not assignable to type 'Foo[]'.
  Type '{ name: string; bar?: string; }' is not assignable to type 'Foo'.
    Property 'bar' is optional in type '{ name: string; bar?: string; }' but required in type 'Foo'.

This error does not occur when assigning to a class property instead of a const:

TS Playground

class FooFactory {
  private foos: Foo[];

  constructor(foos) {
    this.foos = f;
  }
}

const common: Partial<Foo> = { bar: 'foobar' };

const fs = new FooFactory([
  { ...common, name: 'one' },
  { ...common, name: 'two' },
]); // no error

Playground Link:

Related Issues:

[RyanCavanaugh]

const common: Partial<Foo> = { bar: 'foobar' };
            ~~~~~~~~~~~~~~

When you write a type annotation for a variable declaration, TypeScript validates that the initializer conforms to that type, but otherwise discards information about the initializer. So we later here:

const foos: Foo[] = [
  { ...common, name: 'one' },
  { ...common, name: 'two' },
]; // error

TS doesn't know that you didn't specify name and not bar in common.

You can simply remove that type information to make the error go away.

In the class example, the constructor takes a parameter of type any so naturally there is no error.

See #26064 for discussion of syntax that would allow simultaneous in-place type validation without losing inference information from the initializer.