ES6默认参数：值分配不正确

Question

I'm playing with ES6 Default parameters, and I got a weird behaviour.

Here is an short example of the problem:

function test(firstValue, secondValue=5){
  console.log("firstValue: " + firstValue);
  console.log("secondValue: " + secondValue);
  console.log("----------")
}

test(2, secondValue = 3)
test(secondValue = 3)

And its output:

firstValue: 2
secondValue: 3
----------
firstValue: 3
secondValue: 5
----------

In the second case, I expected firstValue: undefined and secondValue: 3 . Is this behaviour normal. Do I miss something?

Answer 1

When you do

test(2, secondValue = 3)

you are, in effect, doing this:

secondValue = 3
test(2, 3)

The first part ( secondValue = 3 ) creates a global variable called secondValue thanks to The Horror of Implicit Globals .* It has nothing to do with the secondValue parameter in your function. JavaScript doesn't have named arguments. (Eg, you can't say "here's the value for secondValue " when making a call except by putting it in the right position in the argument list. If you want to specify the names when making the call, you can use destructuring as cubbuk points out , which isn't really named arguments but can serve the same sort of purpose.)

The second part (passing 3 into test ) happens because the result of an assignment is the value that was assigned (so secondValue = 3 sets secondValue to 3 and results in the value 3 , which is then passed into test ).

To call test with 3 for secondValue , just do that. Eg:

test(2, 3);

If you want to leave the second one off, the default will be used:

test(2);

Example:

 function test(firstValue, secondValue=5){ console.log("firstValue: " + firstValue); console.log("secondValue: " + secondValue); console.log("----------"); } test(2, 3); test(2); 

---

* (that's a post on my anemic little blog)

Answer 2

You are using a global variable without declaring it. The assignment generates a global variable and does not affect the variables of function test . These local variables are independent form the calling scope.

 function test(firstValue, secondValue=5){ console.log("firstValue: " + firstValue); console.log("secondValue: " + secondValue); console.log("----------") } var secondValue; // with declaration console.log(secondValue); test(2, secondValue = 3) test(secondValue = 3) console.log(secondValue); 

Answer 3

You want to do destructuring I guess:

 function test({firstValue, secondValue = 5} = {}){ console.log("firstValue: " + firstValue); console.log("secondValue: " + secondValue); console.log("----------") } test() // prints firstValue: undefined, secondValue: 5 test({}) // prints firstValue: undefined, secondValue: 5 test({firstValue: 2}) // prints firstValue: 2, secondValue: 5 test({secondValue: 3}) // prints firstValue: undefined, secondValue: 3