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Why does babel rewrite imported function call to (0, fn)(...)?

Given an input file like

import { a } from 'b';

function x () {
  a()
}

babel will compile it to

'use strict';

var _b = require('b');

function x() {
  (0, _b.a)();
}

but when compiled in loose mode the function call is output as _b.a();

I've done some research into where the comma operator is added in the hope there was a comment explaining it. The code responsible for adding it is here.

like image 927
Will Smith Avatar asked Aug 28 '15 15:08

Will Smith


3 Answers

(0, _b.a)() ensures that the function _b.a is called with this set to the global object (or if strict mode is enabled, to undefined). If you were to call _b.a() directly, then _b.a is called with this set to _b.

(0, _b.a)(); is equivalent to

0; // Ignore result
var tmp = _b.a;
tmp();

(the , is the comma operator, see https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Comma_Operator).

like image 197
Rob W Avatar answered Nov 08 '22 23:11

Rob W


The comma operator evaluates each of its operands (from left to right) and returns the value of the last operand.

console.log((1, 2)); // Returns 2 in console
console.log((a = b = 3, c = 4)); // Returns 4 in console

So, let see an example:

var a = {
  foo: function() {
    console.log(this === window);
  }
};

a.foo(); // Returns 'false' in console
(0, a.foo)(); // Returns 'true' in console

Now, in foo method, this is equal to a (because foo is attached to a). So if you call a.foo() directly, it will log false in console.

But, if you were call (0, a.foo)(). The expression (0, a.foo) will evaluate each of its operands (from left to right) and returns the value of the last operand. In other words, (0, a.foo) is equivalent to

function() {
  console.log(this === window);
}

Since this function no longer is attached to anything, its this is the global object window. That's why it log true in console when call (0, a.foo)().

like image 26
Huong Nguyen Avatar answered Nov 09 '22 01:11

Huong Nguyen


Calling a function in this roundabout way:

(throwAwayValueHere, fn)(args);

works like this:

  • The comma expression throwAwayValueHere, fn is evaluated: The comma operator evaluates its first operand, throws away that value, then evaluates its second operand and takes that value as its result.
  • That value is then called as a function, passing in the arguments.

Doing the call that way has an effect in two situations:

1. If the function is on an object property, e.g.:

(throwAwayValueHere, obj.fn)(args);

it calls the function without setting this to obj during the function call; instead, it's set to the default, either the global this value (window on browsers) or undefined in strict mode.

Example:

"use strict";
const obj = {
    value: 42,
    fn: function() {
        console.log(`typeof this = ${typeof this}`);
        if (typeof this === "object") {
            console.log(`this.value = ${this.value}`);
        }
    }
};

// Normal call:
console.log(`obj.fn():`);
obj.fn();

// Indirect call:
console.log(`(0, obj.fn)():`);
(0, obj.fn)();

This is the reason Babel is doing it there: In the original code, the call was simply a(), which calls a with the default this value. Doing (0, _b.a)() does the same thing even though a is a property of _b.

2. If the function is eval, it makes it an indirect eval which means it's evaluated as though at global scope, rather than eval's default behavior of running arbitrary code from a string in local scope, giving it access to all in-scope variables.

Example:

"use strict";

let a = "global a";

function directEval() {
    let a = "local a";
    eval("console.log(`a = ${a}`);");
}

function indirectEval() {
    let a = "local a";
    (0,eval)("console.log(`a = ${a}`);");
}

console.log("direct:");
directEval();
console.log("indirect:");
indirectEval();
like image 2
T.J. Crowder Avatar answered Nov 09 '22 00:11

T.J. Crowder