C++ int with preceding 0 changes entire value

Question

I have this very strange problem where if I declare an int like so

int time = 0110;

and then display it to the console the value returned is 72. However when I remove the 0 at the front so that int time = 110; the console then displays 110 like expected.

Two things I'd like to know, first of all why it does this with a preceding 0 at the start of the int and is there a way to stop it so that 0110 at least equals 110?
Secondly is there any way to keep it so that 0110 returns 0110?
If you take a crack guess at the variable name I'm trying to do operations with 24hr time, but at this point any time before 1000 is causing problems because of this.

Thanks in advance!

possible duplicate of [Is 0 a decimal literal or an octal literal?](http://stackoverflow.com/questions/6895522/is-0-a-decimal-literal-or-an-octal-literal) — b4hand, Mar 29 '15 at 05:24
possible duplicate of [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) — phuclv, Mar 29 '15 at 05:37

Vlad from Moscow · Accepted Answer · 2015-03-29T05:30:11.817

6

An integer literal that starts from 0 defines an octal integer literal. Now in C++ there are four categories of integer literals

integer-literal:
    decimal-literal integer-suffixopt
    octal-literal integer-suffixopt
    hexadecimal-literal integer-suffixopt
    binary-literal integer-suffixopt

And octal-integer literal is defined the following way

octal-literal:
    0 octal-literal
    opt octal-digit

That is it starts from 0.

Thus this octal integer literal

corresponds to the following decimal number

8^2 + 8^1

that is equal to 72.

You can be sure that 72 in octal representation is equivalent to 110 by running the following simple program

#include <iostream>
#include <iomanip>

int main() 
{
    std::cout << std::oct << 72 << std::endl;

    return 0;
}

The output is

edited Mar 29 '15 at 05:30

answered Mar 29 '15 at 05:18

Vlad from Moscow

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How would I then save it as a variable? – SteppingHat Mar 30 '15 at 12:32
@SteppingHat You can use a variable of any integral type that can accomodate this value.. The value of the variable will not depend on how the integer literal is written in the source code. – Vlad from Moscow Mar 30 '15 at 12:38

score 3 · Answer 2 · answered Feb 01 '19 at 14:55

It is because of Integer Literals. Placing a 0 before number means its a octal number. For binary it is 0b, for hexadecimal it is 0x or 0X. You don't need to write any thing for decimal. See the code bellow.

#include<stdio.h>

int main()
{
    int binary = 0b10;
    int octal=010;
    int decimal = 10;
    int hexa = 0x10;
    printf("%d %d %d %d\n", octal, decimal, hexa, binary);
}

For more information visit tutorialspoint.

score 2 · Answer 3 · answered Mar 29 '15 at 05:15

The compiler is interpreting the leading zero as an octal number. The octal value of "110" is 72 in decimal. There's no need for the leading zero if you're just storing an int value.

You're trying to store "time" as it appears on a clock. That's actually more complicated than a simple int. You could store the number of minutes since midnight instead.

score 1 · Answer 4 · answered Mar 29 '15 at 05:13

1

Zero at the start means the number is in octal. Without it is decimal.

answered Mar 29 '15 at 05:13

Ed Heal

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C++ int with preceding 0 changes entire value

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