I'm trying to parse a binary file in the browser. I have 4 bytes that represent a 32-bit signed integer.
Is there a straight forward way of converting this to a dart int, or do I have to calculate the inverse of two's complement manually?
Thanks
Edit: Using this for manually converting:
int readSignedInt() {
int value = readUnsignedInt();
if ((value & 0x80000000) > 0) {
// This is a negative number. Invert the bits and add 1
value = (~value & 0xFFFFFFFF) + 1;
// Add a negative sign
value = -value;
}
return value;
}
You can use ByteArray from typed_data library.
import 'dart:typed_data';
int fromBytesToInt32(int b3, int b2, int b1, int b0) {
final int8List = new Int8List(4)
..[3] = b3
..[2] = b2
..[1] = b1
..[0] = b0;
return int8List.asByteArray().getInt32(0);
}
void main() {
assert(fromBytesToInt32(0x00, 0x00, 0x00, 0x00) == 0);
assert(fromBytesToInt32(0x00, 0x00, 0x00, 0x01) == 1);
assert(fromBytesToInt32(0xF0, 0x00, 0x00, 0x00) == -268435456);
}
Place the 4 bytes in a ByteArray and extract the Int32 like this:
import 'dart:scalarlist';
void main() {
Int8List list = new Int8List(4);
list[0] = b0;
list[1] = b1;
list[2] = b2;
list[3] = b3;
int number = list.asByteArray().getInt32(0);
}
John
I'm not exactly sure what you want, but maybe this code sample might get you ideas:
int bytesToInteger(List<int> bytes) {
var value = 0;
for (var i = 0, length = bytes.length; i < length; i++) {
value += bytes[i] * pow(256, i);
}
return value;
}
So let's say we have [50, 100, 150, 250] as our "4 bytes", the ending result is a 32-bit unsigned integer. I have a feeling this isn't exactly what you are looking for, but it might help you.
Related
I am validating DPDK receive functionality & for this I'm shooting a pcap externally &
added code in l2fwd to dump received packets to pcap, the l2fwd dumped pcap have all the packets from shooter but some of them are not in sequence.
Shooter is already validated.
DPDK version in use-21.11
link of the pcap used : https://wiki.wireshark.org/uploads/__moin_import__/attachments/SampleCaptures/tcp-ecn-sample.pcap
Out of order packets are random. For the first run I saw no jumbled packets but was able to replicate the issue on second run with the 2nd,3rd,4th packets jumbled having order 3,4,2.
Below is snipped from l2fwd example & our modifications as //TESTCODE..
/* Read packet from RX queues. 8< */
for (i = 0; i < qconf->n_rx_port; i++) {
portid = qconf->rx_port_list[i];
nb_rx = rte_eth_rx_burst(portid, 0,
pkts_burst, MAX_PKT_BURST);
port_statistics[portid].rx += nb_rx;
for (j = 0; j < nb_rx; j++) {
m = pkts_burst[j];
// TESTCODE_STARTS
uint8_t* pkt = rte_pktmbuf_mtod(m, uint8_t*);
dump_to_pcap(pkt, rte_pktmbuf_pkt_len(m));
// TESTCODE_ENDS
rte_prefetch0(rte_pktmbuf_mtod(m, void *));
l2fwd_simple_forward(m, portid);
}
}
/* >8 End of read packet from RX queues. */
Below is code for dump_to_pcap
static int
dump_to_pcap(uint8_t* pkt, int pkt_len)
{
static FILE* fp = NULL;
static int init_file = 0;
if (0 == init_file) {
printf("Creating pcap\n");
char pcap_filename[256] = { 0 };
char Two_pcap_filename[256] = { 0 };
currentDateTime(pcap_filename);
sprintf(Two_pcap_filename,".\\Rx_%d_%s.pcap", 0, pcap_filename);
printf("FileSName to Create: %s\n", Two_pcap_filename);
fp = fopen(Two_pcap_filename, "wb");
if (NULL == fp) {
printf("Unable to open file\n");
fp = NULL;
}
else {
printf("File create success..\n");
init_file = 1;
typedef struct pcap_file_header1 {
unsigned int magic; // a 32-bit "magic number"
unsigned short version_major; //a 16-bit major version number
unsigned short version_minor; //a 16-bit minor version number
unsigned int thiszone; //a 32-bit "time zone offset" field that's actually not used, so ou can (and probably should) just make it 0
unsigned int sigfigs; //a 32-bit "time stamp accuracy" field that's not actually used,so you can (and probably should) just make it 0;
unsigned int snaplen; //a 32-bit "snapshot length" field
unsigned int linktype; //a 32-bit "link layer type" field
}dumpFileHdr;
dumpFileHdr file_hdr;
file_hdr.magic = 2712847316; //0xa1b2c3d4;
file_hdr.version_major = 2;
file_hdr.version_minor = 4;
file_hdr.thiszone = 0;
file_hdr.sigfigs = 0;
file_hdr.snaplen = 65535;
file_hdr.linktype = 1;
fwrite((void*)(&file_hdr), sizeof(dumpFileHdr), 1, fp);
//printf("Pcap Header written\n");
}
}
typedef struct pcap_pkthdr1 {
unsigned int ts_sec; /* time stamp */
unsigned int ts_usec;
unsigned int caplen; /* length of portion present */
unsigned int len; /* length this packet (off wire) */
}dumpPktHdr;
dumpPktHdr pkt_hdr;
static int ts_sec = 1;
pkt_hdr.ts_sec = ts_sec++;
pkt_hdr.ts_usec = 0;
pkt_hdr.caplen = pkt_hdr.len = pkt_len;
if (NULL != fp) {
fwrite((void*)(&pkt_hdr), sizeof(dumpPktHdr), 1, fp);
fwrite((void*)(pkt), pkt_len, 1, fp);
fflush(fp);
}
return 0;
}
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));
}
I am trying to play around with a manual encryption in CBC mode but still use Crypto++, just to know can I do it manually.
The CBC algorithm is (AFAIK):
Presume we have n block K[1]....k[n]
0. cipher = empty;
1. xor(IV, K1) -> t1
2. encrypt(t1) -> r1
3. cipher += r1
4. xor (r1, K2) -> t2
5. encrypt(t2) -> r2
6. cipher += r2
7. xor(r2, K3)->t3
8. ...
So I tried to implement it with Crypto++. I have a text file with alphanumeric characters only. Test 1 is read file chunk by chunk (16 byte) and encrypt them using CBC mode manually, then sum up the cipher. Test 2 is use Crypto++ built-in CBC mode.
Test 1
char* key;
char* iv;
//Iterate in K[n] array of n blocks
BSIZE = 16;
std::string vectorToString(vector<char> v){
string s ="";
for (int i = 0; i < v.size(); i++){
s[i] = v[i];
}
return s;
}
vector<char> xor( vector<char> s1, vector<char> s2, int len){
vector<char> r;
for (int i = 0; i < len; i++){
int u = s1[i] ^ s2[i];
r.push_back(u);
}
return r;
}
vector<char> byteToVector(byte *b, int len){
vector<char> v;
for (int i = 0; i < len; i++){
v.push_back( b[i]);
}
return v;
}
string cbc_manual(byte [n]){
int i = 0;
//Open a file and read from it, buffer size = 16
// , equal to DEFAULT_BLOCK_SIZE
std::ifstream fin(fileName, std::ios::binary | std::ios::in);
const int BSIZE = 16;
vector<char> encryptBefore;
//This function will return cpc
string cpc ="";
while (!fin.eof()){
char buffer[BSIZE];
//Read a chunk of file
fin.read(buffer, BSIZE);
int sb = sizeof(buffer);
if (i == 0){
encryptBefore = byteToVector( iv, BSIZE);
}
//If i == 0, xor IV with current buffer
//else, xor encryptBefore with current buffer
vector<char> t1 = xor(encryptBefore, byteToVector((byte*) buffer, BSIZE), BSIZE);
//After xored, encrypt the xor result, it will be current step cipher
string r1= encrypt(t1, BSIZE).c_str();
cpc += r1;
const char* end = r1.c_str() ;
encryptBefore = stringToVector( r1);
i++;
}
return cpc;
}
This is my encrypt() function, because we have only one block so I use ECB (?) mode
string encrypt(string s, int size){
ECB_Mode< AES >::Encryption e;
e.SetKey(key, size);
string cipher;
StringSource ss1(s, true,
new StreamTransformationFilter(e,
new StringSink(cipher)
) // StreamTransformationFilter
); // StringSource
return cipher;
}
And this is 100% Crypto++ made solution:
Test 2
encryptCBC(char * plain){
CBC_Mode < AES >::Encryption encryption(key, sizeof(key), iv);
StreamTransformationFilter encryptor(encryption, NULL);
for (size_t j = 0; j < plain.size(); j++)
encryptor.Put((byte)plain[j]);
encryptor.MessageEnd();
size_t ready = encryptor.MaxRetrievable();
string cipher(ready, 0x00);
encryptor.Get((byte*)&cipher[0], cipher.size());
}
Result of Test 1 and Test 2 are different. In the fact, ciphered text from Test 1 is contain the result of Test 2. Example:
Test 1's result aaa[....]bbb[....]ccc[...]...
Test 2 (Crypto++ built-in CBC)'s result: aaabbbccc...
I know the xor() function may cause a problem relate to "sameChar ^ sameChar = 0", but is there any problem relate to algorithm in my code?
This is my Test 2.1 after the 1st solution of jww.
static string auto_cbc2(string plain, long size){
CBC_Mode< AES >::Encryption e;
e.SetKeyWithIV(key, sizeof(key), iv, sizeof(iv));
string cipherText;
CryptoPP::StringSource ss(plain, true,
new CryptoPP::StreamTransformationFilter(e,
new CryptoPP::StringSink(cipherText)
, BlockPaddingSchemeDef::NO_PADDING
) // StreamTransformationFilter
); // StringSource
return cipherText;
}
It throw an error:
Unhandled exception at 0x7407A6F2 in AES-CRPP.exe: Microsoft C++
exception: CryptoPP::InvalidDataFormat at memory location 0x00EFEA74
I only got this error when use BlockPaddingSchemeDef::NO_PADDING, tried to remove BlockPaddingSchemeDef or using BlockPaddingSchemeDef::DEFAULT_PADDING, I got no error . :?
StringSource ss1(s, true,
new StreamTransformationFilter(e,
new StringSink(cipher)));
This uses PKCS padding by default. It takes a 16-byte input and produces a 32-byte output due to padding. You should do one of two things.
First, you can use BlockPaddingScheme::NO_PADDING. Something like:
StringSource ss1(s, true,
new StreamTransformationFilter(e,
new StringSink(cipher)
BlockPaddingScheme::NO_PADDING));
Second, you can process blocks manually, 16 bytes at a time. Something like:
AES::Encryption encryptor(key, keySize);
byte ibuff[<some size>] = ...;
byte obuff[<some size>];
ASSERT(<some size> % AES::BLOCKSIZE == 0);
unsigned int BLOCKS = <some size>/AES::BLOCKSIZE;
for (unsigned int i=0; i<BLOCKS; i==)
{
encryptor.ProcessBlock(&ibuff[i*16], &obuff[i*16]);
// Do the CBC XOR thing...
}
You may be able to call ProcessAndXorBlock from the BlockCipher base class and do it in one shot.
int main(){
int input;
int bin = 0, i = 1;
print("Please input a number");
input = num.parse(stdin.readLineSync());
while(input > 0)
{
bin = bin + (input % 2)*i;
input = input/2;
i = i * 10;
}
return 0;
}
It returned infinite numbers.
You just need to take care of double to int conversion: input = (input/2).floor()
See this working code:
void main() {
int input;
int bin = 0, i = 1;
input = 5;
while(input > 0)
{
bin = bin + (input % 2)*i;
input = (input/2).floor();
i = i * 10;
}
print(bin);
}
Here is a version of the above function that:
uses the integer division
with a ternary conditional operator, avoids the conversion to string.
sets the most significant bit to the left (bin = (dec % 2) + bin; which is what most people expects, but is not what the original snippet did)
String dec2bin(int dec) {
var bin = '';
while (dec > 0) {
bin = (dec % 2 == 0 ? '0' : '1') + bin;
dec ~/= 2;
}
return bin;
}
P.S: But, of course, one can simply write:
var bin = num.toRadixString(2);
There is no real need to write your own dec2bin function.
As the int result can become easily big in length and the max int value in dart is 2e53 (and much less if you compile to web). it's better to change the approach and return as a String.
String dec2bin(int decimal) {
String bin = '';
while (decimal > 0) {
bin = bin + (decimal % 2).toString();
decimal = (decimal / 2).floor();
}
return bin;
}
print(dec2bin(132070242815));
Result: 1111111110011111111111111111110101111
//number is always in int
static String decimalToBinary(int number) {
return number.toRadixString(2);
}
//binary is always in string
static int binaryToDecimal(String binary) {
return int.parse(binary, radix: 2);
}
How to get a 32 bit number in objective c when an byte array is passed to it, similarly as in java where,
ByteBuffer bb = ByteBuffer.wrap(truncation);
return bb.getInt();
Where truncation is the byte array.
It returns 32 bit number.. Is this possible in objective c?
If the number is encoded in little-endian within the buffer, then use:
int32_t getInt32LE(const uint8_t *buffer)
{
int32_t value = 0;
unsigned length = 4;
while (length > 0)
{
value <<= 8;
value |= buffer[--length];
}
return value;
}
If the number is encoded in big-endian within the buffer, then use:
int32_t getInt32BE(const uint8_t *buffer)
{
int32_t value = 0;
for (unsigned i = 0; i < 4; i++)
{
value <<= 8;
value |= *buffer++;
}
return value;
}
UPDATE If you are using data created on the same host then endianness is not an issue, in which case you can use a union as a bridge between the buffer and integers, which avoids some unpleasant casting:
union
{
uint8_t b[sizeof(int32_t)];
int32_t i;
} u;
memcpy(u.b, buffer, sizeof(u.b));
// value is u.i
Depending on the endianness:
uint32_t n = b0 << 24 | b1 << 16 | b2 << 8 | b3;
or
uint32_t n = b3 << 24 | b2 << 16 | b1 << 8 | b0
Not sure if you just want to read 4 bytes and assign that value to an integer. This case:
int32_t number;
memcpy(&number, truncation, sizeof(uint32_t));
About endianess
From your question (for me) was clear that the bytes were already ordered correctly. However if you have to re-order these bytes, use ntohl() after memcpy() :
number=ntohl(number);