includefile(include/header)

COMMENT(manpage, section, releasedate, archive, short name)
manpage(FBB::BigInt)(3bobcat)(_CurYrs_)(libbobcat-dev__CurVers_-x.tar.gz)
                    (Big Integers)

manpagename(FBB::BigInt)(Arithmetic on Integers of Unlimited Size)

manpagesynopsis()
    bf(#include <bobcat/bigint>)nl()
    Linking option: tt(-lbobcat -lcrypto) 

manpagedescription()

This class is defined as a wrapper class around the em(openSSL) tt(BN) series
of functions, offering members to perform arithmetic on integral values of
unlimited sizes. Members are offered to generate primes and to perform all
kinds of common arithmetic operations on tt(BigInt) objects. Also, conversions
to characters and standard numerical value types are offered.

Below, the phrase em(the object) may also refer to the object's value. The
context in which this occurs will make clear that the object's value rather
than the object as-is is referred to.

Various constructors accept tt(BIGNUM) arguments. Type tt(BIGNUM) is the type
containing an integer of unlimited precision as defined by OpenSSL.

Signs of tt(BigInt) are handled in a special way. Whether a tt(BigInt) is
negative or positive is determined by its sign-flag, and not by a sign bit as
is the case with tt(int) typed values. Since tt(BigInt) values have unlimited
precision shifting values to the left won't change their signs.

Operators return either a reference to the current (modified) object or return
a tt(BigInt) object containing the computed value. The rule followed here was
to implement the operators analogously to the way the operators work on
tt(int) type values and variables. E.g., tt(operator+()) returns a tt(BigInt)
value whereas tt(operator+=()) returns a tt(BigInt &) reference.

All members modifying their objects return a reference to the current
(modified) object. All members not modifying the current object return a
tt(BigInt) object. If both members exists performing the same
functionality the name of the member returning a tt(BigInt) object ends
in a tt(c) (const) (e.g., tt(addMod) and tt(addModc)).

Almost all operators, members and constructors (except for the default
constructor) throw tt(Exception) exceptions on failure.

manpagesection(INHERITS FROM)
    -

manpagesection(TYPE)

    The class bf(BigInt) defines the type tt(Word), which is equal to the type
tt(BN_ULONG) used by tt(OpenSSL) to store integral values of unlimited
precision. A tt(Word) is an tt(unsigned long), which is, depending on the
architecture, usually 64 or 32 bits long. 

manpagesection(ENUMERATIONS)
    bf(Msb)nl()
    This (most significant bit) enumeration is used when generating a
        cryptographically strong random number. Its values are:
    itemization(
    itb(MSB_UNKNOWN)
       The most significant bit may be 0 or 1.
    itb(MSB_IS_ONE)
        The most significant bit is guaranteed to be 1.
    itb(TOP_TWO_BITS_ONE)
        The two most significant bits are guaranteed to be 1, resulting in a
        product of two values each containing tt(nBits) having tt(2 * nBits)
        bits.
    )

    bf(Lsb)nl()
    This (least significant bit) enumeration is used when generating random
        numbers, ensuring that the resulting value is either odd or even.
    itemization(
    itb(EVEN)
        The random value will be an even value;
    itb(ODD)
        The random value will be an odd value.
    )

manpagesection(CONSTRUCTORS)

    itemization(
    itb(BigInt())
        The default constructor initializes a tt(BigInt) value to 0.
    itb(explicit BigInt(BIGNUM const &value))
        This constructor initializes a tt(BigInt) from a tt(const BIGNUM). 
    itb(explicit BigInt(BIGNUM const *value))
        This constructor initializes a tt(BigInt) from a pointer to a 
        tt(const BIGNUM).
    itb(BigInt(Type value))
        This constructor is defined as a member template. Any type that can be
converted using a static cast to an tt(unsigned long) can be used as argument
to this constructor. Promotion is allowed, so in many situations where
tt(BigInt)s are expected a plain numerical value can be used as well.
    itb(BigInt(char const *bigEndian, size_t length, bool negative = false))
        This constructor initializes a tt(BigInt) from tt(length)
big-endian encoded bytes stored in tt(bigEndian). This constructor interprets
the tt(char) values pointed at by tt(bigEndian) as unsigned values. Use this
constructor to reconstruct a tt(BigInt) object from the data made available by
the tt(bigEndian) member. If the number represents a negative value, then
provide a third argument tt(true).
    itb(explicit BigInt(std::string const &bigEndian, bool negative = false))
        This constructor initializes a tt(BigInt) from the bytes stored in
tt(bigEndian), which must be big-endian encoded. This constructor interprets
the tt(char) values stored in tt(bigEndian) as unsigned values. If the number
that is stored in tt(bigEndian) represents a negative value, then provide a
second argument tt(true).
    )

    The standard copy constructor is available, the move constructor is not
available. 


manpagesection(MEMBER FUNCTIONS)
    itemization(
    itb(BigInt &addMod(BigInt const &rhs, BigInt const &mod) )
       tt(Rhs) is added (modulo tt(mod)) to the current object.

    itb(BigInt addModc(BigInt const &rhs, BigInt const &mod) )
       The sum (modulo tt(mod)) of the current object and tt(rhs) is returned.

    itb(BigInt::Word at(size_t index) const)
       Returns the tt(Word) at tt(index). E.g., on a 32 bit architecture, if
        the bf(BigInt) value equals 2+sups(33), then tt(at(0)) returns 0, and
        tt(at(1)) returns 2. If tt(index) equals or exceeds the value returned
        by tt(nWords) an tt(FBB::Exception) is thrown.

    itb(BIGNUM const &bignum() const)
       A reference to the tt(BIGNUM) value maintained by the current
        tt(BigInt) object is returned.

    itb(char *bigEndian() const) The value represented by the current object
        is stored in a series of tt(char) typed values in big-endian order. If
        a value consists of 5 tt(char)s the eight most significant bits will
        be stored in the tt(char) having index value 0, the eight least
        significant bits will be stored in the tt(char) having index value
        4. When needed simply swap tt(char[i]) with tt(char[j]) (i = 0
        .. nBytes/2, j = nBytes-1 .. nBytes/2) to convert to little-endian
        order. The return value consists of a series of tt(sizeInBytes()) (see
        below) dynamically allocated tt(char) values. The caller of
        tt(bigEndian) owns the allocated memory and should eventually delete
        it again using tt(delete[]). Note that the current object's em(sign)
        cannot be inferred from the return value.

    itb(BigInt &clearBit(size_t index))
       The current object's bit at index position tt(index) is cleared.

    itb(BigInt clearBit(size_t index) const)
       A copy of the current object having its bit at index position
        tt(index) cleared.

    itb(long long diophantus(long long *factor1, long long *factor2, 
                             long long value1, long long value2))
       The integral solution of tt(factor1 * value1 + factor2 * value2 = gcd)
        is computed. The function returns the greatest common divisor
        (tt(gcd)) of tt(value1) and tt(value2), and returns their
        multiplication factors in, respectively, tt(*factor1) and
        tt(*factor2). The solution is not unique: another solution is obtained
        by adding tt(k * value2) to tt(factor1) and subtracting tt(k * value1)
        from tt(factor2). For values exceeding tt(std::numeric_limits<long,
        long>::max()) the next member can be used.

    itb(BigInt diophantus(BigInt *factor1, BigInt *factor2, 
                            BigInt const &value1, BigInt const &value2))
       The integral solution of tt(factor1 * value1 + factor2 * value2 = gcd)
        is computed. The function returns the greatest common divisor
        (tt(gcd)) of tt(value1) and tt(value2), and returns their
        multiplication factors in, respectively, tt(*factor1) and
        tt(*factor2). The solution is not unique: another solution is obtained
        by adding tt(k * value2) to tt(factor1) and subtracting tt(k * value1)
        from tt(factor2).

    itb(BigInt &div(BigInt *remainder, BigInt const &rhs))
       The current object is divided by tt(rhs). The division's remainder
        is returned in tt(*remainder).

    itb(BigInt divc(BigInt *remainder, BigInt const &rhs) const)
       The quotient of the current object and tt(rhs) is returned. The
        division's remainder is returned in tt(*remainder).

    itb(int compare(BigInt const &rsh) const)
       Using signed values, if the current object is smaller than tt(rhs) -1
        is returned; if they are equal 0 is returned; if the current object is
        larger than tt(ths) 1 is returned (see also tt(uCompare)). 

    itb(BigInt &exp(BigInt const &exponent))
       The current object is raised to the power tt(exponent).

    itb(BigInt expc(BigInt const &exponent) const)
       The current object raised to the power tt(exponent) is returned.

    itb(BigInt &expMod(BigInt const &exponent, BigInt const &mod))
       The current object is raised to the power tt(exponent) modulo tt(mod).

    itb(BigInt expModc(BigInt const &exponent, BigInt const &mod) const)
       The current object raised to the power tt(exponent) modulo tt(mod) is
        returned.

    itb(BigInt &gcd(BigInt const &rhs))
       The greatest common divisor (gcd) of the current object and tt(rhs) is
        assigned to the current object. To compute the least common multiple
        (lcm) the following relationship can be used:
       verb(
    lcm(a, b) = a * b / a.gcd(b)
        )

    itb(BigInt gcdc(BigInt const &rhs) const)
       The greatest common divisor (gcd) of the current object and tt(rhs) is
        returned. To compute the least common multiple (lcm) the following
        relationship can be used: 
       verb(
    lcm(a, b) = a * b / a.gcd(b)
        )

    itb(bool hasBit(size_t index))
        tt(True) is returned if the bit at index position tt(index) has been
        set, tt(false) otherwise.

    itb(BigInt &inverseMod(BigInt const &mod))
       The inverse of the current object modulo tt(mod) is assigned to the
        current object. This is the value tt(ret) for which the following
        expression holds true:
       verb( 
        (*this * ret) % mod = 1 
       )

    itb(BigInt inverseModc(BigInt const &mod) const)
       This  inverse of the current object modulo tt(mod) is returned.

    itb(bool isNegative() const)
       Returns tt(true) if the current object contains a negative value.

    itb(bool isOdd() const)
       Returns tt(true) if the current object is an odd value.

    itb(bool isOne() const)
       Returns tt(true) if the current object equals one (1).

    itb(BigInt &isqrt())
       The current object's integer square root value is assigned to the
        current object. The integer square root of a value tt(x) is the
        biggest integral value whose square does not exceed tt(x). E.g.,
        tt(isqrt(17) == 4). An tt(Exception) exception is thrown if the current
        object's value is smaller than one.

    itb(BigInt isqrtc() const)
        The integer square root of the current object is returned. An
        tt(Exception) exception is thrown if the current object's value is smaller
        than one.  

    itb(bool isZero() const)
       Returns tt(true) if the current object equals zero (0).

    itb(BigInt &lshift()) 
       The current object's bits are shifted one bit to the left. The object's
        sign remains unaltered.

    itb(BigInt lshiftc()) 
       The current object's bits shifted one bit to the left are returned. The
        object's sign will be equal to the current object's sign.

    itb(BigInt &lshift(size_t nBits)) 
       The current object's bits are shifted tt(nBits) to the left. The
        object's sign remains unaltered.

    itb(BigInt lshiftc(size_t nBits) const) 
       The current object's bits shifted tt(nBits) bit to the left are
        returned. The object's sign will be equal to the current object's
        sign.

    itb(BigInt &maskBits(size_t lowerNBits))
        The  current object's tt(lowerNBits) lower bits are kept, its
        higher order bits are cleared. The object's sign is not affected.

    itb(BigInt maskBitsc(size_t lowerNBits) const)
       A copy of the current object is returned having all but its
        tt(lowerNBits) lower bits cleared. The sign of the returned object
        will be equal to the current object's sign.

    itb(size_t maxWordIndex() const)
       Returns the maximum tt(Word)-index that can be used with the tt(at)
        and tt(setWord) members for the current bf(BigInt) value.

    itb(BigInt &mulMod(BigInt const &rhs, BigInt const &mod)) 
       The current object is multiplied  (modulo tt(mod)) by tt(rhs).

    itb(BigInt mulModc(BigInt const &rhs, BigInt const &mod) const) 
       The current object multiplied (modulo tt(mod)) by tt(rhs) is returned.

    itb(BigInt &negate()) 
       The current object's value is negated (i.e., the value changes its
        sign).
        
    itb(BigInt negatec() const) 
       The  negated value of the current object is returned.
        
    itb(BigInt &rshift()) 
       The current object's bits are shifted one bit to the right. The object's
        sign remains unaltered.

    itb(BigInt rshiftc()) 
       The current object's bits shifted one bit to the right are returned. The
        object's sign will be equal to the current object's sign.

    itb(BigInt &rshift(size_t nBits)) 
       The current object's bits are shifted tt(nBits) to the right. The
        object's sign remains unaltered.

    itb(BigInt rshiftc(size_t nBits) const) 
       The current object's bits shifted tt(nBits) bit to the right are
        returned. The object's sign will be equal to the current object's
        sign.

    itb(BigInt &setBit(size_t index))
       The bit at index position tt(index) is set. 

    itb(BigInt setBitc(size_t index) const) 
       A copy of the current object is returned having its bit at index
        position tt(index) set.

    itb(BigInt &setBit(size_t index, bool value))
       The bit at index position tt(index) is set to tt(value).

    itb(BigInt setBitc(size_t index, bool value) const)
       A copy of the current object is returned having its bit at index
        position tt(index) set to tt(value).

    itb(BigInt &setNegative(bool negative))
       The current object's sign will be set to negative if the function's
        argument is tt(true), it will be set to positive if the function's
        argument is tt(false).

    itb(BigInt setNegativec(bool negative) const)
       A copy of the current object is return having a negative sign if the
        function's argument is tt(true) and a positive sign if the function's
        argument is tt(false).

    itb(void setWord(size_t index, BigInt::Word value))
       Assigns tt(value) to the tt(Word) at tt(index). E.g., on a 32 bit
        architecture, if the bf(BigInt) value equals 2+sups(33), then
        after tt(setWord(1, 1)) the value has become 2+sups(32). If tt(index)
        exceeds the value returned by tt(nWords) an tt(FBB::Exception) is
        thrown.

    itb(size_t size() const)
       The number of bytes required to store the current tt(BIGNUM) value is
        returned. 

    itb(size_t sizeInBits() const)
       The number of bits required to represent the current tt(BIGNUM) value
        is returned. 

    itb(size_t sizeInBytes() const)
       The number of bytes required to store the current tt(BIGNUM) value is
        returned (returns the same value as the tt(size) memeber does).

    itb(size_t constexpr sizeOfWord() const)
       bf(BigInt) values are stored in units of `words', which are unsigned
        long values. These values may consist of, e.g., 32 or 64 bits. The
        number of bytes occupied by a `word' is returned: 4 for a 32 bit
        value, 8 for a 64 bit value, and possibly other values, depending on
        specific architecture peculiarities. The value returned by this
        member, therefore, is architecture dependent.

    itb(BigInt &sqr())
      The current object's value is squared.

    itb(BigInt sqrc() const)
      The square of the current object is returned.

    itb(BigInt &sqrMod(BigInt const &mod) const)
      The current object's value is squared modulo tt(mod).

    itb(BigInt sqrModc(BigInt const &mod) const)
      The square (modulo tt(mod)) of the current object is returned.

    itb(BigInt &subMod(BigInt const &rhs, BigInt const &mod)) 
       tt(Rhs) is subtracted modulo tt(mod) from the current object.

    itb(BigInt subModc(BigInt const &rhs, BigInt const &mod) const) 
      The difference (modulo tt(mod)) of the current object and tt(rhs) is
        returned.

    itb(void swap(BigInt &other))
       The current object swaps its value with tt(other).

    itb(BigInt &tildeBits())
       All the bits in
        the bytes of the current object and the sign of the current object
        are toggled.
       So, after
            verb(
        Bigint b(5);
        b.tildeBits();
            )
       tt(b) contains the value -250. Also see the discussion with
        tt(operator~()) below.

    itb(BigInt tildeBitsc() const)
       A copy of the current object whose bits are toggled is returned.

    itb(BigInt &tildeInt())
       The `tilde' operation is performed on the current object using the
        standard tt(int) semantics. E.g., ~5 results in -6.  Also see the
        discussion with tt(operator~()) below.

    itb(BigInt tildeIntc() const)
       A copy of the current object is returned to which the `tilde' operation
        has been performed using the standard tt(int) semantics.

    itb(unsigned long ulong() const) 
       The absolute value stored in the current object is returned as an
        unsigned long. If it cannot be represented by an unsigned long it
        returns tt(0xffffffffL).

    itb(int uCompare(BigInt const &rsh) const)
       Using absolute values, if the current object is smaller than tt(rhs) -1
        is returned; if they are equal 0 is returned; if the current object is
        larger than tt(ths) 1 is returned (see also tt(uCompare)).
    )

manpagesection(OVERLOADED OPERATORS)

Except for some operators all operators perform their intuitive
operations. Where that isn't completely true an explanatory remark is
provided. E.g., tt(operator*()) multiplies two tt(BigInt)s, possibly promoting
one of the operands; tt(operator*=()) multiplies the lhs by the rhs
tt(BigInt), possibly promoting the rhs operand.

    Here are the available operators:

    bf(Unary operators:)

    itemization(
    itb(bool operator bool() const)
       Returns tt(true) if the tt(BigInt) value is unequal zero, otherwise
        tt(false) is returned.

    itb(BigInt &operator++())
    itb(BigInt operator++(int))
    itb(BigInt &operator--())
    itb(BigInt operator--(int))
    itb(BigInt operator-())
    itb(int operator[](size_t idx) const)
       With tt(BigInt) objects it returns the bit-value of the object's
        tt(idx)th bit as the value 0 or 1.

    itb(BigInt::Bit operator[](size_t idx))
       With non-const tt(BigInt) objects it returns a reference to the
        bit-value of the object's tt(idx)th bit. When used as em(lvalue)
        assigning a 0 or non-zero value to the operator's return value will
        either clear or set the bit.  Likewise, the following arithmetic
        assignment operators may be used: binary or (tt(|=)), binary and
        (tt(&=)) or binary xor (tt(^=)). When used as em(rvalue) the value of
        the object's tt(idx)th bit is returned as a tt(bool) value. When
        inseerted into a tt(std::ostream) the bit's value is displayed as 0 or
        1. 

    itb(BigInt operator~())
       This operator is em(not) implemented as it cannot be implemented so
        that it matches the actions of this operator when applied to tt(int)
        type values.

        When used on tt(int) values this operator toggles all the tt(int)'s
        bits. E.g., ~5 represents -6, and ~-6 again equals five. The -6 is the
        result of the sign bit of tt(int) values. The obvious implementation
        of tt(BigInt::operator~()) is to toggle all the value's bits and to
        toggle its sign bit. For 5 this would result in -250: 5, being 101
        (binary), fits in one byte, so ~5 becomes 11111010 (binary), which is
        250. Its sign must be reversed as well, so it becomes -250.  This
        clearly differs from the value represented by the tt(int) constant ~5:
        when constructing tt(BigInt(~5)), the value -6 is obtained.

       It is possible to change the implementation. E.g., after
            verb(
        Bigint b(5);
        b = ~b;
            )
       tt(~b) could be implemented so that it results in the value -6. But
        this too leads to unexpected results. While tt(5 & ~5 == 0), this
        would no longer hold true for tt(BigInt) objects: Assuming tt(b)
        contains 5 then tt(b & ~b) would expand to (binary) tt(101 &
        (negative)110) which equals (binary) 100.

       Since either implementation produces unexpected results
        tt(BigInt::operator~()) was not implemented. Instead two members are
        offered: tt(tildeBits()), toggling all the bits of all the
       tt(BigInt) bytes and toggling its sign (so
            verb(
        Bigint b(5);
        b.tildeBits();
            )
       changes tt(b)'s value into -250), and tt(tildeInt()) changing the
        object's value into the value that would have been obtained if a
       tt(BigInt) was a mere tt(int) (so    
        verb(
        Bigint b(5);
        b.tildeInt();
            )
       changes tt(b)'s value into -6).
    )

    bf(Binary operators:)

    itemization(
    itb(BigInt operator*(BigInt const &lhs, BigInt const &rhs))
    itb(BigInt operator/(BigInt const &lhs, BigInt const &rhs))
       This operator returns the quotient of the tt(lhs) object divided by the
        tt(rhs) object. The remainder is lost (The member tt(div) performs the
        division and makes the remainder available as well).
    itb(BigInt operator%(BigInt const &lhs, BigInt const &rhs))
    itb(BigInt operator+(BigInt const &lhs, BigInt const &rhs))
    itb(BigInt operator-(BigInt const &lhs, BigInt const &rhs)) 
    itb(BigInt operator<<(BigInt const &lhs, size_t nBits))
        See also the tt(lshift) members. If tt(lhs) is positive, 
    itb(BigInt operator>>=(BigInt const &lhs, size_t nBits))
       See also the tt(rshift) members.
    itb(BigInt operator&(BigInt const &lhs, BigInt const &rhs)) 
        This operator returns a tt(BigInt) value consisting of the
        tt(bit_and)-ed bits and sign flags of lhs and rhs
        operands. Consequently, if one operand is positive, the resulting
        value will be positive. 
    itb(BigInt operator|(BigInt const &lhs, BigInt const &rhs)) 
        This operator returns a tt(BigInt) value consisting of the
        tt(bit_or)-ed bits and sign flags of lhs and rhs
        operands. Consequently, if either operand is negative, the result will
        be negative. 
    itb(BigInt operator^(BigInt const &lhs, BigInt const &rhs)) 
        This operator returns a tt(BigInt) value consisting of the
        tt(bit_xor)-ed bits and sign flags of lhs and rhs
        operands. Consequently, if exactly one operand is negative, the result
        will be negative. 
    )

    bf((Arithmetic) assignment operator(s):)

    itemization(
    itb(BigInt &operator=(BigInt const &rhs))
    itb(BigInt &operator*=(BigInt const &rhs))
    itb(BigInt &operator/=(BigInt const &rhs))
       This operator assigns the result of the (integer) division of the current
        tt(BigInt) object by tt(ths) to the current object. The remainder is
        lost. The member tt(div) divides and makes the remainder available as
        well.
    itb(BigInt &operator%=(BigInt const &rhs))
    itb(BigInt &operator+=(BigInt const &rhs))
    itb(BigInt &operator-=(BigInt const &rhs))
    itb(BigInt &operator<<=(size_t nBits))
        See also the tt(lshift) members.
    itb(BigInt &operator>>=(size_t nBits))
        See also the tt(rshift) members.
    itb(BigInt &operator&=(BigInt const &rhs)) 
        This operator tt(bit_and)s the bits and sign flags of the current
        object and the rhs operand.
    itb(BigInt &operator|=(BigInt const &rhs)) 
        This operator tt(bit_or)s the bits and sign flags of the current
        object and the rhs operand.
    itb(BigInt &operator^=(BigInt const &rhs)) 
        This operator tt(bit_xor)s the bits and sign flags of the current
        object and the rhs operand.
    )

    Note that the move operator is not available

manpagesection(STATIC MEMBERS)

    All members returning a tt(BigInt) computed from a set of arguments and
not requiring an existing tt(BigInt) object are defined as static members.

    itemization(
    itb(BigInt fromText(std::string text, int mode = 0))
       This member converts a textual representation of a number to a
        tt(BigInt) value. Conversion continues until the end of tt(text) or
        until a character outside of an expected range is encountered.  

       The expected range may be preset by specifying tt(mode) as tt(ios::dec,
        ios::oct,) or tt(ios::hex) or (the default) the expected range is
        determined by tt(fromText) itself by inspecting the characters in
        tt(text).

        By default if tt(text) contains hexadecimal characters then
        tt(fromText) assumes that the number is represented as a hexadecimal
        value (e.g., tt("abc") is converted to the (decimal) value 2748); if
        tt(text) starts with 0 and contains only characters in the range 0
        until (including) 7 then tt(fromText) assumes the number is
        represented as an octal value (e.g., tt("01234") is converted to the
        (decimal) value 668). Otherwise a decimal value is assumed. 

        If the text does not represent a valid numerical value (of the
        given extraction mode) then a tt(FBB::Exception) exception is thrown
        (tt(fromText: text does not represent a BigInt value)).

    itb(BigInt rand(size_t size, Msb msb = MSB_IS_ONE, Lsb lsb = ODD))
       This member returns a cryptographically strong pseudo-random number
        of tt(size) bits. The most significant bit(s) can be controlled by
        tt(msb) (by default bf(MSB_IS_ONE)), the least significant bit can be
        controlled by tt(lsb) (by default bf(ODD)). Before calling this
        member the random number generator must have been seeded.

        From the bf(RAND_add)(3ssl) man-page:

       OpenSSL makes sure that the PRNG state is unique for each thread. On
        systems that provide tt(/dev/urandom), the randomness device is used
        to seed the PRNG transparently. However, on all other systems, the
        application is responsible for seeding the PRNG by calling
        bf(RAND_add)(3ssl), bf(RAND_egd)(3ssl), bf(RAND_load_file)(3ssl), or
        bf(RAND_seed)(3ssl).

    itb(BigInt randRange(BigInt const &max))
       This member returns a cryptographically strong pseudo-random
        number in the range tt(0 <= number < max). Before calling this
        member the random number generator must have been seeded (see also
        bf(rand), described above).

    itb(BigInt setBigEndian(std::string const &bytes)) 
       The tt(bytes.length()) bytes of tt(bytes) are used to compute a
        tt(BigInt) object which is returned by this function. The characters
        in tt(bytes) are interpreted as a series of bytes in big-endian
        order. See also the member function tt(bigEndian()) above. The
        returned tt(BigInt) has a positive value.

    itb(BigInt prime(size_t nBits, 
                            BigInt const *mod = 0, BigInt const *rem = 0,
                            PrimeType primeType = ANY))
       This member returns a prime number of tt(bBits) bits. If both tt(mod)
        and tt(rem) are non-zero, the condition prime % mod == rem.
        (E.g., use tt(prime % mod == 1) in order to suit a given
        generator). The parameter tt(primeType) can be tt(ANY), tt((prime - 1)
        / 2) may or may not be a prime. If it is tt(SAFE) then  tt((prime - 1)
        / 2) will be a (so-called em(safe)) prime.

    itb(BigInt pseudoRand(size_t size, Msb msb = MSB_IS_ONE, Lsb lsb =
            ODD)) 
       This member returns a potentially predictable pseudo-random number of
        tt(size) bits. The most significant bit(s) can be controlled by
        tt(msb) (by default bf(MSB_IS_ONE)), the least significant bit can be
        controlled by tt(lsb) (by default bf(ODD)). It can be used for
        non-cryptographic purposes and for certain purposes in cryptographic
        protocols, but usually not for key generation.

    itb(BigInt pseudoRandRange(BigInt const &max))
       This member returns a potentially predictable  pseudo-random
       number in the range tt(0 <= number < max).
    )

manpagesection(FREE FUNCTIONS IN THE FBB NAMESPACE)

    itemization(
    itb(std::ostream &operator<<(ostream &out, BigInt const &value))
        Inserts tt(value) into the provided tt(ostream). If the tt(hex)
manipulator has been inserted into the stream before inserting the tt(BigInt)
value the value will be displayed as a hexadecimal value (without a leading
tt(0x)); if the tt(oct) manipulator has been inserted the value will be
represented as an octal value (starting with a 0). The value will be
displayed as a decimal value if the tt(dec) manipulator is active. If the
tt(BigInt) value is negative its value will be preceded by a minus
character. 
    itb(std::istream &operator>>(istream &in, BigInt &value))
       Extracts tt(value) from the provided tt(istream). Depending on the
        currently set extraction mode (tt(dec, oct,) or tt(hex)) the matching
        set of characters will be extracted from tt(in) and converted to a
        number which is stored in tt(value). Extraction stops at EOF or at the
        first character outside of the range of characters matching the
        extraction mode. if no numerical characters were extracted the
        stream's tt(failbit) is set. The extracted value may be preceded
        by a minus character, resulting in an extracted negative value.
    )

manpagesection(EXAMPLE)
        verb(
#include <iostream>
#include <bobcat/bigint>

using namespace std;
using namespace FBB;

int main()
{
    BigInt value(BigInt::prime(100));
    BigInt mod(BigInt::rand(50));
    BigInt inverse(value.inverseModc(mod));

    cout << '(' << value << " * " << inverse << ") % " << mod << " = " <<
             (    value       *      inverse     ) %      mod << endl;
}
        ) 

manpagefiles()
    em(bobcat/bigint) - defines the class interface

manpageseealso()
    bf(bobcat)(7), bf(diffiehellman)(3bobcat), 
    bf(RAND_add)(3ssl), bf(RAND_egd)(3ssl),
    bf(RAND_load_file)(3ssl), bf(RAND_seed)(3).

manpagebugs()

    Sep/Oct 2013: due to a change in library handling by the linker
(cf. lurl(http://fedoraproject.org/wiki/UnderstandingDSOLinkChange) and
lurl(https://wiki.debian.org/ToolChain/DSOLinking)) libraries that are
indirectly required are no longer automatically linked to your program. With
bf(BigInt) this is tt(libcrypto), which requires programs to link to both
tt(bobcat) and tt(crypto).

includefile(include/trailer)
