How is eval working overhere or how is it behaving?

Question

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 

Answer 1

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.

Answer 2

You are expecting that concatenating the number 0 and the string '1' would be concatenated as strings, but then evaluated as a number after being concatenated, and then converted again to give a string result.

That doesn't really make sense if you think of it. There is no reason that the string would be evaluated as a number, and besides, if it was evaluated as a number then the result would also be a number, not a string.

The number is converted to a string so that it can be concatenated to the other string, so the result is the same as for '0'+'1', which is the string '01'. There is no conversion to number and back to string after that.