bit-shift

问题 I have a question in my assignment / project that adds 1 to an unsigned integer only using bit-shifting and logical operators . There shouldn't be any "+", "-", "*", or "/" symbols in the function. I am trying from last days but no success yet. So far I've tried the following: int A = (((B&C)<<1)^(B^C)) Can anybody help me to solve this.? You can help me in any programming language. 回答1: unsigned int i = ...; unsigned int mask = 1; while (i & mask) { i &= ~mask; mask <<= 1; } i |= mask; 回答2:

问题 Let consider the following program test.c : #include <stdio.h> struct test { unsigned int a:5; }; int main () { unsigned int i; struct test t = {1}; for (i = 0; i < t.a << 1; i++) printf("%u\n", i); return 0; } When compiled with gcc -Wsign-compare test.c the following warning is produced (tested with gcc 4.8.1): test.c:9:19: warning: comparison between signed and unsigned integer expressions [-Wsign-compare] for (i = 0; i < t.a << 1; i++) ^ clang -Wsign-compare test.c produces the following

问题 I'm trying to make a 8-bit Sequential Multiplier on Quartus II. I did all the simulations of all blocks, but one is showing error on the VWF simulation. The sum_reg block it's doing a infinite shift in a very small time interval. In the "dark blue" part of waveform simulation, on o_DOUT, it's when the shift gones infinite until the MSB goes to the LSB. The image below shows what happens in the dark blue part of the simulation: Someone know what's happen? Below the code: Sum register(where the

问题 It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. Closed 6 years ago . I have a latitude value as double and I want to perform a Bitwise AND operation on it followed by right shift of bits. Following is my line of code: pBuffer[1]=(latitude_decimal_degrees & 0xFF0000) >> 16; However

问题 This question already has answers here : How can I multiply and divide using only bit shifting and adding? (13 answers) Division by a constant using shifts and adds/subtracts (3 answers) Closed 6 years ago . I'm interested how to do this, because I just found out you can do integer multiplication easily, by using left shift operators: x * 25 = (x << 4) + (x << 3) + x Where the sum of base 2 powers equals to 25: 2^4+2^3+2^0 = 25 How does x / 25 work out with right shifts? Edit: I'm not going

问题 If I have an unsigned integer between 0 - 16 and I want to write it to a binary file without writing a whole byte to it, how would one shift bits to achieve it? 0-16 means I only need 4 bits, so I should be able to store 2 different numbers in a single byte right? The following code writes 1 number to 1 byte: std::ofstream file; file.open("test.bin", std::ios::binary|std::ios::out); char oneByteNum = (char)fourByteNum; // Converting from some 4 byte integer to a 1 byte char file.write(

问题 This question already has answers here : Arithmetic right shift gives bogus result? (2 answers) Is right shift undefined behavior if the count is larger than the width of the type? (2 answers) Closed 4 years ago . Can't understand behavior of this bit shift: int container = 1; cout<<(container>>32)<<endl; If it's logical shift the output should be 0, but it's 1 instead, as if it was cyclic shift. When looking at disassembly I see that command used is SAR. Please explain this behavior to me.

问题 Assume you have a binary tree which classifies images. Each node is a different binary test. When an image is fed to the tree, a unique path through the tree is generated. A path is described as a binary word which is as long as the depth of the tree. For instance, for a 2-stage binary tree, an example of path would be (0,1) ((left,right) thus ending in the second leaf of the tree from the left). We also assume that for any image being fed to the tree, all node tests are executable. Thus, we

问题 Given a two floating point Numbers A and B, which are command line arguments, I must create methods to do bitwise operations on them. Including And, Or, Not, xOr, floating Point addition etc... How can I access each bit in C? I need to access the specific 1's and 0's. Any idea how? 回答1: Here's an example using unions, as Keith suggests. /* -std=c99 */ #include <stdio.h> int main(int argc, char * argv[]) { /* note, I'm assuming an int has the same size of a float here */ assert( sizeof

问题 This question already has answers here : What is going on with bitwise operators and integer promotion? (4 answers) C++ Unexpected Integer Promotion (2 answers) Closed 4 months ago . I was a bit messing around with uint8_t and was curious what happens when I outflow bits to the left and found that uint8_t i = 234; uint8_t j = (i << 1); auto k = (i << 1); std::cout << (int)j << std::endl; std::cout << k << std::endl; prints out 212 468 and not the expected 212 212 It seems like << does promote