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Accepted
Milenko Markovic
Asked: 2025-01-06 18:57:26 +0800 CST

如何使用非空断言重构我的代码？对象可能“未定义”

  • 772

我有问题

export const tfModules = (moduleIds: readonly string[], mapping: Map<string, [string, string]>): string[] =>
  moduleIds.map(m => mapping.get(m).at(0))
    .filter(m => m !== undefined)
    .map(m => m!);

得到

error TS2532: Object is possibly 'undefined'.

使用非空断言来表明值不能为空或未定义的最简单方法是什么？

typescript

1 个回答

  1. Best Answer

    您可以mapping.get(m)!：

    export const tfModules = (moduleIds: readonly string[], mapping: Map<string, [string, string]>): string[] =>
  moduleIds.map(m => mapping.get(m)!.at(0)).filter(m => m !== undefined)

    操场

    没有非空断言的更快版本：

    export const tfModules = (moduleIds: readonly string[], mapping: Map<string, [string, string]>): string[] =>
  moduleIds.reduce((r, m) => {
    const found = mapping.get(m);
    found && r.push(found[0]);
    return r;
  }, [] as string[]);

    操场

    • 3

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