了解條件類型的類型推斷

Question

我很難理解為什么 TypeScript 在這里將program.options推斷為ProgramAOptions | ProgramBOptions ProgramAOptions | ProgramBOptions 。 因此它無法編譯代碼，因為optA在ProgramBOptions中不存在。 您能否解釋或指出解釋此行為的文檔？

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'.*/
    }
}

Answer 1

通過為每對名稱和選項顯式聲明接口，我將通過以下方式解決您的問題：

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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