I have a byte array, as follows:
byte[] array = new byte[] { 0xAB, 0x7B, 0xF0, 0xEA, 0x04, 0x2E, 0xF3, 0xA9};
The task is to find the quantity of occurrences '0xA' in it.
Could you advise what to do? The answer is 6.
So from your comment, you want the total count of appearances of the bit pattern 1010 in the bytes in your array.
For a given byte b, the count is the sum of
(b & 0x0A) == 0x0A ? 1 : 0
(b & 0x14) == 0x14 ? 1 : 0
(b & 0x28) == 0x28 ? 1 : 0
(b & 0x50) == 0x50 ? 1 : 0
(b & 0xA0) == 0xA0 ? 1 : 0
(left as an exercise: what is this doing?)
Put this in a function, call it for each byte in the array, sum the results.
If you treat the entire array as a single bit-string:
0xAB, 0x7B, 0xF0, 0xEA, 0x04, 0x2E, 0xF3, 0xA9 is then:
10101011 01111011 11110000 11101010 00000100 00101110 11110011 10101001
==== ==== ====
==== ==== ====
This has 1010 occurring 6 times.
If you don't try to match across byte boundaries, you could try something like the following (tested in Perl and translated by hand):
int counter = 0;
for (int i = 0; i < array.length; ++i)
{
for (int bits = 0xA0, mask = 0xF0; bits >= 0x0A; bits >>= 1, mask >>= 1)
{
if ((array[i] & mask) == bits)
++counter;
}
}
To match across byte boundaries, you have to shift the bits in from the next byte. Try something like this (tested in Perl and translated by hand):
int counter = 0;
byte tester = array[0];
for (int i = 1; i < array.length + 1; ++i)
{
byte nextByte = i < array.length ? array[i] : 0;
for (int bit = 0; bit < 8; ++bit)
{
if ((tester & 0xF0) == 0xA0)
++counter;
tester <<= 1;
if ((nextByte & 0x80) != 0)
tester |= 1;
nextByte <<= 1;
}
}
Both programs count 6 as there are no 1010 sequences across byte-boundaries in this example.
Related
I need to implement my own packets to send over UDP. I decided that I would do this by sending a char buffer which has the sequence number, checksum, size, and the data of the packet which is bytes from a file. The string i'm sending separates each field by a semicolon. Then, when I receive the string (which is my packet) I want to extract each felid, use them accordingly (the sequence number, size, and checksum) and write the bytes to a file. So far I have wrote a method to create 100 packets, and I'm trying to extract and write the bytes to a file (I'm not doing it in the receiver yet, first I'm testing the parsing in the sender). For some reason, the bytes written to my file are incorrect and I'm getting "JPEG DATATSTREAM CONTAINS NO IMAGE" error when I try to open it.
struct packetNode{
char packet[1052]; // this is the entire packet data including the header
struct packetNode *next;
};
This is how I'm creating my packets:
//populate initial window of size 100
for(i = 0; i < 100; i++){
memset(&data[0], 0, sizeof(data));
struct packetNode *p; // create packet node
p = (struct packetNode *)malloc(sizeof(struct packetNode));
bytes = fread(data, 1, sizeof(data), fp); // read 1024 bytes from file into data buffer
int b = fwrite(data, 1, bytes, fpNew);
printf("read: %d\n", bytes);
memset(&p->packet[0], 0, sizeof(p->packet));
sprintf(p->packet, "%d;%d;%d;%s", s, 0, numPackets, data); // create packet
//calculate checksum
int check = checksum8(p->packet, sizeof(p->packet));
sprintf(p->packet, "%d;%d;%d;%s", s, check, numPackets, data); //put checksum in packet
s++; //incremenet sequence number
if(i == 0){
head = p;
tail = p;
tail->next = NULL;
}
else{
tail->next = p;
tail = p;
tail->next = NULL;
}
}
fclose(fp);
and this is where I parse and write the bytes to a file:
void test(){
FILE *fpNew = fopen("test.jpg", "w");
struct packetNode *ptr = head;
char *tokens;
int s, c, size;
int i = 0;
char data[1024];
while(ptr != NULL){
memset(&data[0], 0, sizeof(data));
tokens = strtok(ptr->packet,";");
s = atoi(tokens);
tokens = strtok(NULL, ";");
c = atoi(tokens);
tokens = strtok(NULL, ";");
size = atoi(tokens);
tokens = strtok(NULL, ";");
if(tokens != NULL)
strcpy(data, tokens);
printf("sequence: %d, checksum: %d, size: %d\n", s,c,size);
int b = fwrite(data, 1, sizeof(data), fpNew);
ptr = ptr->next;
i++;
}
fclose(fpNew);
}
Since there is transfer of binary data, a JPEG stream, this data cannot be treated as a string. It's better to go all binary. For instance, instead of
sprintf(p->packet, "%d;%d;%d;%s", s, 0, numPackets, data); // create packet
you would do
sprintf(p->packet, "%d;%d;%d;", s, 0, numPackets);
memcpy(&p->packet[strlen(p->packet)], data, bytes);
but this leads to parsing problems: we would need to change this:
tokens = strtok(NULL, ";");
if(tokens != NULL)
strcpy(data, tokens);
to something like this:
tokens += 1 + ( size < 10 ? 1 : size < 100 ? 2 : size < 1000 ? 3 : size < 10000 ? 4 : 5 );
memcpy(data, tokens, sizeof(data));
#Binary Protocol
It's easier to use a binary packet:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#pragma push(pack,1)
typedef struct Packet {
int seq, maxseq, size;
unsigned short cksum;
unsigned char payload[];
} Packet;
#pragma pop(pack)
typedef struct PacketNode{
struct PacketNode * next;
Packet packet;
} PacketNode;
PacketNode * allocPacketNode(int maxPayloadSize) {
void * ptr = malloc(sizeof(PacketNode) + maxPayloadSize); // FIXME: error checking
memset(ptr, 0, sizeof(PacketNode) + maxPayloadSize); // mallocz wouldn't cooperate
return (PacketNode*) ptr;
}
PacketNode * prepare(FILE * fp, int fsize, int chunksize)
{
PacketNode * head = allocPacketNode(chunksize);
PacketNode * pn = head;
int rd, seq = 0;
int maxseq = fsize / chunksize + ( fsize % chunksize ? 1 : 0 );
while ( ( rd = fread(pn->packet.payload, 1, chunksize, fp ) ) > 0 )
{
printf("read %d bytes\n", rd);
pn->packet.seq = seq++;
pn->packet.maxseq = maxseq;
pn->packet.size = rd + sizeof(Packet);
pn->packet.cksum = 0;
pn->packet.cksum = ~checksum(&pn->packet, pn->packet.size);
if ( rd == chunksize )
pn = pn->next = allocPacketNode(chunksize);
}
return head;
}
int checksum(unsigned char * data, int len)
{
int sum = 0, i;
for ( i = 0; i < len; i ++ )
sum += data[i];
if ( sum > 0xffff )
sum = (sum & 0xffff) + (sum>>16);
return sum;
}
void test( PacketNode * ptr ) {
FILE *fpNew = fopen("test.jpg", "w");
while (ptr != NULL)
{
printf("sequence: %d/%d, checksum: %04x, size: %d\n",
ptr->packet.seq,
ptr->packet.maxseq,
ptr->packet.cksum,
ptr->packet.size - sizeof(Packet)
);
int b = fwrite(ptr->packet.payload, ptr->packet.size - sizeof(Packet), 1, fpNew);
ptr = ptr->next;
}
fclose(fpNew);
}
void fatal( const char * msg ) { printf("FATAL: %s\n", msg); exit(1); }
int main(int argc, char** argv)
{
if ( ! argv[1] ) fatal( "missing filename argument" );
FILE * fp = fopen( argv[1], "r" );
if ( ! fp ) fatal( "cannot open file" );
fseek( fp, 0, SEEK_END );
long fsize = ftell(fp);
fseek( fp, 0, SEEK_SET );
printf("Filesize: %d\n", fsize );
test( prepare(fp, fsize, 1024) );
}
The #pragma push(pack,1) changes how the compiler aligns fields of the struct. We want them to be compact, for network transport. Using 1 is byte-aligned. The #pragma pop(pack) restores the previous setting of the pack pragma.
A note on the checksum method
First we calculate the sum of all the bytes in the packet:
int sum = 0, i;
for ( i = 0; i < len; i ++ )
sum += data[i];
Since the packet uses an unsigned short (16 bits, max value 65535 or 0xffff) to store the checksum, we make sure that the result will fit:
if ( sum > 0xffff ) // takes up more than 16 bits.
Getting the low 16 bits of this int is done using sum & 0xffff, masking out everything but the low 16 bits. We could simply return this value, but we would loose the information from higher checksum bits. So, we will add the upper 16 bits to the lower 16 bits. Accessing the higher 16 bits is done by shifting the int to the right 16 bits, like so: sum >> 16. This is the same as sum / 65536, since 65536 = 216 = 1 << 16.
sum = (sum & 0xffff) + (sum>>16); // add low 16 bits and high 16 bits
I should note that network packet checksums are usually computed 2 bytes (or 'octets' as they like to call them there) at a time. For that, the data should be cast to an unsigned short *, and len should be divided by 2. However! len may be odd, so in that case we'll need to take special care of the last byte. For instance, assuming that the maximum packet size is even, and that the len argument is always <= max_packet_size:
unsigned short * in = (unsigned short *) data;
if ( len & 1 ) data[len] = 0; // make sure last byte is 0
len = (len + 1) / 2;
The rest of the checksum method can remain the same, except that it should operate on in instead of data.
I have a FixMessage and I want to calculate the checksum manually.
8=FIX.4.2|9=49|35=5|34=1|49=ARCA|52=20150916-04:14:05.306|56=TW|10=157|
The body length here is calculated:
8=FIX.4.2|9=49|35=5|34=1|49=ARCA|52=20150916-04:14:05.306|56=TW|10=157|
0 + 0 + 5 + 5 + 8 + 26 + 5 + 0 = 49(correct)
The checksum is 157 (10=157). How to calculate it in this case?
You need to sum every byte in the message up to but not including the checksum field. Then take this number modulo 256, and print it as a number of 3 characters with leading zeroes (e.g. checksum=13 would become 013).
Link from the FIX wiki: FIX checksum
An example implementation in C, taken from onixs.biz:
char *GenerateCheckSum( char *buf, long bufLen )
{
static char tmpBuf[ 4 ];
long idx;
unsigned int cks;
for( idx = 0L, cks = 0; idx < bufLen; cks += (unsigned int)buf[ idx++ ] );
sprintf( tmpBuf, "%03d", (unsigned int)( cks % 256 ) );
return( tmpBuf );
}
Ready-to-run C example adapted from here
8=FIX.4.2|9=49|35=5|34=1|49=ARCA|52=20150916-04:14:05.306|56=TW|10=157|
#include <stdio.h>
void GenerateCheckSum( char *buf, long bufLen )
{
unsigned sum = 0;
long i;
for( i = 0L; i < bufLen; i++ )
{
unsigned val = (unsigned)buf[i];
sum += val;
printf("Char: %02c Val: %3u\n", buf[i], val); // print value of each byte
}
printf("CheckSum = %03d\n", (unsigned)( sum % 256 ) ); // print result
}
int main()
{
char msg[] = "8=FIX.4.2\0019=49\00135=5\00134=1\00149=ARCA\00152=20150916-04:14:05.306\00156=TW\001";
int len = sizeof(msg) / sizeof(msg[0]);
GenerateCheckSum(msg, len);
}
Points to Note
GenerateCheckSum takes the entire FIX message except CheckSum field
Delimiter SOH is written as \001 which has ASCII value 1
static void Main(string[] args)
{
//10=157
string s = "8=FIX.4.2|9=49|35=5|34=1|49=ARCA|52=20150916-04:14:05.306|56=TW|";
byte[] bs = GetBytes(s);
int sum=0;
foreach (byte b in bs)
sum = sum + b;
int checksum = sum % 256;
}
//string to byte[]
static byte[] GetBytes(string str)
{
byte[] bytes = new byte[str.Length * sizeof(char)];
System.Buffer.BlockCopy(str.ToCharArray(), 0, bytes, 0, bytes.Length);
return bytes;
}
Using BodyLength[9] and CheckSum[10] fields.
BodyLength is calculated starting from field starting after BodyLenght and
before CheckSum field.
CheckSum is calculated from ‘8= upto SOH before the checksum field.
Binary value of each character is calculated and compared to the LSB of the calculated value to the checksum value.
If the checksum has been calculated to be 274 then the modulo 256 value is 18 (256 + 18 = 274). This value would be transmitted a 10=018 where
"10="is the tag for the checksum field.
In Java there is a method from QuickFixJ.
String fixStringMessage = "8=FIX.4.29=12535=81=6090706=011=014=017=020=322=837=038=4.39=054=155=ALFAA99=20220829150=0151=06020=06021=06022=F9014=Y";
int checkSum = quickfix.MessageUtils.checksum(fixStringMessage);
System.out.prinln(checkSum);
Output: 127
Hope it can help you.
I'm familiar with WideCharToMultiByte and MultiByteToWideChar conversions and could use these to do something like:
UTF8 -> UTF16 -> 1252
I know that iconv will do what I need, but does anybody know of any MS libs that will allow this in a single call?
I should probably just pull in the iconv library, but am feeling lazy.
Thanks
Windows 1252 is mostly equivalent to latin-1, aka ISO-8859-1: Windows-1252 just has some additional characters allocated in the latin-1 reserved range 128-159. If you are ready to ignore those extra characters, and stick to latin-1, then conversion is rather easy. Try this:
#include <stddef.h>
/*
* Convert from UTF-8 to latin-1. Invalid encodings, and encodings of
* code points beyond 255, are replaced by question marks. No more than
* dst_max_len bytes are stored in the destination array. Returned value
* is the length that the latin-1 string would have had, assuming a big
* enough destination buffer.
*/
size_t
utf8_to_latin1(char *src, size_t src_len,
char *dst, size_t dst_max_len)
{
unsigned char *sb;
size_t u, v;
u = v = 0;
sb = (unsigned char *)src;
while (u < src_len) {
int c = sb[u ++];
if (c >= 0x80) {
if (c >= 0xC0 && c < 0xE0) {
if (u == src_len) {
c = '?';
} else {
int w = sb[u];
if (w >= 0x80 && w < 0xC0) {
u ++;
c = ((c & 0x1F) << 6)
+ (w & 0x3F);
} else {
c = '?';
}
}
} else {
int i;
for (i = 6; i >= 0; i --)
if (!(c & (1 << i)))
break;
c = '?';
u += i;
}
}
if (v < dst_max_len)
dst[v] = (char)c;
v ++;
}
return v;
}
/*
* Convert from latin-1 to UTF-8. No more than dst_max_len bytes are
* stored in the destination array. Returned value is the length that
* the UTF-8 string would have had, assuming a big enough destination
* buffer.
*/
size_t
latin1_to_utf8(char *src, size_t src_len,
char *dst, size_t dst_max_len)
{
unsigned char *sb;
size_t u, v;
u = v = 0;
sb = (unsigned char *)src;
while (u < src_len) {
int c = sb[u ++];
if (c < 0x80) {
if (v < dst_max_len)
dst[v] = (char)c;
v ++;
} else {
int h = 0xC0 + (c >> 6);
int l = 0x80 + (c & 0x3F);
if (v < dst_max_len) {
dst[v] = (char)h;
if ((v + 1) < dst_max_len)
dst[v + 1] = (char)l;
}
v += 2;
}
}
return v;
}
Note that I make no guarantee about this code. This is completely untested.
I have used the following code for converting the bigint in decimal to bytearray (raw data), but I'm getting wrong result.
What is the mistake here?
I'm trying this in Apple Mac ( for Iphone app)
COMP_BYTE_SIZE is 4
Is there any bigendian/ little endian issue, please Help.
void bi_export(BI_CTX *ctx, bigint *x, uint8_t *data, int size)
{
int i, j, k = size-1;
check(x);
memset(data, 0, size); /* ensure all leading 0's are cleared */
for (i = 0; i < x->size; i++)
{
for (j = 0; j < COMP_BYTE_SIZE; j++)
{
comp mask = 0xff << (j*8);
int num = (x->comps[i] & mask) >> (j*8);
data[k--] = num;
if (k < 0)
{
break;
}
}
}
Thanks.
The argument size is at least x->size*4, ie. the target array is big enough? Also use
comp mask = (comp)0xff << (j*8);
num should be cast to uint8_t before copy
data[k--] = (uint8_t)num;
Say I have a large number (integer or float) like 12345 and I want it to look like 12,345.
How would I accomplish that?
I'm trying to do this for an iPhone app, so something in Objective-C or C would be nice.
Here is the answer.
NSNumber* number = [NSNumber numberWithDouble:10000000];
NSNumberFormatter *numberFormatter = [[NSNumberFormatter alloc] init];
[numberFormatter setNumberStyle:kCFNumberFormatterDecimalStyle];
[numberFormatter setGroupingSeparator:#","];
NSString* commaString = [numberFormatter stringForObjectValue:number];
[numberFormatter release];
NSLog(#"%# -> %#", number, commaString);
Try using an NSNumberFormatter.
This should allow you to handle this correctly on an iPhone. Make sure you use the 10.4+ style, though. From that page:
"iPhone OS: The v10.0 compatibility mode is not available on iPhone OS—only the 10.4 mode is available."
At least on Mac OS X, you can just use the "'" string formatter with printf(3).
$ man 3 printf
`'' Decimal conversions (d, u, or i) or the integral portion
of a floating point conversion (f or F) should be
grouped and separated by thousands using the non-mone-
tary separator returned by localeconv(3).
as in printf("%'6d",1000000);
Cleaner C code
// write integer value in ASCII into buf of size bufSize, inserting commas at tousands
// character string in buf is terminated by 0.
// return length of character string or bufSize+1 if buf is too small.
size_t int2str( char *buf, size_t bufSize, int val )
{
char *p;
size_t len, neg;
// handle easy case of value 0 first
if( val == 0 )
{
a[0] = '0';
a[1] = '\0';
return 1;
}
// extract sign of value and set val to absolute value
if( val < 0 )
{
val = -val;
neg = 1;
}
else
neg = 0;
// initialize encoding
p = buf + bufSize;
*--p = '\0';
len = 1;
// while the buffer is not yet full
while( len < bufSize )
{
// put front next digit
*--p = '0' + val % 10;
val /= 10;
++len;
// if the value has become 0 we are done
if( val == 0 )
break;
// increment length and if it's a multiple of 3 put front a comma
if( (len % 3) == 0 )
*--p = ',';
}
// if buffer is too small return bufSize +1
if( len == bufSize && (val > 0 || neg == 1) )
return bufSize + 1;
// add negative sign if required
if( neg == 1 )
{
*--p = '-';
++len;
}
// move string to front of buffer if required
if( p != buf )
while( *buf++ = *p++ );
// return encoded string length not including \0
return len-1;
}
I did this for an iPhone game recently. I was using the built-in LCD font, which is a monospaced font. I formatted the numbers, ignoring the commas, then stuck the commas in afterward. (The way calculators do it, where the comma is not considered a character.)
Check out the screenshots at RetroJuJu. Sorry--they aren't full-sized screenshots so you'll have to squint!
Hope that helps you (it's in C) :
char* intToFormat(int a)
{
int nb = 0;
int i = 1;
char* res;
res = (char*)malloc(12*sizeof(char));
// Should be enough to get you in the billions. Get it higher if you need
// to use bigger numbers.
while(a > 0)
{
if( nb > 3 && nb%3 == 0)
res[nb++] = ',';
// Get the code for the '0' char and add it the position of the
// number to add (ex: '0' + 5 = '5')
res[nb] = '0' + a%10;
nb++;
a /= 10;
}
reverse(&res);
return res;
}
There might be a few errors I didn't see (I'm blind when it comes to this...)
It's like an enhanced iToA so maybe it's not the best solution.
Use recursion, Luke:
#include <stdio.h>
#include <stdlib.h>
static int sprint64u( char* buffer, unsigned __int64 x) {
unsigned __int64 quot = x / 1000;
int chars_written;
if ( quot != 0) {
chars_written = sprint64u( buffer, quot);
chars_written += sprintf( buffer + chars_written, ".%03u", ( unsigned int)( x % 1000));
}
else {
chars_written = sprintf( buffer, "%u", ( unsigned int)( x % 1000));
}
return chars_written;
}
int main( void) {
char buffer[ 32];
sprint64u( buffer, 0x100000000ULL);
puts( buffer);
return EXIT_SUCCESS;
}