Understanding type inference with conditional types

Question

I have trouble understanding why TypeScript infers program.options here as ProgramAOptions | ProgramBOptions. Thus it cannot compile code since optA does not exist in ProgramBOptions. Can you explain or point me to documentation which explains this behaviour?

type ProgramName = 'a' | 'b';

type ProgramAOptions = {
    optA: number;
};

type ProgramBOptions = {
    optB: number;
};

type Program<T extends ProgramName> = {
    name: T;
    options: T extends 'a' ? ProgramAOptions : ProgramBOptions;
};name

function test(p: Program<ProgramName>) : void
{
    if (p.name === 'a')
    {
        p.options.optA = 10; /* this line would not compile with error:

        error TS2339: Property 'optA' does not exist on type 'ProgramAOptions | ProgramBOptions'.
        Property 'optA' does not exist on type 'ProgramBOptions'.*/
    }
}

Guerric P · Accepted Answer · 2021-12-14 17:27:55Z

2

Here is how I'd solve your problem, by explicitly declaring interfaces for every couple of names and options:

type ProgramName = 'a' | 'b';

interface ProgramAOptions {
    optA: number;
};

interface ProgramBOptions {
    optB: number;
};

type ProgramOptions = ProgramAOptions | ProgramBOptions;

interface Program {
    name: ProgramName;
    options: ProgramOptions;
}

interface ProgramA extends Program {
    name: 'a';
    options: ProgramAOptions;
}

interface ProgramB extends Program {
    name: 'b';
    options: ProgramBOptions;
}

type Programs = ProgramA | ProgramB;

function test(p: Programs): void {
    if (p.name === 'a') {
        p.name // is of type "a"
        p.options.optA = 10; // Works
    }
}

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answered Dec 14, 2021 at 17:27

Guerric P

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2 Comments

Kuba Wyrostek Over a year ago

This works great indeed, +1, nevertheless I would love to understand why 'conditional types' way does not.

Guerric P Over a year ago

Because type narrowing works on single variables and on union types. In your case you have a single generic type in which you provided yourself the T type so you give no room to type narrowing

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