What does an extra 0 in front of an int value mean?

Question

Inspiring from a obfuscated piece of code, I have a small question regarding to assign value to an integer:

#include <iostream>
#include <cstdio>

int main() {
    int i = 0101;
    std::cout << i << "\n";
}

And the output was 65, and I have no idea where 65 came from? Any idea?

The linked duplicate is closed and generally of low quality. This question and answer however, are plain and straight-forward. I'm re-opening this with the intention to use as canonical dupe for octal literal FAQs. — Lundin, Jan 11 '19 at 07:30

score 12 · Accepted Answer · answered Jan 18 '11 at 06:01

12

It specifies an octal (base-8) number: 0101 == 1 * (8 * 8) + 1 == 65.

answered Jan 18 '11 at 06:01

user541686

@:Lambert: Thanks a lot ;)! It was really interesting. – roxrook Jan 18 '11 at 06:05
@Chan: You're welcome. :) I actually learned it myself a few hours ago, when learning about the grammar of the D language! – user541686 Jan 18 '11 at 06:06
@Chan: The full expansion of 0101, if the extra detail helps you understand: 101 in base 8 is (each digit from left to right) `1 * (8 ** 2) + 0 * (8 ** 1) + 1 * (8 ** 0)`; this reduces to the expression in the answer. – Fred Nurk Jan 18 '11 at 06:37

Nawaz · Answer 2 · 2011-01-18T06:17:54.130

0

Lambert already explained that. So let me tell you what else you can do.

You can write hexadecimal integer:

int main() {
    int i = 0x101; //0x specifies this (i.e 101) is hexadecimal integer
    std::cout << i << "\n"; //prints 257 (1 * 16 * 16 + 1)
}

Output:

edited Jan 18 '11 at 06:17

answered Jan 18 '11 at 06:10

Nawaz

2 Answers2