parity

9 bit protocol on UART in embedded Linux

一笑奈何 提交于 2019-12-06 12:18:34
I am trying to force a 9-bit protocol on a UART in embedded Linux. Currently I am testing this out on the am335x_evm board. I am planning on doing this using stick parity. Ideally I was hoping I would not need to actually modify any of the code for the omap-serial.c driver. The reason for the 9-bit protocol is to support some legacy hardware that uses it. The parity bit needs to be 1 for the address portion of the message, 0 for the data portion, then 1 again for the termination byte. I was planning on having a process running in user space that would interface with the UART through standard

One instruction to clear PF (Parity Flag) — get odd number of bits in result register

房东的猫 提交于 2019-12-06 08:13:17
问题 In x86 assembly, is it possible to clear the Parity Flag in one and only one instruction, working under any initial register configuration? This is equivalent to creating a result register with an odd number of bits, with any operation that sets flags (expressly excluding mov ). For contrast, setting the parity flag can be done in one instruction: cmp bl, bl And there are many ways to clear the parity flag with two instructions: and bl, 0 or bl, 1 However, the one-instruction method remains

Bit parity code for odd number of bits

落爺英雄遲暮 提交于 2019-12-05 04:31:15
I am trying to find the parity of a bitstring so that it returns 1 if x has an odd # of 0's. I can only use basic bitwise operations and what I have so far passes most of the tests, but I'm wondering 2 things: Why does x ^ (x + ~1) work? I stumbled upon this, but it seems to give you 1 if there are an odd number of bits and something else if even. Like 7^6 = 1 because 7 = 0b0111 Is this the right direction of problem solving for this? I'm assuming my problem is stemming from the first operation, specifically (x + ~1) because it would overflow certain 2's complement numbers. Thanks Code: int

Even parity of a unsigned int [duplicate]

天涯浪子 提交于 2019-12-04 16:13:33
This question already has answers here : How to count the number of set bits in a 32-bit integer? (53 answers) Closed 5 years ago . /*A value has even parity if it has an even number of 1 bits. *A value has an odd parity if it has an odd number of 1 bits. *For example, 0110 has even parity, and 1110 has odd parity. *Return 1 iff x has even parity. */ int has_even_parity(unsigned int x) { } I'm not sure where to begin writing this function, I'm thinking that I loop through the value as an array and apply xor operations on them. Would something like the following work? If not, what is the way to

How to judge if there is even 1s in a number's binary representation using C? [duplicate]

◇◆丶佛笑我妖孽 提交于 2019-12-04 13:38:37
问题 This question already has answers here : Computing the Parity (2 answers) Closed 6 years ago . There are already questions about counting how many 1 s are there in a number, but this question is about judging whether there is even or odd number of 1s. Any loop or conditional (including switch) statements are not allowed. Also, division, multiplication or modulus operators should be avoided. To be more specific, we may assume that it's a 32-bits unsigned integer. Actually I've got an

One instruction to clear PF (Parity Flag) — get odd number of bits in result register

岁酱吖の 提交于 2019-12-04 12:21:11
In x86 assembly, is it possible to clear the Parity Flag in one and only one instruction, working under any initial register configuration? This is equivalent to creating a result register with an odd number of bits, with any operation that sets flags (expressly excluding mov ). For contrast, setting the parity flag can be done in one instruction: cmp bl, bl And there are many ways to clear the parity flag with two instructions: and bl, 0 or bl, 1 However, the one-instruction method remains elusive. Not possible. None of the PF-changing commands can unconditionally produce an odd-parity result

How to judge if there is even 1s in a number's binary representation using C? [duplicate]

假装没事ソ 提交于 2019-12-03 08:25:30
This question already has answers here : Computing the Parity (2 answers) There are already questions about counting how many 1 s are there in a number, but this question is about judging whether there is even or odd number of 1s. Any loop or conditional (including switch) statements are not allowed. Also, division, multiplication or modulus operators should be avoided. To be more specific, we may assume that it's a 32-bits unsigned integer. Actually I've got an implementation already, but I cannot work out the reason why it works. Any proof of its correctness or any new idea would be very

non-boost asio checking for errors c++

匆匆过客 提交于 2019-12-02 17:51:07
问题 Is there a way to check for errors in a non-boost asio program using tcp? And is there a way to add an error to the connection randomly? I made a simple Echo Server in C++ for which i now have to generate random errors, but the problem is that i have no clue how. Or if that helps more, i need to check for two dimensional parity (DATA,ACK,NAK). Hope you can help me. Client: #include <iostream> #include <asio\include\asio.hpp> #include <spdlog\spdlog.h> using namespace std; using namespace asio

How to add even parity bit on 7-bit binary number

*爱你&永不变心* 提交于 2019-12-02 08:05:57
I am continuing from my previous question. I am making a c# program where the user enters a 7-bit binary number and the computer prints out the number with an even parity bit to the right of the number. I am struggling. I have a code, but it says BitArray is a namespace but is used as a type. Also, is there a way I could improve the code and make it simpler? namespace BitArray { class Program { static void Main(string[] args) { Console.WriteLine("Please enter a 7-bit binary number:"); int a = Convert.ToInt32(Console.ReadLine()); byte[] numberAsByte = new byte[] { (byte)a }; BitArray bits = new

Parity of a number (Assembly 8086)

与世无争的帅哥 提交于 2019-12-01 08:09:45
Im trying to give a one digit number, and know if the parity is odd or even, for example, give 9 and print that is an odd number. This is what I have: assume cs:cseg,ds:dseg,ss:sseg cseg segment start: mov ax, dseg mov ds, ax mov ah, 01h ; Here, im adding a number int 21h jp even jnp odd even: mov ah,09 lea dx,par int 21h jmp exit odd: mov ah,09 lea dx,odd1 int 21h jmp salir salir: mov ax,4C00h int 21h cseg ends dseg segment byte even Db 'Even number$' odd11 Db 'Odd number$' dseg ends sseg segment stack db 100h dup(?) sseg ends end start Thanks! And sorry for my bad english. To test if a