I have the following code:
type Foo<T extends string = string> = `bar-${T}-foo`; const v: Foo = 'bar-g-foo' That works as expected, but it doesn't force the structure. The following is also valid:
const v: Foo = 'bar--foo' How can I force the usage of T?
The problem here is to check for an empty string. Read more about this problem here: How to write a string type that does not contain an empty string in TypeScript
Based on the stack from above you could do something like this:
type Character = 'a' | 'b' | 'c' | ... ; // Here all possible 1 letter strings type NonEmptyString = `${Character}${string}`; type Foo<T extends string = NonEmptyString> = `bar-${T}-foo`; const v: Foo = 'bar-a-foo' // ✔ const x: Foo = 'bar--foo' // ❌ Type '"bar--foo"' is not assignable to type 
Character union with all the characters you want to allow, but with types only (no functions to get better inference), I don't think you'll do better than this.
Type x = "". More about this here: catchts.com/type-negationIt might not be the best approach, becasue of the reliance on the function assigning the value of generic type dynamically, but depending on your usage this could also be vallid
type NonEmptyString<T extends string> = T extends "" ? never : T; function fn<T extends string>(a: `bar-${NonEmptyString<T>}-foo`) { // ... } fn("bar--foo"); // Error fn("bar-g-foo"); // No error fn("bar-gg-foo"); // No error 
type Foo<T extends string = string> = T extends "" ? never : `bar-${T}-foo`;would work (or perhaps without the default), but sadly, no. With the default, the type just ends up beingstring, and without the default, you have to explicitly provide the type parameter, and while doing that works, those aren't the ergonomics you want. :-|