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 i

Answer 1

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

Answer 2

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

0110

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

110

Answer 3

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.

Answer 4

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.