How to calculate binary checksum? - ios

I'm working on an app of hardware communication that I send or require data from an external hardware. I have the require data part done.
And I just find out I could use some help to calculate the checksum.
A package is created as NSMutableData, then it will be converted in to Byte Array before sending out.
A package looks like this:
0x1E 0x2D 0x2F DATA checksum
I'm thinking I can convert hex into binary to calculate them one by one. But I don't know if it's a good idea. Please let me know if this is the only way to do it, or there are some built in functions I don't know.
Any suggestions will be appreciated.
BTW, I just found the code for C# from other's post, I'll try to make it work in my app. If I can, I'll share it with you. Still any suggestions will be appreciated.
package org.example.checksum;
public class InternetChecksum {
/**
* Calculate the Internet Checksum of a buffer (RFC 1071 - http://www.faqs.org/rfcs/rfc1071.html)
* Algorithm is
* 1) apply a 16-bit 1's complement sum over all octets (adjacent 8-bit pairs [A,B], final odd length is [A,0])
* 2) apply 1's complement to this final sum
*
* Notes:
* 1's complement is bitwise NOT of positive value.
* Ensure that any carry bits are added back to avoid off-by-one errors
*
*
* #param buf The message
* #return The checksum
*/
public long calculateChecksum(byte[] buf) {
int length = buf.length;
int i = 0;
long sum = 0;
long data;
// Handle all pairs
while (length > 1) {
// Corrected to include #Andy's edits and various comments on Stack Overflow
data = (((buf[i] << 8) & 0xFF00) | ((buf[i + 1]) & 0xFF));
sum += data;
// 1's complement carry bit correction in 16-bits (detecting sign extension)
if ((sum & 0xFFFF0000) > 0) {
sum = sum & 0xFFFF;
sum += 1;
}
i += 2;
length -= 2;
}
// Handle remaining byte in odd length buffers
if (length > 0) {
// Corrected to include #Andy's edits and various comments on Stack Overflow
sum += (buf[i] << 8 & 0xFF00);
// 1's complement carry bit correction in 16-bits (detecting sign extension)
if ((sum & 0xFFFF0000) > 0) {
sum = sum & 0xFFFF;
sum += 1;
}
}
// Final 1's complement value correction to 16-bits
sum = ~sum;
sum = sum & 0xFFFF;
return sum;
}
}

When I post this question a year ago, I was still quite new to Objective-C. It turned out to be something very easy to do.
The way you calculate checksum is based on how checksum is defined in your communication protocol. In my case, checksum is just the sum of all the previous bytes sent or the data you want to send.
So if I have a NSMutableData *cmd that has five bytes:
0x10 0x14 0xE1 0xA4 0x32
checksum is the last byte of 0x10+0x14+0xE1+0xA4+0x32
So the sum is 01DB, checksum is 0xDB.
Code:
//i is the length of cmd
- (Byte)CalcCheckSum:(Byte)i data:(NSMutableData *)cmd
{ Byte * cmdByte = (Byte *)malloc(i);
memcpy(cmdByte, [cmd bytes], i);
Byte local_cs = 0;
int j = 0;
while (i>0) {
local_cs += cmdByte[j];
i--;
j++;
};
local_cs = local_cs&0xff;
return local_cs;
}
To use it:
Byte checkSum = [self CalcCheckSum:[command length] data:command];
Hope it helps.

Related

How to convert Uint8List to decimal number in Dart?

I have an Uint8List data list, for example:
Uint8List uintList = Uint8List.fromList([10, 1]);
How can I convert these numbers to a decimal number?
int decimalValue = ??? // in this case 265
Mees' answer is the correct general method, and it's good to understand how to do bitwise operations manually.
However, Dart does have a ByteData class that has various functions to help parse byte data for you (e.g. getInt16, getUint16). In your case, you can do:
Uint8List uintList = Uint8List.fromList([10, 1]);
int decimalValue = ByteData.view(uintList.buffer).getInt16(0, Endian.little);
print(decimalValue); // Prints: 266.
From what I understand of your question, you want decimalValue to be an integer where the least significant byte is (decimal)10, and the byte after that to be 1. This would result in the value 1 * 256 + 10 = 266. If you meant the bytes the other way around, it would be 10 * 256 + 1 = 2560 + 1 = 2561.
I don't actually have any experience with dart, but I assume code similar to this would work:
int decimalValue = 0;
for (int i = 0; i < uintList.length; i++) {
decimalValue = decimalValue << 8; // shift everything one byte to the left
decimalValue = decimalValue | uintList[i]; // bitwise or operation
}
If it doesn't produce the number you want it to, you might have to iterate through the loop backwards instead, which requires changing one line of code:
for (int i = uintList.length-1; i >= 0; i--) {

How can I generate check sum code in dart?

I want to use PayMaya EMV Merchant Presented QR Code Specification for Payment Systems everything is good except CRC i don't understand how to generate this code.
that's all exist about it ,but i still can't understand how to generate this .
The checksum shall be calculated according to [ISO/IEC 13239] using the polynomial '1021' (hex) and initial value 'FFFF' (hex). The data over which the checksum is calculated shall cover all data objects, including their ID, Length and Value, to be included in the QR Code, in their respective order, as well as the ID and Length of the CRC itself (but excluding its Value).
Following the calculation of the checksum, the resulting 2-byte hexadecimal value shall be encoded as a 4-character Alphanumeric Special value by converting each nibble to an Alphanumeric Special character.
Example: a CRC with a two-byte hexadecimal value of '007B' is included in the QR Code as "6304007B".
This converts a string to its UTF-8 representation as a sequence of bytes, and prints out the 16-bit Cyclic Redundancy Check of those bytes (CRC-16/CCITT-FALSE).
int crc16_CCITT_FALSE(String data) {
int initial = 0xFFFF; // initial value
int polynomial = 0x1021; // 0001 0000 0010 0001 (0, 5, 12)
Uint8List bytes = Uint8List.fromList(utf8.encode(data));
for (var b in bytes) {
for (int i = 0; i < 8; i++) {
bool bit = ((b >> (7-i) & 1) == 1);
bool c15 = ((initial >> 15 & 1) == 1);
initial <<= 1;
if (c15 ^ bit) initial ^= polynomial;
}
}
return initial &= 0xffff;
}
The CRC for ISO/IEC 13239 is this CRC-16/ISO-HDLC, per the notes in that catalog. This implements that CRC and prints the check value 0x906e:
import 'dart:typed_data';
int crc16ISOHDLC(Uint8List bytes) {
int crc = 0xffff;
for (var b in bytes) {
crc ^= b;
for (int i = 0; i < 8; i++)
crc = (crc & 1) != 0 ? (crc >> 1) ^ 0x8408 : crc >> 1;
}
return crc ^ 0xffff;
}
void main() {
Uint8List msg = Uint8List.fromList([0x31, 0x32, 0x33, 0x34, 0x35, 0x36, 0x37, 0x38, 0x39]);
print("0x" + crc16ISOHDLC(msg).toRadixString(16));
}

CRC Calculation Of A Mostly Static Data Stream

Background:
I have a section of memory, 1024 bytes. The last 1020 bytes will always be the same. The first 4 bytes will change (serial number of a product). I need to calculate the CRC-16 CCITT (0xFFFF starting, 0x1021 mask) for the entire section of memory, CRC_WHOLE.
Question:
Is it possible to calculate the CRC for only the first 4 bytes, CRC_A, then apply a function such as the one below to calculate the full CRC? We can assume that the checksum for the last 1020 bytes, CRC_B, is already known.
CRC_WHOLE = XOR(CRC_A, CRC_B)
I know that this formula does not work (tried it), but I am hoping that something similar exists.
Yes. You can see how in zlib's crc32_combine(). If you have two sequences A and B, then the pure CRC of AB is the exclusive-or of the CRC of A0 and the CRC of 0B, where the 0's represent a series of zero bytes with the length of the corresponding sequence, i.e. B and A respectively.
For your application, you can pre-compute a single operator that applies 1020 zeros to the CRC of your first four bytes very rapidly. Then you can exclusive-or that with the pre-computed CRC of the 1020 bytes.
Update:
Here is a post of mine from 2008 with a detailed explanation that #ArtemB discovered (that I had forgotten about):
crc32_combine() in zlib is based on two key tricks. For what follows,
we set aside the fact that the standard 32-bit CRC is pre and post-
conditioned. We can deal with that later. Assume for now a CRC that
has no such conditioning, and so starts with the register filled with
zeros.
Trick #1: CRCs are linear. So if you have stream X and stream Y of
the same length and exclusive-or the two streams bit-by-bit to get Z,
i.e. Z = X ^ Y (using the C notation for exclusive-or), then CRC(Z) =
CRC(X) ^ CRC(Y). For the problem at hand we have two streams A and B
of differing length that we want to concatenate into stream Z. What
we have available are CRC(A) and CRC(B). What we want is a quick way
to compute CRC(Z). The trick is to construct X = A concatenated with
length(B) zero bits, and Y = length(A) zero bits concatenated with B.
So if we represent concatenation simply by juxtaposition of the
symbols, X = A0, Y = 0B, then X^Y = Z = AB. Then we have CRC(Z) =
CRC(A0) ^ CRC(0B).
Now we need to know CRC(A0) and CRC(0B). CRC(0B) is easy. If we feed
a bunch of zeros to the CRC machine starting with zero, the register
is still filled with zeros. So it's as if we did nothing at all.
Therefore CRC(0B) = CRC(B).
CRC(A0) requires more work however. Taking a non-zero CRC and feeding
zeros to the CRC machine doesn't leave it alone. Every zero changes
the register contents. So to get CRC(A0), we need to set the register
to CRC(A), and then run length(B) zeros through it. Then we can
exclusive-or the result of that with CRC(B) = CRC(0B), and we get what
we want, which is CRC(Z) = CRC(AB). Voila!
Well, actually the voila is premature. I wasn't at all satisfied with
that answer. I didn't want a calculation that took a time
proportional to the length of B. That wouldn't save any time compared
to simply setting the register to CRC(A) and running the B stream
through. I figured there must be a faster way to compute the effect
of feeding n zeros into the CRC machine (where n = length(B)). So
that leads us to:
Trick #2: The CRC machine is a linear state machine. If we know the
linear transformation that occurs when we feed a zero to the machine,
then we can do operations on that transformation to more efficiently
find the transformation that results from feeding n zeros into the
machine.
The transformation of feeding a single zero bit into the CRC machine
is completely represented by a 32x32 binary matrix. To apply the
transformation we multiply the matrix by the register, taking the
register as a 32 bit column vector. For the matrix multiplication in
binary (i.e. over the Galois Field of 2), the role of multiplication
is played by and'ing, and the role of addition is played by exclusive-
or'ing.
There are a few different ways to construct the magic matrix that
represents the transformation caused by feeding the CRC machine a
single zero bit. One way is to observe that each column of the matrix
is what you get when your register starts off with a single one in
it. So the first column is what you get when the register is 100...
and then feed a zero, the second column comes from starting with
0100..., etc. (Those are referred to as basis vectors.) You can see
this simply by doing the matrix multiplication with those vectors.
The matrix multiplication selects the column of the matrix
corresponding to the location of the single one.
Now for the trick. Once we have the magic matrix, we can set aside
the initial register contents for a while, and instead use the
transformation for one zero to compute the transformation for n
zeros. We could just multiply n copies of the matrix together to get
the matrix for n zeros. But that's even worse than just running the n
zeros through the machine. However there's an easy way to avoid most
of those matrix multiplications to get the same answer. Suppose we
want to know the transformation for running eight zero bits, or one
byte through. Let's call the magic matrix that represents running one
zero through: M. We could do seven matrix multiplications to get R =
MxMxMxMxMxMxMxM. Instead, let's start with MxM and call that P. Then
PxP is MxMxMxM. Let's call that Q. Then QxQ is R. So now we've
reduced the seven multiplications to three. P = MxM, Q = PxP, and R =
QxQ.
Now I'm sure you get the idea for an arbitrary n number of zeros. We
can very rapidly generate transformation matrices Mk, where Mk is the
transformation for running 2k zeros through. (In the
paragraph above M3 is R.) We can make M1 through Mk with only k
matrix multiplications, starting with M0 = M. k only has to be as
large as the number of bits in the binary representation of n. We can
then pick those matrices where there are ones in the binary
representation of n and multiply them together to get the
transformation of running n zeros through the CRC machine. So if n =
13, compute M0 x M2 x M3.
If j is the number of one's in the binary representation of n, then we
just have j - 1 more matrix multiplications. So we have a total of k
j - 1 matrix multiplications, where j <= k = floor(logbase2(n)).
Now we take our rapidly constructed matrix for n zeros, and multiply
that by CRC(A) to get CRC(A0). We can compute CRC(A0) in O(log(n))
time, instead of O(n) time. We exclusive or that with CRC(B) and
Voila! (really this time), we have CRC(Z).
That's what zlib's crc32_combine() does.
I will leave it as an exercise for the reader as to how to deal with
the pre and post conditioning of the CRC register. You just need to
apply the linearity observations above. Hint: You don't need to know
length(A). In fact crc32_combine() only takes three arguments:
CRC(A), CRC(B), and length(B) (in bytes).
Below is example C code for an alternative approach for CRC(A0). Rather than working with a matrix, a CRC can be cycled forward n bits by muliplying (CRC ยท ((2^n)%POLY)%POLY . So the repeated squaring is performed on an integer rather than a matrix. If n is constant, then (2^n)%POLY can be pre-computed.
/* crcpad.c - crc - data has a large number of trailing zeroes */
#include <stdio.h>
#include <stdlib.h>
typedef unsigned char uint8_t;
typedef unsigned int uint32_t;
#define POLY (0x04c11db7u)
static uint32_t crctbl[256];
void GenTbl(void) /* generate crc table */
{
uint32_t crc;
uint32_t c;
uint32_t i;
for(c = 0; c < 0x100; c++){
crc = c<<24;
for(i = 0; i < 8; i++)
/* assumes twos complement */
crc = (crc<<1)^((0-(crc>>31))&POLY);
crctbl[c] = crc;
}
}
uint32_t GenCrc(uint8_t * bfr, size_t size) /* generate crc */
{
uint32_t crc = 0u;
while(size--)
crc = (crc<<8)^crctbl[(crc>>24)^*bfr++];
return(crc);
}
/* carryless multiply modulo crc */
uint32_t MpyModCrc(uint32_t a, uint32_t b) /* (a*b)%crc */
{
uint32_t pd = 0;
uint32_t i;
for(i = 0; i < 32; i++){
/* assumes twos complement */
pd = (pd<<1)^((0-(pd>>31))&POLY);
pd ^= (0-(b>>31))&a;
b <<= 1;
}
return pd;
}
/* exponentiate by repeated squaring modulo crc */
uint32_t PowModCrc(uint32_t p) /* pow(2,p)%crc */
{
uint32_t prd = 0x1u; /* current product */
uint32_t sqr = 0x2u; /* current square */
while(p){
if(p&1)
prd = MpyModCrc(prd, sqr);
sqr = MpyModCrc(sqr, sqr);
p >>= 1;
}
return prd;
}
/* # data bytes */
#define DAT ( 32)
/* # zero bytes */
#define PAD (992)
/* DATA+PAD */
#define CNT (1024)
int main()
{
uint32_t pmc;
uint32_t crc;
uint32_t crf;
uint32_t i;
uint8_t *msg = malloc(CNT);
for(i = 0; i < DAT; i++) /* generate msg */
msg[i] = (uint8_t)rand();
for( ; i < CNT; i++)
msg[i] = 0;
GenTbl(); /* generate crc table */
crc = GenCrc(msg, CNT); /* generate crc normally */
crf = GenCrc(msg, DAT); /* generate crc for data */
pmc = PowModCrc(PAD*8); /* pmc = pow(2,PAD*8)%crc */
crf = MpyModCrc(crf, pmc); /* crf = (crf*pmc)%crc */
printf("%08x %08x\n", crc, crf);
free(msg);
return 0;
}
Example C code using intrinsic for carryless multiply, pclmulqdq == _mm_clmulepi64_si128:
/* crcpadm.c - crc - data has a large number of trailing zeroes */
/* pclmulqdq intrinsic version */
#include <stdio.h>
#include <stdlib.h>
#include <intrin.h>
typedef unsigned char uint8_t;
typedef unsigned int uint32_t;
typedef unsigned long long uint64_t;
#define POLY (0x104c11db7ull)
#define POLYM ( 0x04c11db7u)
static uint32_t crctbl[256];
static __m128i poly; /* poly */
static __m128i invpoly; /* 2^64 / POLY */
void GenMPoly(void) /* generate __m12i8 poly info */
{
uint64_t N = 0x100000000ull;
uint64_t Q = 0;
for(size_t i = 0; i < 33; i++){
Q <<= 1;
if(N&0x100000000ull){
Q |= 1;
N ^= POLY;
}
N <<= 1;
}
poly.m128i_u64[0] = POLY;
invpoly.m128i_u64[0] = Q;
}
void GenTbl(void) /* generate crc table */
{
uint32_t crc;
uint32_t c;
uint32_t i;
for(c = 0; c < 0x100; c++){
crc = c<<24;
for(i = 0; i < 8; i++)
/* assumes twos complement */
crc = (crc<<1)^((0-(crc>>31))&POLYM);
crctbl[c] = crc;
}
}
uint32_t GenCrc(uint8_t * bfr, size_t size) /* generate crc */
{
uint32_t crc = 0u;
while(size--)
crc = (crc<<8)^crctbl[(crc>>24)^*bfr++];
return(crc);
}
/* carryless multiply modulo crc */
uint32_t MpyModCrc(uint32_t a, uint32_t b) /* (a*b)%crc */
{
__m128i ma, mb, mp, mt;
ma.m128i_u64[0] = a;
mb.m128i_u64[0] = b;
mp = _mm_clmulepi64_si128(ma, mb, 0x00); /* p[0] = a*b */
mt = _mm_clmulepi64_si128(mp, invpoly, 0x00); /* t[1] = (p[0]*((2^64)/POLY))>>64 */
mt = _mm_clmulepi64_si128(mt, poly, 0x01); /* t[0] = t[1]*POLY */
return mp.m128i_u32[0] ^ mt.m128i_u32[0]; /* ret = p[0] ^ t[0] */
}
/* exponentiate by repeated squaring modulo crc */
uint32_t PowModCrc(uint32_t p) /* pow(2,p)%crc */
{
uint32_t prd = 0x1u; /* current product */
uint32_t sqr = 0x2u; /* current square */
while(p){
if(p&1)
prd = MpyModCrc(prd, sqr);
sqr = MpyModCrc(sqr, sqr);
p >>= 1;
}
return prd;
}
/* # data bytes */
#define DAT ( 32)
/* # zero bytes */
#define PAD (992)
/* DATA+PAD */
#define CNT (1024)
int main()
{
uint32_t pmc;
uint32_t crc;
uint32_t crf;
uint32_t i;
uint8_t *msg = malloc(CNT);
GenMPoly(); /* generate __m128 polys */
GenTbl(); /* generate crc table */
for(i = 0; i < DAT; i++) /* generate msg */
msg[i] = (uint8_t)rand();
for( ; i < CNT; i++)
msg[i] = 0;
crc = GenCrc(msg, CNT); /* generate crc normally */
crf = GenCrc(msg, DAT); /* generate crc for data */
pmc = PowModCrc(PAD*8); /* pmc = pow(2,PAD*8)%crc */
crf = MpyModCrc(crf, pmc); /* crf = (crf*pmc)%crc */
printf("%08x %08x\n", crc, crf);
free(msg);
return 0;
}

adler32 checksum in objective c [closed]

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I am working on a app which sends data to server with user location info. Server accept this data based on checksum calculation, which is written in java.
Here is the code written in Java:
private static final String CHECKSUM_CONS = "1217278743473774374";
private static String createChecksum(double lat, double lon) {
int latLon = (int) ((lat + lon) * 1E6);
String checkSumStr = CHECKSUM_CONS + latLon;
byte buffer[] = checkSumStr.getBytes();
ByteArrayInputStream bais = new ByteArrayInputStream(buffer);
CheckedInputStream cis = new CheckedInputStream(bais, new Adler32());
byte readBuffer[] = new byte[50];
long value = 0;
try {
while (cis.read(readBuffer) >= 0) {
value = cis.getChecksum().getValue();
}
} catch (Exception e) {
LOGGER.log(Level.SEVERE, e.getMessage(), e);
}
return String.valueOf(value);
}
I tried looking for help to find out how to write objective c equivalent of this. Above function uses adler32 and I don't have any clue about that. Please help.
Thanks for your time.
The answers shown here by #achievelimitless and #user3275097 are incorrect.
First off, signed integers should not be used. The modulo operator on negative numbers is defined differently in different languages, and should be avoided when possible. Simply use unsigned integers instead.
Second, the loops will quickly overflow the 16-bit accumulators, which will give the wrong answer. The modulo operations can be deferred, but they must be done before overflow. You can calculate how many loops you can do safely by assuming that all of the input bytes are 255.
Third, because of the second point, you should not use 16-bit types. You should use at least 32-bit types to avoid having to do the modulo operation very often. You still need to limit the number of loops, but the number gets much bigger. For 32-bit unsigned types, the maximum number of loops is 5552. So the basic code looks like:
#define MOD 65521
#define MAX 5552
unsigned long adler32(unsigned char *buf, size_t len)
{
unsigned long a = 1, b = 0;
size_t n;
while (len) {
n = len > MAX ? MAX : len;
len -= n;
do {
a += *buf++;
b += a;
} while (--n);
a %= MOD;
b %= MOD;
}
return a | (b << 16);
}
As noted by #Sulthan, you should simply use the adler32() function provided in zlib, which is already there on Mac OS X and iOS.
On basis of definition of adler32 checksum as mentioned in wikipedia,
Objective C implementation would be like this:
static NSNumber * adlerChecksumof(NSString *str)
{
NSMutableData *data= [[NSMutableData alloc]init];
unsigned char whole_byte;
char byte_chars[3] = {'\0','\0','\0'};
for (int i = 0; i < ([str length] / 2); i++)
{
byte_chars[0] = [str characterAtIndex:i*2];
byte_chars[1] = [str characterAtIndex:i*2+1];
whole_byte = strtol(byte_chars, NULL, 16);
[data appendBytes:&whole_byte length:1];
}
int16_t a=1;
int16_t b=0;
Byte * dataBytes= (Byte *)[data bytes];
for (int i=0; i<[data length]; i++)
{
a+= dataBytes[i];
b+=a;
}
a%= 65521;
b%= 65521;
int32_t adlerChecksum= b*65536+a;
return #(adlerChecksum);
}
Here str would be your string as mentioned in your question..
So when you want to calculate checksum of some string just do this:
NSNumber * calculatedChkSm= adlerChecksumof(#"1217278743473774374");
Please Let me know if more info needed

UDP reliable data service implementation

I'm trying to implement a simple data transfer using UDP. I have a problem for the checksum, given a packet containing the data, how should I implement the checksum? also any idea how to implement the timeouts so it will trigger the retransmission ? Thanks
Why not try Reliable UDP, see http://en.wikipedia.org/wiki/Reliable_User_Datagram_Protocol
It has a standard.
here's one approach for the internet checksum
unsigned short checkSum() {
unsigned long sum = 0;
int i;
for(i=0; i < your packet length ; i++) {
sum += (your packet data[i] & 0xFFFF);
}
while (sum >> 16) {
sum = (sum & 0xFFFF) + (sum >> 16);
}
sum = ~sum;
return ((unsigned short) sum);
}
for the retransmission, you can set alarm to trigger timeout
when packet is loss. you can do something using
signal (SIGALRM, timeout function);
Hope it helps!

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