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A little JS [ES3] task: In each case, implement `x` so that the expression is true. Do not modify the expressions though. Post your solutions as a fork of this gist. ***Remember: ES3 only.
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/x/==x |
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x() === x()() === x()()() |
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x > 2 && x < 2 |
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delete x.a && x.a; |
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x[x]==x |
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typeof new x < typeof x |
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x - 1 === x + 1 |
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Nice quiz
expression with delete still didn't solved