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Javascript can create atomic bombs (programming.dev)
submitted 3 years ago by to c/programmerhumor@lemmy.ml
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[–] 6 points 3 years ago (1 child) 
var bomb = []
for(var i = 0; i === -1; i++){
  bomb.push(i)
}
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  • [–] 1 point 3 years ago (2 children)

    what does 3 equals signs do

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  • [–] 4 points 3 years ago (1 child)

    Identity. "A is literally B" instead of "A equals B". This is necessary here in JS because if A is the string "-1" and B is the integer -1, JS evaluates A==B as true because reasons

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  • [–] 2 points 3 years ago* (2 children)

    because if A is the string “-1” and B is the integer -1, JS evaluates A==B as true because reasons

    Interesting. If it were the other way around, I think I would have been fine with it (i.e. == used for comparison with type like any other language and === without type). But as it stands now I would hate it if I had to write in JS (but I don't so it's fine).

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  • [–] 3 points 3 years ago

    It's not that bad, honestly, just something you get used to. When I switch to C++ after a while, I sometimes write === and when I switch back to JS after some time, I occasionally forget to use ===.

    In C++ it's obviously an error and for JS I have my IDE set to warn me about ==. I think I've used == in JS (and PHP) intentionally once in the last ~5 years.

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  • [–] 1 point 2 years ago

    Honestly, I think it actually makes some sense this way around. To me, in JS "==" is kinda "is like" while "===" is "is exactly". Or, put another way, "equals" versus idk, "more-equals". I mean, "===" is a much stronger check of equivalence than normal "==", so I think it deserves to be the one with the extra "="

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  • [–] 3 points 3 years ago* (last edited 3 years ago) (2 children)

    2 equal signs will coerce the second operand into the type of first operand then do a comparison of it can. so 1 == "1" is true. this leads to strange bugs.

    3 equal signs do not do implicit type conversion, cuts down on weird bugs. 1==="1" is false.

    edit: it appears to be more complicated than that for double equals and the position of operands don't matter. https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Equality

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  • [–] 1 point 3 years ago (1 child)

    Doesn't it widen the types regardless of position? Meaning 1 == "1" will be compared as strings, not numbers.

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  • [–] 1 point 3 years ago (1 child)

    https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Equality

    mdn goes into it more and it's way more involved than I thought, looks like order of operand doesn't matter. see the number to string section

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  • [–] 1 point 3 years ago

    It seems it is that way, which is weird. You should always convert to the widest type, meaning string for comparing numbers and strings. I just checked that 1 == "01" is true, which means that "01" gets cast to an integer. And according to the document it might be that for example 1 == "a" would basically be interpreted as 1 === NaN which is false.

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