I am trying to print previously set integer with following code:
int n=015;
cout << n;
The output I am getting is 13. Why would it be?
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6because when there is a leading zero it is in octal not in decimal and shouldn't that be 13? – Biruk Abebe Apr 22 '16 at 19:57
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1If you dont believe him try `int n = 09;` and [watch it not compile](http://coliru.stacked-crooked.com/a/659f3ef552e06a37). – Borgleader Apr 22 '16 at 19:59
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You may want to read more about the c++ syntax [integer_literal](http://en.cppreference.com/w/cpp/language/integer_literal) – gdlmx Apr 22 '16 at 20:02
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The output should be `13`, not `14`. – Keith Thompson Apr 22 '16 at 20:02
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@gdlmx: In early versions of C (K&R1, 1978), the digits `8` and `9` were permitted in octal constants. Both `09` and `011` would have had the value `9`. Fortunately that particular misfeature was removed. – Keith Thompson Apr 22 '16 at 20:03
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Ohhh....I see its an octal representation.thanks.@bkvnet – Hamim Apr 22 '16 at 20:04