CODEWITHSUNDEEP
javascriptecmascript-6default-parameters
Mortz
Mortz

Reputation: 4939

Why doesn't this ES6 default argument value give the desired result?

I found this ES6 code in a book:

let value = 5;

function getValue() {
    return value++;
}

function add(first, second = getValue()) {
    return first + second;
}

console.log(add(1, 1));     // 2
console.log(add(1));        // 6

Why does running console.log(add(1)); return 6 as the value, which means it is taking the parameter second as 5 although my code explicitly specifies that the getValue function should return value++ - which means the first time it is run, the getValue function should be returning 6 and add should be returning 7. I am running this code in the Firefox console - am I missing something?

Upvotes: 2

Views: 99

Answers (5)

Andrew Li
Andrew Li

Reputation: 57982

Default arguments are evaluated at call time, meaning the function getValue invoked every time you invoke add - not when the code is initially run. Since you're using postfix increment, getValue will return 5 the first time you invoke add, then 6, 7, etc. Postfix increment returns the previous value, then increments, for example:

var x = 5
var foo = x++;

Here, foo is given the value 5, then x is incremented. Thus, in your example, the getValue function is actually returning 5 instead 6, then incrementing value when you first invoke it. If you want 6, 7, 8, use prefix increment which returns the value after incrementing:

++x;

This will increment value, then return that incremented value. You could even use compound addition:

x += 1;

This explicitly reassigns x before you access it again. Further reading on those operators here.

Upvotes: 1

Farhad Azarbarzinniaz
Farhad Azarbarzinniaz

Reputation: 739

Hi this is called 'Default value' for parameters. It's mean that you can set Default value for each parameters in function definition .

function add(first, second = getValue()) {
    return first + second;
}

in this case 'second = getValue()', 'getValue()' is defulat value for 'second 'parameter.

when you do 

console.log(add(1, 1));     // 2

because
first==>1 second==>1

console.log(add(1));        // 6

because first==>1 second==>getValue() ===> 6

Upvotes: -1

Sotiris Kiritsis
Sotiris Kiritsis

Reputation: 3346

I believe your problem is not related to ecmascript-6, but in not understand the ++ operator correctly.

According to the documentation:

Unary operator. Adds one to its operand. If used as a prefix operator (++x), returns the value of its operand after adding one; if used as a postfix operator (x++), returns the value of its operand before adding one.

Replacing value++ with either ++value or value + 1 should solve your issue.

Upvotes: 1

Abdullah Danyal
Abdullah Danyal

Reputation: 1146

Because you use the postfix expression, if you use prefix expression then you will get 7.

Airthmetic operators

"If used postfix, with operator after operand (for example, x++), then it returns the value before incrementing. If used prefix with operator before operand (for example, ++x), then it returns the value after incrementing".

Upvotes: 1

Lorenzo Catalano
Lorenzo Catalano

Reputation: 477

value++

gets the value before increasing it,you should do either

++value

or

value+=1;
return value

Upvotes: 1

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