In es2015, `const func = foo => bar` makes `func` a named function, how do you bypass this?

Question

Is there any way to get around this behavior?

> foo => bar;
[Function]
> const func = foo => bar;
undefined
> func
[Function: func]

Answer 1

The simplest way to avoid this without affecting your code structure is probably to use the identity function to wrap the definition:

const id = x => x;

const func = id( foo => bar );
console.log(func.name) // undefined

If you don't want to declare a helper function, you can inline it as well, or use an IIFE:

const func1 = (f=>f)( foo => bar );
const func2 = (()=> foo => bar )();

but basically anything except the grouping operator (a parenthesised expression) will do:

const func3 = (0, foo => bar); // comma operator, known from indirect eval
const func4 = [foo => bar][0]; // member access on array literal
const func5 = true ? foo => bar : null; // ternary operator

_{(And while you're at it, don't forget to add an appropriate comment that explains why you're doing this)}

Of course, that might not prevent a debugger or console from inferring some other, internal name for the function.