Float Data Type

[anhome]

As i know float has 8 digit, how about if i want n-digit ?
for example:
in float, PI = 3.142857 -->8 digit
how about if want 10 digit? or 20 digit ?
wish somebody could solve this for me. i would be very thankful.

[giblit]

Double can use 16. If you want any larger than that you better make your own class of a "BigInt" type.

[anhome]

what is the content of the class ?can u tell me?

[ENIGMAx]

^^ There's some stuff on it around the 'net in case you'd want to learn about how it works and write it yourself, but you could try using source that someone else already wrote. Like for instance http://cpp-bigint.sourceforge.net/

I remember back when I learned this in a class at my University and a homework assignment was to write code for a BigInt and then put it to use.

[giblit]

Just create a new class for your custom number to start try something like

class BigInt
{
public:
    BigInt( void );
    BigInt( what ever you want here string might be easiest? );
};

Example using string:

class BigInt
{
public:
    BigInt( void );
    BigInt( const std::string& );
    BigInt& operator+=( const BigInt& );
private:
    std::string value;
};

BigInt::BigInt( void ) : value( "0" ) {}
BigInt::BigInt( const std::string &value ) : value( value ) {}
BigInt& BigInt::operator+= ( const BigInt &rhs )
{
    std::string temp( rhs.value ) , result( "" );
    std::reverse( value.begin() , value.end() );
    std::reverse( temp.begin() , temp.end() );
    short sum( 0u ) , carry( 0u ) , remainder( 0u );

    while( value.size() != temp.size() ) value.size() > temp.size() ? temp += '0' : value += '0';

    for( auto i( 0u ); i < value.size(); ++i )
    {
        sum = ( value[ i ] - '0' ) + ( temp[ i ] - '0' ) + carry;
        remainder = sum % 10;
        carry = ( sum - remainder ) / 10;
        result += remainder + '0';
    }
    if( carry > 0 ) result += carry + '0';
    std::reverse( result.begin() , result.end() );
    value = result;

    return( *this );
}

The code may not be very neat or well written though so good luck =p