The three least significant bits carry the user’s authority level (0-5). The instruction size is one word, but the bandwidth of the system is only 1/2 word. Take, for example, an 8-bit system with 2 byte words. Here's the documentation: The authority byte is split into two fields. 1 To understand the difference between byte- and word-addressable, understand that a byte is always 8 bits, while a word may differ from system to system. In this case, no C program will even come close (think 20x slower or more). How to work with the bits in a byte Ask Question Asked 9 years, 11 months ago Modified 9 years, 11 months ago Viewed 10k times 6 I have a single byte which contains two values. Note that if the byte-to-count mapping is inlined right next to this instruction then the whole data table will be in the CPU cache so it will be really fast. If it is compiled C it will probably generate a lot of memory accesses that will slow it down even beyond the larger number of clock cycles it requires. That feels like a bit (ha) of a waste, and it is. ![]() We also talk about them being true and false. ![]() Generally, we represent them with the numerals 1 and 0. A binary number is made up of elements called bits where each bit can be in one of the two possible states. We'll explore how computers can use bits and bytes to represent all types of information in this unit. It might represent a number, a letter, or a program instruction. It might even represent part of an audio recording or a pixel in an image. A byte represents different types of information depending on the context. C s normal storage options would mean devoting an entire char (8 bits) or int (16 bits) to keeping track of such tiny values. Binary is a base-2 number system that uses two mutually exclusive states to represent information. A byte represents different types of information depending on the context. The problem with Kernighan's solution is that even when hand-coded in assembly it is slower than my algorithm. When dealing with microcontrollers in particular, you will regularly have a small collection of sensors or switches that are providing on/off values. Note that the shift instructions in Art's solution are very CPU expensive. I dont know will you be able to do this in LAD, though. If any of these intergers is different from 0, then some bits are active. Given a non-negative number n and two values l and r.The problem is to check whether all the bits are set or not in the range l to r in the binary representation of n. This is the absolute fastest way to do this and will be faster than any mathematical computation. Load this 64 bits as two double words, as long integer. If the number has many bytes in it (say an array pointed to by ebp), then you loop through the bytes (where ecx is the number of bytes in the array containing the number): sub ecx, 1 If the number is a single byte then the answer is one instruction: mov eax, Conversion between bits and bytes is a simple calculation: divide by 8 8 to convert from bits to bytes or multiply by 8 8 to convert from bytes to bits. The fastest way to do this calculation is with a table array edx where the bl register contains a byte value. So, can this be done in better than O( n) bit tests? No.Ĭan it be done in fewer than O( n) add/shift/mask operations? Yes. ![]() I need to see if a particular bit is set. If n is the number of bits then even a simple bit mask ( &) is already an O( n) operation. Ask Question Asked 5 years, 10 months ago Viewed 22k times 11 In C, I have a 32 bit value which I am storing in an int. What's tricky about this is when you aren't careful about the definition of n and let n be "the number of bit shifting/masking instructions" or some such. Any algorithm must examine all of the bits in a number. V &= v - 1 // clear the least significant bit setīrian Kernighan's method goes through as many iterations as there are Unsigned int c // c accumulates the total bits set in v But I usually write: /* Return type (8/16/32/64 int size) is specified by argument size.Bit Twiddling Hacks presents a number of methods, including this one: Counting bits set, Brian Kernighan's way unsigned int v // count the number of bits set in v Yeah, I know I don't "have" to do it this way.
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