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64 lines
1.5 KiB
64 lines
1.5 KiB
/*
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* A random test based on a Slashdot comment :-)
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*/
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var res;
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/*===
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0 number
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===*/
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/* {} + [] parses as an empty block statement (!) followed by unary plus
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* and an array, and is equivalent to:
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*
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* {};
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* +[];
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*
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* where +[] evaluates to ToNumber([]):
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*
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* [] -> ToPrimitive([], hint Number)
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* -> [].[[DefaultValue]] (hint = Number)
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* -> Object.prototype.valueOf() -> returns [] itself, not primitive
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* -> Array.prototype.toString()
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* -> ""
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* -> 0
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*
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* The coercion of an empty string or a string consisting only of
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* whitespace is 0 and not a NaN (!); see E5 Section 9.3.1.
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*/
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res = eval('{}+[]');
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print(res, typeof res);
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/*===
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[object Object] string
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[object Object] string
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===*/
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/* [] + {} parses as a binary addition.
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*
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* Each side is first coerced using ToPrimitive():
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*
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* ToPrimitive({})
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* -> {}.[[DefaultValue]] (no hint -> Number)
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* -> Object.prototype.valueOf() -> returns [] itself, not pritimive
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* -> Object.prototype.toString()
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* -> "[object Object]"
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*
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* ToPrimitive([])
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* -> [].[[DefaultValue]] (no hint -> Number)
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* -> Object.prototype.valueOf() -> returns {} itself, not primitive
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* -> Array.prototype.toString()
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* -> ""
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*
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* So, [] + {} should yield "" + "[object Object]" == "[object Object].
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*
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* Same goes for ({} + []) which is a binary addition expression because
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* of the parens.
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*/
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res = eval('[]+{}');
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print(res, typeof res);
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res = eval('({}+[])');
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print(res, typeof res);
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