[javascript] Can (a== 1 && a ==2 && a==3) ever evaluate to true?

If you take advantage of how == works, you could simply create an object with a custom toString (or valueOf) function that changes what it returns each time it is used such that it satisfies all three conditions.

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const a = {_x000D_
  i: 1,_x000D_
  toString: function () {_x000D_
    return a.i++;_x000D_
  }_x000D_
}_x000D_
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if(a == 1 && a == 2 && a == 3) {_x000D_
  console.log('Hello World!');_x000D_
}
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The reason this works is due to the use of the loose equality operator. When using loose equality, if one of the operands is of a different type than the other, the engine will attempt to convert one to the other. In the case of an object on the left and a number on the right, it will attempt to convert the object to a number by first calling valueOf if it is callable, and failing that, it will call toString. I used toString in this case simply because it's what came to mind, valueOf would make more sense. If I instead returned a string from toString, the engine would have then attempted to convert the string to a number giving us the same end result, though with a slightly longer path.