bit-shift

问题 I\'m a bit confused because I wanted to initialize a variable of type unsigned long whose size is 8 bytes on my system (on every modern system I suppose). When I want to assign 1 << 63 to the variable, I get a compiler warning however and the number is in fact 0. When I do 1 << 30 I get the expected result of 2 ^ 30 = 1073741824 . Yet when I do 1 << 31 , I get the result of 2 ^ 64 (I think; actually this shouldn\'t be possible) which prints 18446744071562067968 . Can anyone explain this

问题 I recently faced a strange behavior using the right-shift operator. The following program: #include <cstdio> #include <cstdlib> #include <iostream> #include <stdint.h> int foo(int a, int b) { return a >> b; } int bar(uint64_t a, int b) { return a >> b; } int main(int argc, char** argv) { std::cout << \"foo(1, 32): \" << foo(1, 32) << std::endl; std::cout << \"bar(1, 32): \" << bar(1, 32) << std::endl; std::cout << \"1 >> 32: \" << (1 >> 32) << std::endl; //warning here std::cout << \"(int)1 >

问题 Does endianness matter at all with the bitwise operations? Either logical or shifting? I\'m working on homework with regard to bitwise operators, and I can not make heads or tails on it, and I think I\'m getting quite hung up on the endianess. That is, I\'m using a little endian machine (like most are), but does this need to be considered or is it a wasted fact? In case it matters, I\'m using C. 回答1: Endianness only matters for layout of data in memory. As soon as data is loaded by the

问题 Consider the following code: int i = 3 << 65; I would expect that the result is i==0 , however the actual result is i==6 . With some testing I found that with the following code: int i, s; int a = i << s; int b = i << (s & 31); the values of a and b are always the same. Does the C standard say anything about shifting more than 32 bits (the width of type int ) or is this unspecified behavior? 回答1: From my WG12/N1124 draft ( not the standard, but Good Enough For Me), there's the following block

问题 I was looking at code from Mozilla that add a filter method to Array and it had a line of code that confused me. var len = this.length >>> 0; I have never seen >>> used in JavaScript before. What is it and what does it do? 回答1: It doesn't just convert non-Numbers to Number, it converts them to Numbers that can be expressed as 32-bit unsigned ints. Although JavaScript's Numbers are double-precision floats(*), the bitwise operators ( << , >> , & , | and ~ ) are defined in terms of operations on

问题 This question already has answers here : What are bitwise shift (bit-shift) operators and how do they work? (9 answers) Closed last year . I am trying to understand the shift operators and couldn\'t get much. When I tried to execute the below code System.out.println(Integer.toBinaryString(2 << 11)); System.out.println(Integer.toBinaryString(2 << 22)); System.out.println(Integer.toBinaryString(2 << 33)); System.out.println(Integer.toBinaryString(2 << 44)); System.out.println(Integer

问题 Multiplication and division can be achieved using bit operators, for example i*2 = i<<1 i*3 = (i<<1) + i; i*10 = (i<<3) + (i<<1) and so on. Is it actually faster to use say (i<<3)+(i<<1) to multiply with 10 than using i*10 directly? Is there any sort of input that can\'t be multiplied or divided in this way? 回答1: Short answer: Not likely. Long answer: Your compiler has an optimizer in it that knows how to multiply as quickly as your target processor architecture is capable. Your best bet is

问题 I\'ve been attempting to learn C in my spare time, and other languages (C#, Java, etc.) have the same concept (and often the same operators) ... What I\'m wondering is, at a core level, what does bit-shifting ( << , >> , >>> ) do, what problems can it help solve, and what gotchas lurk around the bend? In other words, an absolute beginner\'s guide to bit shifting in all its goodness. 回答1: The bit shifting operators do exactly what their name implies. They shift bits. Here's a brief (or not-so

问题 In C bitwise left shift operation invokes Undefined Behaviour when the left side operand has negative value. Relevant quote from ISO C99 (6.5.7/4) The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros. If E1 has an unsigned type, the value of the result is E1 × 2 E2 , reduced modulo one more than the maximum value representable in the result type. If E1 has a signed type and nonnegative value, and E1 × 2 E2 is representable in the result type, then that

问题 I have C code in which I do the following. int nPosVal = +0xFFFF; // + Added for ease of understanding int nNegVal = -0xFFFF; // - Added for valid reason Now when I try printf (\"%d %d\", nPosVal >> 1, nNegVal >> 1); I get 32767 -32768 Is this expected? I am able to think something like 65535 >> 1 = (int) 32767.5 = 32767 -65535 >> 1 = (int) -32767.5 = -32768 That is, -32767.5 is rounded off to -32768. Is this understanding correct? 回答1: It looks like your implementation is probably doing an