![]() That's where the expression in the return statement comes from. int rejThreshold (orig > 3) & 0x1F > is the 'shift right' operator. by 8-5 ), and set the three upper bits to zero using bitwise AND operation, like this: byte orig. The result is then zero if the bit we're checking was zero, or non-zero if it was one. To get a value of the five most significant bits in a byte as an integer, shift the byte to the right by 3 (i.e. The single one bit in the mask will basically let the equivalent bit in the value flow through to the result. You can see that all the zero bits in the mask result in the equivalent result bits being zero. If you want to set the Nth bit of a number to 1: mask 1 << n if n is 3, mask results in 00001000 bytevalue bytevalue or mask. Decimal BinaryĤ 0000 0100 (the bitmask for the third-from-least bit).Ġ000 0100 (the result of the AND operation). Let's say you have the value 13 and you want to see if the third-from-least-significant bit is set. The expressions most significant bit first and least significant bit at last are indications on the ordering of the sequence of the bits in the bytes sent over a wire in a serial transmission protocol or in a stream (e.g. The relevant table is: AND | 0 1įor a given char value, we use the single-bit bit masks to check if a bit is set. The AND operation & will set a bit in the target only if both bits are set in the tewo sources. The simplified AND-only version is below. If you want to know how the bitwise operators work, see here. The number one mantra in optimisation is "Measure, don't guess!" Just make sure you benchmark any solution you're given, including this one (a). And if you make it static and suggest inlining, or force it inline as a macro definition, you can even bypass the cost of a function call. Do not pass in an invalid n, that'll be undefined behaviour.Īt insane optimisation level -O3, gcc gives us: isNthBitSet: pushl %ebp No error checking is done since you cited speed as the most important consideration. If those assumption are incorrect, it simply comes down to expanding and/or re-ordering the mask array. This assumes 8-bit bytes (not a given in C) and the zeroth bit being the highest order one. Many systems used bank switching approaches to allow programs to access more memory than the processor could natively address.The following function can do what you need: int isNthBitSet (unsigned char c, int n).For example SDRAM uses the same address lines twice to send a "row address" and a "column address". Some busses multiplex different signals on the same lines.This requires further addressing which may or may not be handled by bits of the main address bus. Some busses allow the movement of varying size data units.a 'bit' is atomic: the smallest unit of storage A bit stores just a 0 or 1 'In the computer it's all 0's and 1's'. In this section, we'll learn how bits and bytes encode information. The "word size" of the processor is not nessacerally the same as the width of the memory data bus or the smallest addressable unit of memory. At the smallest scale in the computer, information is stored as bits and bytes.Many processors do not have a dedicated IO map, so parts of the memory address space may need to be used for things other than memory.Some of the ways in which real systems can be more complicated than this include. The reality is more complicated and to give a definitive answer requires more information. This however reflects a rather oversimplified worldview. So we can address a 524288 bit (65536 octet) memory." "16 bits can address 2^16 memory locations, each location is 8 bits. There are two sides to this, what your instructor probablly wants you to tell him and what the reality is.įirst what your instructor probablly wants you to tell him. That step eliminates confusion and makes problems like this trivial. Of what? 8-bit words, making it (still) 64ki bytes of addressable RAM. A memory management unit might use 16-bit addresses and have 20-bit hardware address, so the CPU needs to switch and map things to make use of the actual 20-bit address range of RAM chips that can be addressed.īe sure to specify units in your answers. mentioned by others are not relevant here. (If an instruction wanted 1 byte, it would dispatch the address with the least bit missing, fetch both bytes in that step, then look at the least bit of the desired address to decide which half to use.) It only needs 32ki addresses and always gets 2 bytes with each location. But it has 15 bits of address bus and 16 bits of data bus. It uses 16-bit addresses in its instructions, and like your example has 64ki. The word size here may or may not match the CPU computation unit size, and this may or may not match the logical granularity in addressing.įor example, a CPU may advertise a 16-bit bus (for this purpose). There are 16 bits gling out to the memory so it can choose 64ki locations. In context, the word size goes with the address size to describe the memory bus.
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