How do I refer to a class given its name as a string?

Question

In Perl:

package Foo {
    sub new { bless {} }
    sub some_method { 42 }
}
my $f = "Foo"->new;
say ref $f;         # 'Foo'
$f->some_method;

In Python:

class Foo:
    def some_method():
        return 42

f = globals()['Foo']()
print(type(f).__name__)     # 'Foo'
f.some_method

In Ruby:

class Foo
    def some_method
        return 42
    end
end

f = Module.const_get('Foo').new
puts f.class            # 'Foo'
f.some_method

In Javascript:

class Foo {
    some_method() { return 42 }
}

let f = new ?????????????;
console.log(f.constructor.name);    // 'Foo'
f.some_method();

If Foo were a plain function and not a class, this["Foo"] would work, but how do you deal with classes? I tried eval, but the f instance will not exist outside the scope, so that solution is not generally suitable.

Edit to address possible duplicates: Factories, registries and the like only work when I can refer to an existing class in the factory, but if the classes I want to refer to are not known in advance, I cannot use this work-around.

do you mean you want to create an instance of the class based on a string `"foo"` or are you just talking about instantiating a class in a standard way? — Liam, Apr 21 '17 at 09:21
@AnmolMittal: Completely missing the point of the question: The OP wants to refer to Foo by name: `new something['Foo']()` — Cerbrus, Apr 21 '17 at 09:22
Guys, read the damn question. There's ___3___ examples of other languages where he refers to the class by it's name. — Cerbrus, Apr 21 '17 at 09:23
I don't know any of those languages.....For all I know that's how you instantiate a class in Ruby.... — Liam, Apr 21 '17 at 09:23
@Liam: Neither do I, but that doesn't make the snippets incomprehensible. Don't assume those languages _need_ you to wrap a class name in quotes. That's just silly. — Cerbrus, Apr 21 '17 at 09:24
but why you want to make it more complex when you can directly access the class by "Foo" ?? — Anmol Mittal, Apr 21 '17 at 09:25
@AnmolMittal because sometimes you can only know the name of the class you want to instantiate inside a given scope. This question does make sense, it just has already been answered in other posts. — Logar, Apr 21 '17 at 09:27
@Logar: That dupe target is basically hardcoding a list of classes. I'm pretty sure that's not what the OP wants. — Cerbrus, Apr 21 '17 at 09:28
By the way, what's wrong with new this["Foo"] ? I mean, if you can make it work with a function, I'm pretty sure you can make it work with a class. — Logar, Apr 21 '17 at 09:49
that's not es6, you are using babel compiler, which basically converts class to function — Anmol Mittal, Apr 21 '17 at 09:59
Thanks for the research, this question can be marked as a duplicate. — daxim, Apr 21 '17 at 10:34

score -2 · Answer 1 · edited May 23 '17 at 10:31

-2

What I suggest is use some variable for mapping :-

class Foo{
 a(){}
}

var class_map = {
 "Foo" : Foo
}

var f = new class_map['Foo']()

Here it specifies that

globally-declared classes are globals, but not properties of the global object...

Hence, you have to map explicitly

edited May 23 '17 at 10:31

Community

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answered Apr 21 '17 at 09:43

Anmol Mittal

843
5
12

score -2 · Accepted Answer · edited May 23 '17 at 12:17

-2

Copied the relevant part from https://stackoverflow.com/a/40069309/46395:

class Foo {
    some_method() { return 42 }
}

let f = new(eval('Foo'));
console.log(f.constructor.name);    // 'Foo'
f.some_method();

Please go upvote that original answer.

edited May 23 '17 at 12:17

Community

1
1

answered Apr 21 '17 at 10:32

daxim

39,270
4
65
132

If it's a dupe, CV as dupe, don't answer it. – Cerbrus Apr 21 '17 at 10:53

How do I refer to a class given its name as a string?

2 Answers2