C#

binary-operator-declarator:
type   operator   overloadable-binary-operator   (   type   identifier   ,   type   identifier   )
 
 
decimal-integer-literal:
decimal-digits   integer-type-suffixopt
 
hexadecimal-integer-literal:
0x   hex-digits   integer-type-suffixopt
0X   hex-digits   integer-type-suffixopt

Luc Pattyn wrote:

binary operators

Luc Pattyn wrote:

supporting binary and octal literals

Luc Pattyn wrote:

octal is thing of the past

Luc Pattyn wrote:

Binary notation tends to lack compactness

PIEBALDconsult wrote:

binary operators
... operate on two operands. You know that.

• We are dealing with number representaion in a given base, hence the function should return a string, not an integer.
• What if the input is 64 instead of 8?

CPallini wrote:

should return a string

CPallini wrote:

You are talking about literals

CPallini wrote:

conceptually (and technically) wrong

public static int
Oct
(
    int Value
)
{
    return ( System.Convert.ToInt32 ( Value.ToString() , 8 ) ) ;
}

musefan wrote:

public int Oct(int dec)
{
return ((dec / 8) * 10) + (dec % 8);
}

int decimal5=5;
int octal5=Oct(decimal5);   // returns 5
int decimal12=12;
int octal12=Oct(decimal12); // returns the value of fourtheen
int octalSum=octal5+octal12;
Console.WriteLine(octalSum); // prints "19" i.e. uses invalid characters in the octal world.

int five=ReadAnOctalNumberFromConsole();  // let's assume you enter 5
int twelve=12;
int sum=five+twelve;
WriteToConsoleUsingOctal(sum); // prints "21"

string five="5";
string twelve="12";
string sum=OctalStringArithmetic.Add(five, twelve);
Console.WriteLine(sum); // prints "21" if you did it well

erdinc27 wrote:

nothing on the root