I was going through my company’s code base and found a statement that compares an array with 0 like this:
array > 0;
if we let array = [“1”], which has a single element, the above statement would be true; but if let array = [“1”, “2”] or [], the statement would become false;
Can someone explain the meaning of this statement, why it yields such results, and if it would be useful in any situation?
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Answer
When you use >, the engine will first convert both sides to a primitive, which will call the valueOf, and, if that didn’t return a primitive, then the toString method if it exists. For arrays, only the toString method returns a primitive, so that’s what’s used – and what it does is equivalent to doing .join(',').
console.log(['1', '2'].toString());
Looking at the spec again, after the array has been turned into a primitive, we now have one side that’s a string (which came from the array), and another side that’s a number. So, both sides are then converted to numbers:
d. Let nx be ? ToNumeric(px). e. Let ny be ? ToNumeric(py).
And then the numbers are compared.
In the case of ['1'], you get 1 > 0, which is true.
In the case of ['1', '2'], the resulting string is '1,2', which cannot be converted into a number, so the following runs:
h. If nx or ny is NaN, return undefined.
and when undefined is returned by this algorithm, the whole > evaluates to false.
and if it would be useful in any situation?
For clean, understandable code, it generally wouldn’t. Better to explicitly cast to types whose comparison makes intuitive sense first.