0
276
ViewsTypescipt evaluates conditional types incorrectly
I have a very primitive code that for some reason gives a type error.
type ConfigEntry = {
name?: string,
value?: string
}
function parser1(c: ConfigEntry): Record<string, string> {
const name = c.name;
const value = c.value;
return { name: value }; // fine, error since name can be undefined
}
function parser2(c: ConfigEntry): Record<string, string> | undefined {
const name = c.name;
const value = c.value;
if (name === undefined) return undefined;
else return { name: value }; // compiler still argues that name can be undefined
// but in any circumstances it is not
}
If I understand TS correctly, compiler should examine conditions and evaluate types accordingly.
Also none of the other conditional expressions known to me work as well.
Example is here
3 answers
Answer question0
You are only doing null check for name, while you should be doing it for value instead.
function parser2(c: ConfigEntry): Record<string, string> | undefined {
const name = c.name; // <--- This line is not needed since we are not doing anything with `name`
const value = c.value;
if (value === undefined) return undefined; // <--- Should check `value` here. `name` is not needed because it is unused.
else return { name: value };
}
0
You are checking that name is undefined. However here
return { name: value };
What you are really using is the value variable. The only thing you need to change is checking that value is undefined:
if (value === undefined) return undefined;
else return { name: value };
0
What lead me to confusion is that because of a typo (name vs [name]) TS was highlighting key, not value, that actually had an error
So the correct code will be fixing the typo and specifying the return value as
Record<string, string | undefined>
function parser4(c: ConfigEntry): Record<string, string | undefined> | undefined {
const name = c.name;
const value = c.value;
return name ? { [name]: value } : undefined;
}Thank you all for the help.