了解条件类型的类型推断

Question

我很难理解为什么 TypeScript 将这里的 program.options 推断为 ProgramAOptions | ProgramBOptions。因此它无法编译代码，因为optA 在ProgramBOptions 中不存在。你能解释一下或指向我解释这种行为的文档吗？

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

通过为每对名称和选项显式声明接口，我将通过以下方式解决您的问题：

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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