127 #define B00010110 22 128 #define B10111 23 129 #define B010111 23 130 #define B0010111 23 131 #define B00010111 23 132 #define B11000 24 133 #define B011000 24 134 #define B0011000 24 135 #define B00011000 24 136 #define B11001 25 137 #define B011001 25 ...
l(8)=12+52×Cum_Count1(0)10=52×210=22=(010110)2u(8)=12+52×Cum_Count1(1)10-1=52×1010-1=63=(111111)2Encoding the next two 1s we obtain l(9)=22+42×Cum_Count1(0)10=42×210=30=(011110)2u(9)=22+42×Cum_Count1(1)10-1=42×1010-1=63=(111111)2l(10)=30+34×Cum...
The second number (22) is entered in the right hand margin as 010110 (from top to bottom). Whenever a row with a grape intersects a column with a grape, place a grape onto the grid. We can remove the "input" from the margins now. Move all grapes diagonally left & down until they...
A Hollerith string is a sequence of l2-bit characters; they are encoded as two ASCII characters, containing 6 bits each. The first character contains punches 12,0,2,4,6,8 and the second character contains punches 11, 1,3,5, 7, 9. Interleaving the two characters gives the original 12 ...
B00010110 0x16 #define B00010111 0x17 #define B00011000 0x18 #define B00011001 0x19 #define B00011010 0x1A #define B00011011 0x1B #define B00011100 0x1C #define B00011101 0x1D #define B00011110 0x1E #define B00011111 0x1F #define B00100000 0x20 #define B00100001 0x21 #...
// eg. we have 3 bytes as ( 0100 1101, 0110 0001, 0110 1110) --> (010011, 010110, 000101, 101110) char_array_4[0] = (char_array_3[0] & 0xfc) >> 2; // get first 6 bits of first byte, char_array_4[1] = ((char_array_3[0] & 0x03) << 4) + ((char_array_3[...
Hi, I am trying to compile some piece of code which has global arrays. I have initialized all the global arrays to make sure the compiler does not generate the code to initialize them to 0. However, I am still getting these "memset" call...