We know that 0xaa and 0x55 are hexa-decimal representation . Moreover each character in hexadecimal is represent using 4bits. so , 0xaa is equivalent to 1010 1010 (since, a = 1010 in binary) and 0x55 is equivalent to 0101 0101 (since, 5 = 0101 in binary)

2014年9月29日 · I am surprised by C++'s behavior when applying bit-wise not to an unsigned char. Take the binary value 01010101b, which is 0x55, or 85. Applying bit-wise not on an eight bit representation should ...

2021年9月26日 · .byte 0x55 .byte 0xaa so people sometimes refer to the signature in that manner. Share. Improve this answer.

2015年1月27日 · I can only use the bitwise operators mentioned below to create the described function: /* * allEvenBits - return 1 if all even-numbered bits in word set to 1 * Examples allEvenBits(0xFFFFFFFE) ...

2018年9月21日 · Magic number is a word of 0xaa55 not a word of 0x55aa. A word is 16-bits (2 bytes). Because the x86 processors are little endian they are stored in memory as the byte 0x55 followed by a byte of 0xaa. The magic number is the last 2 bytes of the boot sector. Not sure what makes you think that 0xaa or 0x55 can't fit in a byte (both are less than 256)

2016年10月11日 · There is nothing magical or mystical about that combination. Implementers needed a means by which to determine if the first sector of a device was bootable (boot signature) and that combination occurring in the last two bytes …

2018年12月25日 · The boot signature is ALWAYS a 0x55 byte followed by a 0xaa byte. How you display it in a debugger or dump utility may be different if they output the data as WORD's instead of bytes. Because the x86 is little endian if you display (or define) it as a word it will be 0xaa55 . The low byte 0x55 will be stored in memory followed by the 0xaa in the.

2013年7月23日 · 0x65 && 0x55 Since both operands 0x65 and 0x55 evaluate to true the whole expression evaluates to true and thus expands to 1, which applies to c99 but other answers in the linked thread explain how it applies before c99 as well.

Finally, 0xAA is 10101010 and 0x55 is 01010101. When these masks and shifts are done, the result is: 76543210 Voila! this is the reversed order. Notice that for each pair of shifts, the bit masks are inverses of each other, and the number of bits being shifted are the same as the length of the sequences of 1 bits in the mask.

2014年1月15日 · 0x55 = 01010101 0x33 = 00110011 0x0f = 00001111 0xff = 11111111 Let's go with an 8 bit example: a = 01101010. a & 0x55 = 01000000; (a >> 1) & 0x55 = 00010101 If we add these two together, we get the bit count of each two bit pair. This result is at most 0x10, as the mask has only one bit set for each addition.