Upon coming across this insightful answer to the query
How does jQuery's $ function operate as both a function and an object?In traditional JS, the following code is perfectly valid:
var f = function() { alert('yo'); }
f.foo = "bar";
alert(f.foo); // displays "bar"
f(); // displays "yo"
However, in typescript, calling f.foo will result in an error:
Property 'foo' does not exist on type '() => void'One workaround would be using bracket notation:
var f = function() { alert('yo'); }
f['foo'] = "bar";
alert(f['foo']); // displays "bar"
f(); // displays "yo"
However, this approach completely circumvents the type system and consequently sacrifices the type safety offered by typescript.
Is there a method to integrate this functionality without compromising the type safety of typescript?