程序员Kevin
8/28/2025
TypeScript Generics Issue: Default Parameter Bypassing Type Check 😤
Hey TypeScript wizards! 👋 I'm stuck on this weird generic type issue and it's driving me nuts. I've got this Example
class that's supposed to enforce a minimum number constraint, but the default constructor is totally bypassing my type checks!
Here's the setup:
type Config = { min?: number; max?: number; }; class Example<C extends Config> { example = true; // This empty default is causing problems! 😫 constructor(private config: Readonly<C> = {} as C) {} public min<const N extends number>(min: N) { return new Example<C & { min: N }>({ ...this.config, min, }); } }
I want to make sure new Example()
can't be assigned to Example<{ min: 1 }>
, but right now TypeScript is letting it slide!
Here's what I've tested:
// These SHOULD work ✅ test(new Example({ min: 1 })); test(new Example().min(1)); test(new Example({}).min(1)); // These correctly fail ❌ test(new Example({})); test(new Example({ min: 2 })); test(new Example().min(2)); // But this one sneaks through! 😡 test(new Example()); // Why no error?!
I tried fixing the test function to catch it:
const test = <C extends { min: 1 }>(example: Example<C>) => example; test(new Example()); // Now it errors, but that's cheating!
The real problem: I need new Example()
itself to be invalid when assigned to Example<{ min: 1 }>
.
Things I've tried:
Any ideas how to make TypeScript properly enforce this? I'm working on a library and this type safety is crucial! 🚨
PS: Here's the Playground link if you want to poke around. Thanks in advance for saving my sanity! 🙏
全栈小王
8/28/2025
Hey there! 👋 Oh man, I feel your pain with this TypeScript generics puzzle! I've wrestled with similar issues when building type-safe libraries, and that sneaky default parameter behavior can be super frustrating. Let me share what I've learned from my own battles with TypeScript's type system. 🛡️
The Core Issue:
The problem is that {} as C
is essentially telling TypeScript "trust me, this empty object is of type C" (that as C
is the culprit!). TypeScript's type system is structural, so an empty object {}
technically satisfies any object type unless we explicitly prevent it.
Here's How to Fix It: We need to make the default parameter type-safe while keeping the flexibility you want. Here's the solution I'd recommend:
class Example<C extends Config = {}> { // Note the default type here! example = true; constructor(private config: Readonly<C> = {} as C) {} public min<const N extends number>(min: N) { return new Example<C & { min: N }>({ ...this.config, min, }); } } // Now this will properly error 🎉 const test = (example: Example<{ min: 1 }>) => example; test(new Example()); // Error: Property 'min' is missing!
Key Improvements:
= {}
as the default generic type parameternew Example()
explicitly an Example<{}>
Example<{ min: 1 }>
Bonus Tip: If you want to make this even more bulletproof, you could use a factory function instead of direct constructor calls:
function createExample<C extends Config>(config?: C) { return new Example(config || {}); } // Now you get proper type inference too! const example = createExample().min(1); // correctly typed
Watch Out For:
as
assertion is always a red flag 🚩 - it's often better to restructure your types than to use type assertions{}
satisfies any object type unless constrainedPersonal Experience: I once spent a whole day debugging a similar issue in a validation library! The solution was similar - being explicit about default generic types made all the difference. TypeScript can be tricky, but when you get it right, the type safety is chef's kiss 🤌
Hope this helps! If you're still running into issues or want to explore alternative approaches, feel free to ask. Happy typing! 💻🎉
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