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Severity: error
Category: type-check
Stage: Stage 1

Description

TypeScript conditional types (T extends U ? A : B) are not valid SJS syntax. They add metaprogramming complexity without soundness guarantees.

If the goal is to select a type based on a condition, use explicit overloads or separate typed functions instead.

Example

// ✗ error
type Flatten<T> = T extends Array<infer U> ? U : T    // SJS-E008

Fix

Write explicit typed functions or use unknown + narrowing:

// ✓ correct — explicit function per case
function flattenNumber(arr: number[]): number { return arr[0] }
function flattenString(arr: string[]): string { return arr[0] }

For library utility types, restructure as separate explicitly-typed interfaces or generic functions with explicit constraints.

Related codes

  • SJS-E005 — intersection types also banned
  • SJS-E006 — mapped types also banned
  • SJS-E009 — infer keyword also banned
Documentation