bit-shift

问题 I want to create a 64-bit barrel shifter in verilog (rotate right for now). I want to know if there is a way to do it without writing a 65 part case statement? Is there a way to write some simple code such as: Y = {S[i - 1:0], S[63:i]}; I tried the code above in Xilinx and get an error: i is not a constant. Main Question: Is there a way to do this without a huge case statment? 回答1: I've simplified some of the rules for clarity, but here are the details. In the statement Y = {S[i - 1:0], S[63

问题 Learning the hard way, I tried to left shift a long long and uint64_t to more than 32 bits on an x86 machine resulted 0 . I vaguely remember to have read somewhere than on a 32 bit machine shift operators only work on the first 32 bits but cannot recollect the source. I would like to know is if Shifting more than 32 bits of a uint64_t integer on an x86 machine is an Undefined Behavior? 回答1: The standard says (6.5.7 in n1570): 3 The integer promotions are performed on each of the operands. The

问题 This question already has answers here : Closed 8 years ago . Possible Duplicate: why is 1>>32 == 1? -1 as an int converted to binary is represented by 32 1's. When I right-shift it 31 times, I get 1 (31 0's and one 1). But when I right-shift it 32 times, I get -1 again. Shouldn't it be equal to 0? 回答1: The Java specification explains the shift operators as follows: If the promoted type of the left-hand operand is int , only the five lowest-order bits of the right-hand operand are used as the

问题 In chapter 2, the section on bitwise operators (section 2.9), I'm having trouble understanding how one of the sample methods works. Here's the method provided: unsigned int getbits(unsigned int x, int p, int n) { return (x >> (p + 1 - n)) & ~(~0 << n); } The idea is that, for the given number x , it will return the n bits starting at position p , counting from the right (with the farthest right bit being position 0). Given the following main() method: int main(void) { int x = 0xF994, p = 4, n

问题 I'm a little confused now by java left shift operation, 1<<31 = 0x80000000 --> this I can understand But 1<<32 = 1 Why is this? 1<<33 = 2 Looks like more shifting values, modulus 32 of the value is taken. Thanks everybody for the replying and giving the quote from JLS. I just want to know more. Any idea of the reason why it's designed in this way? Or is it just some convention? Apparently C doesn't have this quirk? Thanks to @paxdiablo. Looks like C declares this behaviour undefined. I have

问题 I must be absolutely crazy here, but gcc 4.7.3 on my machine is giving the most absurd result. Here is the exact code that I'm testing: #include <iostream> using namespace std; int main(){ unsigned int b = 100000; cout << (b>>b) << endl; b = b >> b; cout << b << endl; b >>= b; cout << b << endl; return 0; } Now, any number that's right shifted by itself should result in 0 ( n/(2^n) == 0 with integer divide , n>1 , and positive/unsigned ), but somehow here is my output: 100000 100000 100000 Am

问题 In all versions of C and C++ prior to 2014, writing 1 << (CHAR_BIT * sizeof(int) - 1) caused undefined behaviour, because left-shifting is defined as being equivalent to successive multiplication by 2 , and this shift causes signed integer overflow: The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros. [...] If E1 has a signed type and nonnegative value, and E1 × 2 E2 is representable in the result type, then that is the resulting value; otherwise,

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 5 years ago . I would be thankful for a good tutorial, that explain for java newbies how in java all the "bit shifting" work. I always stumble across it, but never understood how it works. It should explain all the operations and concepts that are possible with byteshifting/bitmanipulation in java. This is just an example

问题 Note - This is NOT a duplicate of this question - Count the consecutive zero bits (trailing) on the right in parallel: an explanation? . The linked question has a different context, it only asks the purpose of signed() being use. DO NOT mark this question as duplicate. I've been finding a way to acquire the number of trailing zeros in a number. I found a bit twiddling Stanford University Write up HERE here that gives the following explanation. unsigned int v; // 32-bit word input to count

问题 I'm making a multiplier in a very simple assembly language in which I have BEQ, NAND, and ADD to create a SRL. I also have to keep the multiplier under 50 lines (16 used thus far) so hopefully the solution can be thrown in a loop. EDIT: My question is how can I implement an SRL with just a NAND and an ADD Had an idea although it is very inefficient, maybe someone can improve it: Decrement say, a, by 1. Store that value in b. Add b and b and store in c. Beq c with a, if it's true then b is