Clever Geek Handbook

Strange behavior with JavaScript expressions

Can someone please explain to me why this is happening? And please, if anyone knows this behavior, edit the title.

With this code:

 const arr = ['RIPA'], varB = "RIPB";
let _params;
_params && Array.isArray(_params) ? arr.push(..._params) : 
arr.push(_params);

_params && console.log("I will never appear");
varB && console.log("I will appear");

arr.push(varB);
console.log('array',arr);
console.log("Type of the _params --> ", typeof _params);

Output:

array [ 'RIPA', undefined, 'RIPB' ]
Type of the _params -->  undefined

jsBIN: https://jsbin.com/bawepasivo/edit?js,console
repl.it: https://repl.it/GaHX

If _params

- undefined

, as far as possible the execution of the second expression, if the expression &&

returns the first false and the last trusted value.

+3


source to share


3 answers


Your expression runs like this:

(_params && Array.isArray(_params)) ? arr.push(..._params) : arr.push(_params);

But you probably meant the following:



_params && (Array.isArray(_params) ? arr.push(..._params) : arr.push(_params));

You just need to add parentheses.

+2


source 


 false  && false                  ? never executed       : _params is undefined
_params && Array.isArray(_params) ? arr.push(..._params) : arr.push(_params);

another way:



if (_params && Array.isArray(_params)) { // (false && false) === false
    arr.push(..._params); // it will be never executed
} else {
    arr.push(_params); // _params is undefined
}
+1


source 


let _params; // undefined

_params && Array.isArray(_params) ?

- false

, so the called code arr.push(_params);

, which results inarr.push(undefined);

0


source






More articles:


All Articles