How would you describe the modulus function in c++ utilizing equations that don't include the actual modulus function

[dalydir]

If x mod y equals a. I know that a = the remainder of x divided by y.
I've got a mental block and I can't figure out how I'd program mod without the utilization of the mod ("%") function. I have some variables in a program I'm using that I cannot use the mod function directly upon, without it disrupting the equation.

Anyone mind typing out some quick code of how I could substitute the mod function? My numbers are anywhere from 1 to 512 bit numbers I'm trying to mod.

[Ispil]

Well, mathematically I'd see it like this:

a/b = c
floor(c)
a-(b*c) = d

and d would be the remainder.

[dalydir]

I was not aware of the command "floor", thank you for that. Though, I'm still in need of help; boost::multiprecision (for the 512bit integers) doesn't have a matching function to call for "floor" 8(

[Peter87]

For integers you can do it with only simple arithmetic operations.

[dalydir]

I've got to do it with fractions that cannot be simplified until after the modulus is completed. The only way I've found to do that is to do each numerator/denominator individually. Peter87, that's what I'm asking for, what the arithmetic operations would be, though I may have found them.

[Peter87]

Hint: Think about what integer division (division of two integers) gives you in C++.