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).
USAGE¶
string s;
s=clear_bit("_",5);
Because “_” is the highest possible value (63), the variable s will now contain the charcter ”?” which 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”.