question in hugeint class

[obay]

how to make (/)operator?


// Fig. 11.23: Hugeint.cpp
// HugeInt member-function and friend-function definitions.
#include <cctype> // isdigit function prototype
#include <cstring> // strlen function prototype
//#include <math.h> i don't need pow function
#include "Hint.h" // HugeInt class definition
#include<cstdlib>
// default constructor; conversion constructor that converts
// a long integer into a HugeInt object
HugeInt::HugeInt( long value )
{
// initialize array to zero
for ( int i = 0; i <= 29; i++ )
integer[ i ] = 0;

// place digits of argument into array
for ( int j = 29; value != 0 && j >= 0; j-- )
{
integer[ j ] = value % 10;
value /= 10;
} // end for
} // end HugeInt default/conversion constructor

// conversion constructor that converts a character string
// representing a large integer into a HugeInt object
HugeInt::HugeInt( const char *string )
{
// initialize array to zero
for ( int i = 0; i <= 29; i++ )
integer[ i ] = 0;

// place digits of argument into array
int length = strlen( string );

for ( int j = 30 - length, k = 0; j <= 29; j++, k++ )

if ( isdigit( string[ k ] ) )
integer[ j ] = string[ k ] - '0';
} // end HugeInt conversion constructor

// addition operator; HugeInt + HugeInt
HugeInt HugeInt::operator+( const HugeInt &op2 ) const
{
HugeInt temp; // temporary result
int carry = 0;

for ( int i = 29; i >= 0; i-- )
{
temp.integer[ i ] =
integer[ i ] + op2.integer[ i ] + carry;

// determine whether to carry a 1
if ( temp.integer[ i ] > 9 )
{
temp.integer[ i ] %= 10; // reduce to 0-9
carry = 1;
} // end if
else // no carry
carry = 0;
} // end for

return temp; // return copy of temporary object
} // end function operator+

// addition operator; HugeInt + int
HugeInt HugeInt::operator+( int op2 ) const
{
// convert op2 to a HugeInt, then invoke
// operator+ for two HugeInt objects
return *this + HugeInt( op2 );
} // end function operator+

// addition operator;
// HugeInt + string that represents large integer value
HugeInt HugeInt::operator+( const char *op2 ) const
{
// convert op2 to a HugeInt, then invoke
// operator+ for two HugeInt objects
return *this + HugeInt( op2 );
} // end operator+

// overloaded output operator
ostream& operator<<( ostream &output, const HugeInt &num )
{
int i;

for ( i = 0; ( num.integer[ i ] == 0 ) && ( i <= 29 ); i++ )
; // skip leading zeros

if ( i == 30 )
output << 0;
else

for ( ; i <= 29; i++ )
output << num.integer[ i ];

return output;
}
int HugeInt::countd()
{
int i;
for ( i = 0; (integer[ i ] == 0 ) && ( i <= 29 ); i++ );
i=30-i;
return i;
}

HugeInt HugeInt::operator*(HugeInt o)
{
HugeInt hold("0");
for(int i=29,dn=29-o.countd();i>dn;i--)
{
hold=hold + *this*o.integer[i];
return hold;
}
}

HugeInt HugeInt::operator*(const int o)
{
HugeInt temp(1); // temporary result
int carry = 0;
int dn=29-countd();
for ( int i = 29; i >= dn; i-- )
{
//std::cout<<"d = "<<carry<<" ";
temp.integer[ i ] =
integer[ i ] * o + carry;
std::cout<<"d = "<<carry<<" __ ";//zero!
// determine whether to carry a 1
if ( temp.integer[ i ] > 9 )
{
//std::cout<<"d = "<<temp.integer[ i ]<<" ";
carry = temp.integer[ i ]/10;
temp.integer[ i ] %= 10; // reduce to 0-9
} // end if
else // no carry
carry = 0;
} // end for

return temp; // return copy of temporary object
}

bool HugeInt::operator>(HugeInt o)
{
if(countd()>o.countd())
return 1;
else
if(countd()<o.countd())
return 0;
else
for(int i=30-countd();i<30;i++)
{
if(integer[i]>o.integer[i])
return 1;
else
if(integer[i]<o.integer[i])
return 0;
}
return 0;
}

[Aceix]

OP wrote:

how to make (/)operator?

Do you mean overloading, and pls use code tags

[doug4]

I agree about using code tags. You can use the "<>" button on the Format options. It will make your code MUCH easier for us to read.

And what's the deal with declaring functions to return bool and then returning an int? Use the values true and false. That's what they are there for, and it is easier for a reader to know what's going on.