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Asked: 2026-05-24T07:46:32+00:00

eval(0+’1’+3) => 11 (???) When eval(0+’1′) is executed =>1. Iam expecting that 0+’1′ will

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eval(0+'1'+3) => 11 (???)

When eval(0+’1′) is executed =>1. Iam expecting that 0+’1′ will give me 1 & 3 will be considered as string & o/p => 13. But, why is that not happening?

whereas

eval(1+'1'+3) => 113
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1 Answer

  1. Editorial Team

    You are creating the string “013”, which is evaluated as a JavaScript integer literal. Integer literals starting with 0 are interpreted base 8 (octal), so your number is 8 + 3, which is 11.

    Only integer literals starting with a non-zero digit are interpreted base 10.

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