👀 Views: 22 💬 Answers: 1 📅 Created: 2025-06-24

typescript arrays type-inference flatten TypeScript

I'm having a hard time understanding I'm sure I'm missing something obvious here, but I've spent hours debugging this and I'm working on a TypeScript project where I need to flatten a multi-dimensional array that contains mixed types, including numbers, strings, and objects. My current implementation is resulting in type errors because TypeScript need to infer the correct types once the array is flattened. Here’s the array I’m working with: ```typescript const mixedArray: (number | string | { id: number; name: string })[][] = [ [1, 'two', { id: 1, name: 'Object1' }], [3, 'four', { id: 2, name: 'Object2' }], ['five', [6, { id: 3, name: 'Object3' }]], ]; ``` I want to flatten this array into a single-dimensional array while preserving the type information of each element. After trying to use `Array.prototype.flat()` like this: ```typescript const flattenedArray = mixedArray.flat(); ``` I encountered a TypeScript behavior saying that the type of `flattenedArray` is inferred as `(number | string | { id: number; name: string })[] | (number | { id: number; name: string })[]`. It seems that TypeScript is unable to handle the mixed types correctly when flattening. To resolve this, I’ve also tried using a recursive approach: ```typescript function flattenArray(arr: any[]): any[] { return arr.reduce((flat, toFlatten) => { return flat.concat(Array.isArray(toFlatten) ? flattenArray(toFlatten) : toFlatten); }, []); } const flattenedArray = flattenArray(mixedArray); ``` However, the type of `flattenedArray` is still showing as `any[]`, which defeats the purpose of using TypeScript. Is there a better way to flatten the array while keeping the type information intact? Any help or suggestions on how to approach this would be greatly appreciated! I’m using TypeScript version 4.5.2. How would you solve this? I recently upgraded to Typescript 3.11. I'm developing on macOS with Typescript.