0XFF00 is hexidecimal. This is a base-16 system where numbers go from 0-9, then A-F. This signifies going from 0 to 15 in our numbering system (base 10). 10 is actually 16 in base 10, …

Assuming your byte1 is a byte(8bits), When you do a bitwise AND of a byte with 0xFF, you are getting the same byte.

0xff00 & 0xf0f0 is: 0xf000. becomes. 1111 1111 0000 0000 & 1111 0000 1111 0000 ----- 1111 0000 0000 0000 In binary it makes more sense: the result of a bitwise & is 1 only when both …

2015年11月23日 · a[1] = (addr & 0xff00) >> 8; And then do it step by step. addr & 0xff00 This gets the bits 8 to 15 of the value in addr, the result after the operation is 0x0000d300. >> 8 This …

2015年12月3日 · If any bit in VALUE that corresponds to a set bit in 0xFF00 is also set then the condition evaluates to true. If you want to check that all corresponding bits of the left operand …

2010年11月14日 · Apparently (all those defines in uppercase make it hard reading) the functions are swapping the internal byte ordering of values that occupy 2 or 4 bytes.

2010年10月30日 · This kind of code tends to be used to swap things between big endian and little endian format. There is also a little trick that will convert a word in some known format (lets …

2012年8月1日 · Starting at address 0xFF00 inside the address space mapped for the rtl8187 device is a memory range that holds information structured the same way as the rtl818x_csr …

2013年12月19日 · So initially c = 0xff00. c = c + '\xff' In binary is. c = 1111 1111 0000 0000 + 1111 1111 1111 1111 Which yields signed short int. c = 1111 1110 1111 1111 (0xfeff) c and d …

Mask val to get the second byte by &ing with 0xFF00: result is 0xCC00. Shift that result 8 bits to the left: result is 0xCC0000. Shift val 8 bits to the right: result is 0xAABBCC (the LSB has …