ldmud-3.4.1/doc/
ldmud-3.4.1/doc/efun.de/
ldmud-3.4.1/doc/efun/
ldmud-3.4.1/doc/man/
ldmud-3.4.1/doc/other/
ldmud-3.4.1/mud/
ldmud-3.4.1/mud/heaven7/
ldmud-3.4.1/mud/lp-245/
ldmud-3.4.1/mud/lp-245/banish/
ldmud-3.4.1/mud/lp-245/doc/
ldmud-3.4.1/mud/lp-245/doc/examples/
ldmud-3.4.1/mud/lp-245/doc/sefun/
ldmud-3.4.1/mud/lp-245/log/
ldmud-3.4.1/mud/lp-245/obj/Go/
ldmud-3.4.1/mud/lp-245/players/lars/
ldmud-3.4.1/mud/lp-245/room/death/
ldmud-3.4.1/mud/lp-245/room/maze1/
ldmud-3.4.1/mud/lp-245/room/sub/
ldmud-3.4.1/mud/lp-245/secure/
ldmud-3.4.1/mud/morgengrauen/
ldmud-3.4.1/mud/morgengrauen/lib/
ldmud-3.4.1/mud/sticklib/
ldmud-3.4.1/mud/sticklib/src/
ldmud-3.4.1/mudlib/uni-crasher/
ldmud-3.4.1/pkg/
ldmud-3.4.1/pkg/debugger/
ldmud-3.4.1/pkg/diff/
ldmud-3.4.1/pkg/misc/
ldmud-3.4.1/src/autoconf/
ldmud-3.4.1/src/hosts/
ldmud-3.4.1/src/hosts/GnuWin32/
ldmud-3.4.1/src/hosts/amiga/
ldmud-3.4.1/src/hosts/win32/
ldmud-3.4.1/src/ptmalloc/
ldmud-3.4.1/src/util/
ldmud-3.4.1/src/util/erq/
ldmud-3.4.1/src/util/indent/hosts/next/
ldmud-3.4.1/src/util/xerq/
ldmud-3.4.1/src/util/xerq/lpc/
ldmud-3.4.1/src/util/xerq/lpc/www/
ldmud-3.4.1/test/t-030925/
ldmud-3.4.1/test/t-040413/
ldmud-3.4.1/test/t-041124/
SYNOPSIS
        string clear_bit(string str, int n)

DESCRIPTION
        Return the new string where bit n is cleared in string str.
        Note that the old string str is not modified.

        Each character contains 6 bits. So you can store a value
        between 0 and 63 ( 2^6=64) in one character. Starting
        character is the blank character " " which has the value 0.
        The first charcter in the string is the one with the lowest
        bits (0-5).

EXAMPLES
        string s;
        s=clear_bit("_",5);

        Because "_" is the highest possible value (63), the variable s
        will now contain the charcter "?" wich is equal to 31
        (63-2^5=31).

        string s;
        s=clear_bit("?<",3);
        s=clear_bit(s,8);

        s will now contain the string "78". "?" equals 31 and "<"
        equals 28. Now "?<" is equal to 31+28<<6=31+1792=1823 which is
        in binary notation (highest bit on the right side)
        11111000111. Now clearing the bit 3 and bit 8 (bit numbering
        starts with zero) will result in 11101000011. The first 6 bits
        are in decimal notation 23 and the next 6 are equal to 24. Now
        the 23 is the character "7" and 24 is the "8". So the string s
        contains "78".

SEE ALSO
        set_bit(E), next_bit(E), last_bit(E), test_bit(E), count_bits(E),
        and_bits(E), or_bits(E), xor_bits(E), invert_bits(E), copy_bits(E)