Question

Here is code i saw in a C program , i knew this piece of code is to set a bit in the bit ASCII bit map corresponding to the character c.

field[ (c & 0x7f) >> 3 ] |= 1 << (c & 0x07);

field is an array of 16 characters, each character is 8 bits.

for example ’97’ is lower case ‘a’, if we set c to 97, then bit position 97 will be set to 1.

any one know why above code will set bit map corresponding to the character c?
and what are those magic number 0x7f, 0x07, 3 and 1 for?

Answer 1

If your array is 16 bytes long, it has 128 bits (16 x 8). So the first mask (0x7f) guarantees that you are only interested in the first 128 bits. Once you shift it 3 bits right, you have 4 bits left that are used to address your bitfield (the number ((c & 0x7F) >> 3 is a number between 0 and 15). So this part uses the upper 4 bits to address the byte.

Now, you need to address the bit in the byte, so you use the mask 0x07 to limit the value to the range 0 – 7 (corresponding to the bits 0 to 7). You use this number to shift the 1 so many positions.

At the end, you have a bit set in a position 0 to 127 (16 bytes of 8 bits). I hope this helps!