![\"Figure](\"http:\/\/c-for-dummies.com\/blog\/wp-content\/uploads\/2016\/04\/figure1-row_colum_bit_fields.png\")
Figure 1. A byte containing two 3-bit fields.<\/p><\/div>\n
In the figure, you see that three bits are used for both row and column values. Three bits express values from 0 through 7, which is enough to account for 8 rows: Row And the resulting byte would be:<\/p>\n The nifty part is that you don’t have to strain your head to assembly this binary information; the computer does the job for you. You just need to extract specific bits from the byte to obtain or set the Queen’s position. To code the solution, you need two of the bitwise operators used in previous lessons to extra values: bitwise To obtain the row value, the rest of the byte is masked and then the values are shifted right three bits, as illustrated in Figure 2.<\/p>\n Figure 2. Masking out the row value, then shifting that value right.<\/p><\/div>\n The equation To obtain the column, bits are masked as well, but the result need not be shifted, as shown in Figure 3.<\/p>\n Figure 3. Masking the column value.<\/p><\/div>\n After peeling off the rest of the byte with Here is code that extracts a chess piece’s location from an unsigned char<\/em> value. The variable is unsigned<\/em> so that negative results aren’t generated.<\/p>\n Here is a sample run:<\/p>\n Because a char<\/em> variable is (usually) 8-bits wide, I could have made the row and column bit fields 4-bits wide each. That would make manipulation easier, plus it would also make the code more readable to anyone who knows hexadecimal and its relationship to binary.<\/p>\n For example, row 2 column 7 would be 0x27 hexadecimal. Also, the bitwise In next week’s Lesson<\/a>, I cover a function that grabs a bit field of any length from any position within a byte.<\/p>\n","protected":false},"excerpt":{"rendered":" A bit field need not be limited to a single bit. Continue reading 000<\/code> through 111<\/code>. So if the Queen is at row 3, column 8, the binary equivalents would be:<\/p>\n010<\/code>
\nColumn 111<\/code><\/p>\n00010111<\/code><\/p>\n&<\/code> (and) and the right-shift operator, >><\/code>.<\/p>\n![\"Figure](\"http:\/\/c-for-dummies.com\/blog\/wp-content\/uploads\/2016\/04\/figure2-row_mask_shift.png\")
& 0x38<\/code> masks out the three relevant bits. Next, those three bits are shifted to the right three places. The result is 0x02<\/code> (refer to Figure 2), which means that the Queen is at row 3 on the chessboard. (The first row is row zero.)<\/p>\n![\"Figure](\"http:\/\/c-for-dummies.com\/blog\/wp-content\/uploads\/2016\/04\/figure3-column_mask.png\")
& 0x07<\/code>, the remaining value is the column. It just happens to be 7, which is coincidentally the same value as the &<\/code> (and) mask. And the 7 implies column 8 because, again, numbering starts at column zero.<\/p>\n\r\n#include <stdio.h>\r\n\r\nint main()\r\n{\r\n unsigned char board,row,column;\r\n\r\n board = 23;\r\n\r\n row = board & 0x38;\r\n row = row >> 3;\r\n column = board & 0x07;\r\n\r\n printf(\"The Queen is at row %d, column %d\\n\",\r\n row+1,column+1);\r\n\r\n return(0);\r\n}<\/pre>\nThe Queen is at row 3, column 8<\/code><\/pre>\n&<\/code> (and) masks would be 0xF0<\/code> and 0x0F<\/code>. Simple. But not every bit field is exactly 4-bits wide, so rarely will you be so lucky.<\/p>\n