JavaScript && opertor in return statement

I have a piece of JavaScript code as follows:

function main(condition){
    if(condition){
        doSomething();
        return obj;
    }
}

now I want to refactor this code to get rid of the "if" statement. Here is what I want to do

function main(condition){
    var doSomethingAndReturnObj = function(){
        doSomething();
        return obj;
    }
    return condition &&a开发者_JS百科mp; doSomethingAndReturnObj();
}

here is where I need help. The caller of main function expects a return value of "undefined" or an obj. In my refactored code, would my

return condition && doSomethingAndReturnObj();

convert the return value to a true and false type?

Thanks for your replies.

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The short answer is: no, && does not convert things to Boolean values.

&& only continues if it gets a truthy value, returning the first falsy value, or the last value, meaning:

undefined && true == undefined
true && true      == true
true && false     == false
1 && 2 && 3       == 3
1 && 2 && 0 && 4  == 0

So if your condition is falsy, e.g. false or undefined, it will return that exact value. If your condition is truthy, it will return whatever doSomethingAndReturnObj() returns.

falsy values for reference: null, undefined, 0, false, NaN, "". everything else is truthy.

---

JavaScript's && operator is coalescing.

js> 0 && 7
0
js> 1 && 7
7

---

The value returned by doSomethingAndReturnObj() will not be converted to true or false.