The full error:
This expression is not callable.
No constituent of type 'true | CallableFunction' is callable
This is the code that was triggering the error:
public static base(
text,
callB: boolean | CallableFunction = false,
) {
const sw = Swal.fire({
text,
});
if (callB) {
sw.then(callB());
}
}
I changed the type of callB to :
callB: (param: any) => void |boolean = false
and when I remove the callB type definition:
callB= false
I get this error:
This expression is not callable.
Type 'Boolean' has no call signatures
This is what you are looking for:
function base(text: string, callB: false | CallableFunction = false) {
console.log(text);
if (callB) {
callB();
}
}
Essentially, you want to pass a function or false. When it is truthy you want to call the function.
But as per your definition
function base(text: string, callB: boolean | CallableFunction = false) {
console.log(text);
if (callB) {
callB();
}
}
this would be a valid call base('some str', true) where callB is true and not a function that it will try to execute using callB(). Not boolean is not callable, it is? Hence the error.
TS Playground link: https://tsplay.dev/m056qW
The type
boolean | Function
is equivalent to
true | false | Function
Now because a value of type Function is truthy, your check
if (callB)
narrows the type of callB to
true | Function
and true is not callable, hence the error.
It's not clear how your base method is meant to be used so I've outlined two solutions:
If true isn't a valid input, adjust your parameter type accordingly
static base(
text,
callB: false | CallableFunction = false
) {
if (callB) {
callB();
}
}
If true is a case you must handle, use a more precise conditional check such as
if (typeof callB === 'function') {
callB() // OK
}
If you use false | Callablefunction (or boolean | CallableFunction) , you can coerce type by using typeof:
static base(
text,
callB: false | CallableFunction = false
) {
if (typeof callB === "function") {
callB()
}
}