//  ============================================================
//
//  assign.cpp
//
//  Copyright 2002, Dennis Meilicke and Rene Peralta
//
//  ============================================================
//
//  Description:
//
//  Methods for assigning values to a LargeNumber (ln).
//
//  ============================================================
#include <lnv3/ln3impl.h>


//  ============================================================


void
Assign( Base *A, long a )
{
    if( a == 0 )
    {
        A->size = 0;
        A->sign = positive;
    }
    else
    {
        A->size = 1;
        if( a < 0 )
        {
            A->sign = negative;
            A->digit[ 0 ] = ( 0 - a );
        }
        else
        {
            A->sign = positive;
            A->digit[ 0 ] = a;
        }
    }
}


//  ============================================================


void
Assign( Base *A, unsigned long a )
{
    if( a == 0 )
    {
        A->size = 0;
        A->sign = positive;
    }
    else
    {
        A->size = 1;
        A->sign = positive;
        A->digit[ 0 ] = a;
    }
}


//  ============================================================


void
Assign( Base *A, char *a )
//
//  This is a very simple minded character to binary conversion
//  routine.  This thing is probably "slow as a dog", but...
//
//  Possible improvements:
//  1.  Use numbers bigger than 10.  That is, convert groups
//  of numbers all at once.
//  2.  A divide-and-conquer algorithm might work nicely, here.
//  3.  See Knuth for a base-conversion algorithm.  Convert
//  the string to a base-10^n array.  Then use a base conversion 
//  algorithm to convert it to our base.  This is probably what 
//  ln2's algorithm was doing (who could tell?).
//
//  Since this assignment function should not be called all that
//  often, is it neccessary to squeeze every cycle out of this?
//
{
    A->sign = positive;
    A->size = 0;

    sign_t newSign = positive;
    for( ; *a ; a++ )
    {
        if( *a == '-' )
            newSign = negative;
        else if( *a == '+' )
            newSign = positive;
        else
        {
            //(*this) = 10 * (*this) + (list[i] - 48);
            Multiply( A, 10L );
            add( A, (long)( *a - 48 ) );
        }
    }

    A->sign = newSign;
}