What is the reason behind Typescript's discomfort with utilizing a basic object as an interface containing exclusively optional properties?

Question

What is the reason behind Typescript's discomfort with utilizing a basic object as an interface containing exclusively optional properties?

Trying to create a mock for uirouter's StateService has been a bit challenging for me. This is what I have attempted:

beforeEach(() => {
  stateService = jasmine.createSpyObj('StateService', ['go']) as StateService;
}

...

it('calls go', () => {
  // ...
  let options = { location: 'replace', inherit: true };
  expect(stateService.go).toHaveBeenCalledWith('foo', params, options);
});

The error message I receive is:

error TS2345: Argument of type '{ location: string; inherit: boolean; }' is not assignable to parameter of type 'Expected'. Type '{ location: string; inherit: boolean; }' is not assignable to type '{ location?: ExpectedRecursive<boolean | "replace">; relative?: ExpectedRecursive<string | StateObject | StateDeclaration>; ... 8 more ...; source?: ExpectedRecursive<...>; }'. Types of property 'location' are incompatible. Type 'string' is not assignable to type 'ExpectedRecursive<boolean | "replace">'. expect(stateService.go).toHaveBeenCalledWith('foo', params, options);

However, if I place the object directly without assigning it to a variable, everything works smoothly:

it('calls go', () => {
  // ...
  expect(stateService.go).toHaveBeenCalledWith('foo', params, { location: 'replace', inherit: true });
});

I noticed that uirouter's TransitionOptions interface allows for optional properties:

export interface TransitionOptions {
    location?: boolean | 'replace';
    relative?: string | StateDeclaration | StateObject;
    inherit?: boolean;
    notify?: boolean;
    reload?: boolean | string | StateDeclaration | StateObject;
    custom?: any;
    supercede?: boolean;
    reloadState?: StateObject;
    redirectedFrom?: Transition;
    current?: () => Transition;
    source?: 'sref' | 'url' | 'redirect' | 'otherwise' | 'unknown';
}

Even though everything is optional in the interface, Typescript seems to have an issue when I assign this interface to a simple object as a variable. Why does Typescript behave like this?

typescript

Answer 1

Answer №1

The classification of this particular line:

let options = { position: 'modify', include: true };

is

options: { position: text, boolean: true }

This determination is based on the fact that the data type of 'replace' has been extended to string. This causes string to be too broad to assign to position, which should ideally be 'modify' | boolean | undefined.

In order to inform Typescript that 'replace' is not just any string but specifically replace, one can utilize as const.

let options = { position: 'modify' as const, include: true };

Answer 2

The classification of this particular line:

let options = { position: 'modify', include: true };

is

options: { position: text, boolean: true }

This determination is based on the fact that the data type of 'replace' has been extended to string. This causes string to be too broad to assign to position, which should ideally be 'modify' | boolean | undefined.

In order to inform Typescript that 'replace' is not just any string but specifically replace, one can utilize as const.

let options = { position: 'modify' as const, include: true };

Answer 3

Answer №2

Let's simplify this with a similar error in a different context.

function bar(value: "test") {
}

bar("test"); // This is fine

const val1 = "test";
bar(val1); // This works too

let val2 = "test";
bar(val2);  // Compiler error 

let val3:"test" = "test"; 
bar(val3); // No issues here

When you declare a variable and assign a string value, TypeScript assumes it's of type string, not a specific string literal type.

Applying this to your options scenario:

let options1 = { 
    location: 'test',  
    inherit: true 
};

bar(options1.location); // Compilation error

let options2: {location:'test', inherit: boolean} = { 
    location: 'test', 
    inherit: true 
};

bar(options2.location); // This works!

You've encountered an interesting behavior in TypeScript when distinguishing between a string type and a string literal type. Here are some additional intriguing cases:

function bar(value: "test") {
}

function baz(param: {location: 'test'}) {
}

let obj = {
    location: 'test'
}

baz(obj); // Error during compilation
bar(obj.location); // Also leads to a compilation error

baz({ location: 'test'}); // No error.

bar( { location: 'test'}.location); // However, the compiler flags this as an error! (I believe it shouldn't)

Answer 4

Let's simplify this with a similar error in a different context.

function bar(value: "test") {
}

bar("test"); // This is fine

const val1 = "test";
bar(val1); // This works too

let val2 = "test";
bar(val2);  // Compiler error 

let val3:"test" = "test"; 
bar(val3); // No issues here

When you declare a variable and assign a string value, TypeScript assumes it's of type string, not a specific string literal type.

Applying this to your options scenario:

let options1 = { 
    location: 'test',  
    inherit: true 
};

bar(options1.location); // Compilation error

let options2: {location:'test', inherit: boolean} = { 
    location: 'test', 
    inherit: true 
};

bar(options2.location); // This works!

You've encountered an interesting behavior in TypeScript when distinguishing between a string type and a string literal type. Here are some additional intriguing cases:

function bar(value: "test") {
}

function baz(param: {location: 'test'}) {
}

let obj = {
    location: 'test'
}

baz(obj); // Error during compilation
bar(obj.location); // Also leads to a compilation error

baz({ location: 'test'}); // No error.

bar( { location: 'test'}.location); // However, the compiler flags this as an error! (I believe it shouldn't)

Answer 5

Answer №3

The reason behind this is that when you use type inference to declare a variable based on its assigned value, the location property will be generalized as a string. Consequently, this makes it incompatible with the replace value since replace will only refer to the property value and not its data type.

To remedy this issue, you should explicitly specify the variable type during declaration:

function foo(param: TransitionOptions) {
    console.log('foo', param)
}

foo({ location: 'replace', inherit: true })

let bar: TransitionOptions = { location: 'replace', inherit: true }
foo(bar)

Answer 6

The reason behind this is that when you use type inference to declare a variable based on its assigned value, the location property will be generalized as a string. Consequently, this makes it incompatible with the replace value since replace will only refer to the property value and not its data type.

To remedy this issue, you should explicitly specify the variable type during declaration:

function foo(param: TransitionOptions) {
    console.log('foo', param)
}

foo({ location: 'replace', inherit: true })

let bar: TransitionOptions = { location: 'replace', inherit: true }
foo(bar)

What is the reason behind Typescript's discomfort with utilizing a basic object as an interface containing exclusively optional properties?

Answer №1

Answer №2

Answer №3

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