CODEWITHSUNDEEP
javascript
Zed_Blade
Zed_Blade

Reputation: 1081

Javascript string evaluation

There is a portuguese IT company that is looking for some developers in a few areas and just out of curiosity (since I already have a job, thankfully) I went to check the job postings.

When I went to check out the JS developer posting, they provided a piece of JS code that caught my attention. I've worked for some time with JS and I find myself getting back to programming with JS from time to time but to be honest I've never seen anything even similar to the code given.

This is the code:

!(function(x){
    '6D 61 6E'.split(' ').forEach(function(a){
        x+=String.fromCharCode(parseInt(a,16));
    });
    return x;
})('');

I went and wrote this on Chrome's JS console and the output is 'false'. If I understand it correctly, the "strange" code, and according to the ASCII table reads 'm a n', and parseInt is supposed to return an integer based on a hexadecimal radix. It then gets converted once again into a string, this time based on the chars decimal value. To finish it all, we evaluate the return 'x' by "negating it" (not the word I was looking for but can't remember a better one at the time... evaluate maybe?).

Then, why is the output false? If we don't evaluate the return the result is the expected one 'man', but I don't see why we get false on this particular instance.

Anyone care to elaborate?

Upvotes: 2

Views: 180

Answers (1)

nnnnnn
nnnnnn

Reputation: 150030

As you seem to have worked out,

return x;

...will return the string "man". But your question seems to boil down to why !"man" gives false?

From MDN, logical not !:

Returns false if its single operand can be converted to true; otherwise, returns true.

The empty string "" is falsey, so !"" is true, but any other string is truthy, so !"any other string" is false.

Upvotes: 3

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