Question

假设我有以下代码

const publicPath: string | undefined = config.output && config.output.publicPath
invariant(publicPath, "No publicPath for config: ${config}")
// declare const publicPath: string // error

我知道invariant publicPath之后非空（如果是，程序会停止），怎么说打字稿？

即使这不起作用

function invariant<T>(s: T | undefined | null, msg): s is T | never {
    if (!s)
        throw new Error(msg)
    return true
}


const publicPath: string | undefined = '/* ... */'
invariant(publicPath, "No publicPath for config: ${config}")
// publicPath still string

Answer 1

您可以使用更严格的类型声明一个新变量：

const configuredPublicPath: string | undefined = config.output && config.output.publicPath

const publicPath: string = invariant(configuredPublicPath,
                                                     "No publicPath for config: ${config}")

Playground demo link here（请务必点击Options -> strictNullChecks）。

Answer 2

不可能。

你可以这样做：

const publicPath: string | undefined = /* ... */
invariant(publicPath, "No publicPath for config: ${config}")
const publicPath2 = publicPath as string

或者，我建议：

function invariant(s: string | undefined, msg): string {
    if (!s)
        throw new Error(msg)
    return s
}
const publicPath = invariant(config.output && config.output.publicPath, "No publicPath for config: ${config}")

注意：如果配置类型不是所有字符串，请使用泛型：

function invariant<T>(s: T | undefined, msg): T {
    if (!s)
        throw new Error(msg)
    return s
}

Answer 3

您可以使用not-null断言运算符（!）断言变量不为null且未定义：

const publicPath: string | undefined = config.output && config.output.publicPath
invariant(publicPath, "No publicPath for config: ${config}")

const nonNullPublicPath: string = publicPath!

验证后将类型更改为非null

3 个答案: