Is it possible in TypeScript to assert that a const object literal is made in such a way where each key equals its value?
In other words:
// Good
const testIds: KeyEqualsValue = {
foo: 'foo'
} as const
// Bad
const testIds: KeyEqualsValue = {
foo: 'bar' // Error
} as const
Not with a single type, you can do it with a function:
function propAsValue<T extends { [P in keyof T]: P }>(o: T) {
return o;
}
const testIds = propAsValue({
foo: 'foo'
});
const testIds2 = propAsValue({
foo: 'bar'
});
Or with an inline function, if you want to be terse and confuse everyone:
const testIds = (<T extends { [P in keyof T]: P }>(o: T) => o)({
foo: 'foo'
});
Although I am not sure what your use case for this is, you might be better off with using Object.keys.
keyof typeof testIds. Your solution meets these requirements :)
T extends [...] keyof T My mind cannot wrap around how this is not a cyclical derivation.
Yes you can get the object keys with Object.keys then you can loop through the keys and compare the string values. If they do not equal you can throw an error.
