Why it causes an ERROR?

Question

I have coded a simple program in C++ on CODEBLOCK. the program is as follows:

#include <iostream>
using namespace std;

int main()
{
    int num = 09;              //ERROR: Invalid digit 9 in octal constant
    cout << num << endl;

    num = 08;             //ERROR: Invalid digit 9 in octal constant
    cout << num << endl;


    return 0;
}

This code is same as the previous one. But i have changed value and wrote num=09 instead of 015.

I'm agree that if i initialize (int num = 015), it gives output in OCTAL that is 13.

But in the above program where i tried to initialize (int num = 09 and num = 08) it gives ERROR that you can see.

First of all, I wish to know why it generates an ERROR and how?

Second is what is the logic behind it?

Please give me logical reasons with suitable examples if any.

Come on guys, it's a legitimate question and possibly difficult to google for. At least mark it as a dupe instead of downvoting. — Matti Virkkunen, Mar 04 '16 at 17:58
@MattiVirkkunen I googled "c++ zero in front of integer literal" and the answer was contained in the google preview of the first page result. — Weak to Enuma Elish, Mar 04 '16 at 17:59
@JamesRoot: Unless you put in the word "literal" which you might not realize all the results are about zero-padding numbers for output. — Matti Virkkunen, Mar 04 '16 at 18:00
@crashmstr Nice, you found the dupe I couldn't find somehow. — Matti Virkkunen, Mar 04 '16 at 18:01
And a C answer: [How does C Handle Integer Literals with Leading Zeros, and What About atoi?](http://stackoverflow.com/questions/1661369/how-does-c-handle-integer-literals-with-leading-zeros-and-what-about-atoi) — crashmstr, Mar 04 '16 at 18:01
Brother @AlexD I agree with you. But there is again another question if i remove 0 it gives 15 which is in decimal. RIGHT? Why ? — Jalal Jan Khan, Mar 04 '16 at 18:03
@JalalJanKhan: Because it's the leading zero that causes it to be interpreted as octal. Remove the leading zero, it's interpreted as decimal. — Matti Virkkunen, Mar 04 '16 at 18:04
thank you brother @MattiVirkkunen. That is what i was thinking. THANKS ONCE AGAIN — Jalal Jan Khan, Mar 04 '16 at 18:11
@James Root Note: In C, at least, `015` is not a literal. It is an integer constant. C defines 2 types of literal: _string_ and _compound_ and `015` is neither. — chux - Reinstate Monica, Mar 04 '16 at 18:39
Regarding the invalid digit, octal is base 8. That means the valid digits are 0-7, just like valid binary digits are 0-1. So the digits 8 and 9 are not valid octal digits. — dbush, Mar 04 '16 at 19:22

score 6 · Accepted Answer · answered Mar 04 '16 at 17:59

6

Leading zeros cause integers to be interpreted as octal numbers in C/C++. 015 in octal is (1*8) + 5 = 13 in decimal.

answered Mar 04 '16 at 17:59

Matti Virkkunen

63,558
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Detail `0x9` has a leading `0` and is not octal. Some extensions include `0b` as a binary prefix. – chux - Reinstate Monica Mar 04 '16 at 18:42

score 0 · Answer 2 · answered Mar 04 '16 at 18:11

0

Constants in 8-base are defined that way. 15₈ = 13₁₀.

answered Mar 04 '16 at 18:11

Ivan Gritsenko

4,166
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Why it causes an ERROR?

2 Answers2