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);
Related
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--) {
I'm trying to implement a CRC-CCITT calculator in VHDL. I was able to initially do that; however, I recently found out that data is delivered starting at the least-significant byte. In my code, data is transmitted 7 bytes at a time through a frame. So let's say we have the following data: 123456789 in ASCII or 313233343536373839 in hex. The data would be transmitted as such (with the following CRC):
-- First frame of data
RxFrame.Data <= (
1 => x"39", -- LSB
2 => x"38",
3 => x"37",
4 => x"36",
5 => x"35",
6 => x"34",
7 => x"33"
);
-- Second/last frame of data
RxFrame.Data <= (
1 => x"32",
2 => x"31", -- MSB
3 => xx, -- "xx" means irrelevant data, not part of CRC calculation.
4 => xx, -- This occurs only in the last frame, when it specified in
5 => xx, -- byte 0 which bytes contain data
6 => xx,
7 => xx
);
-- Calculated CRC should be 0x31C3
Another example with data 0x4376669A1CFC048321313233343536373839 and its correct CRC is shown below:
-- First incoming frame of data
RxFrame.Data <= (
1 => x"39", -- LSB
2 => x"38",
3 => x"37",
4 => x"36",
5 => x"35",
6 => x"34",
7 => x"33"
);
-- Second incoming frame of data
RxFrame.Data <= (
1 => x"32",
2 => x"31",
3 => x"21",
4 => x"83",
5 => x"04",
6 => x"FC",
7 => x"1C"
);
-- Third/last incoming frame of data
RxFrame.Data <= (
1 => x"9A",
2 => x"66",
3 => x"76",
4 => x"43", -- MSB
5 => xx, -- Irrelevant data, specified in byte 0
6 => xx,
7 => xx
);
-- Calculated CRC should be 0x2848
Is there a concept I'm missing? Is there a way to calculate the CRC with the data being received in reverse order? I am implementing this for CANopen SDO block protocols. Thanks!
CRC calculation algorithm to verify SDO block transfer from CANopen standard
Example code to generate a CRC16 with the bytes read in reverse (last byte first), using a function to do a carryless multiply modulo the CRC polynomial. An explanation follows.
#include <stdio.h>
typedef unsigned char uint8_t;
typedef unsigned short uint16_t;
#define POLY (0x1021u)
/* carryless multiply modulo crc polynomial */
uint16_t MpyModPoly(uint16_t a, uint16_t b) /* (a*b)%poly */
{
uint16_t pd = 0;
uint16_t i;
for(i = 0; i < 16; i++){
/* assumes twos complement */
pd = (pd<<1)^((0-(pd>>15))&POLY);
pd ^= (0-(b>>15))&a;
b <<= 1;
}
return pd;
}
/* generate crc in reverse byte order */
uint16_t Crc16R(uint8_t * b, size_t sz)
{
uint8_t *e = b + sz; /* end of bfr ptr */
uint16_t crc = 0u; /* crc */
uint16_t pdm = 0x100u; /* padding multiplier */
while(e > b){ /* generate crc */
pdm = MpyModPoly(0x100, pdm);
crc ^= MpyModPoly( *--e, pdm);
}
return(crc);
}
/* msg will be processed in reverse order */
static uint8_t msg[] = {0x43,0x76,0x66,0x9A,0x1C,0xFC,0x04,0x83,
0x21,0x31,0x32,0x33,0x34,0x35,0x36,0x37,
0x38,0x39};
int main()
{
uint16_t crc;
crc = Crc16R(msg, sizeof(msg));
printf("%04x\n", crc);
return 0;
}
Example code using X86 xmm pclmulqdq and psrlq, to emulate a 16 bit by 16 bit hardware (VHDL) carryless multiply:
/* __m128i is an intrinsic for X86 128 bit xmm register */
static __m128i poly = {.m128i_u32[0] = 0x00011021u}; /* poly */
static __m128i invpoly = {.m128i_u32[0] = 0x00008898u}; /* 2^31 / poly */
/* carryless multiply modulo crc polynomial */
/* using xmm pclmulqdq and psrlq */
uint16_t MpyModPoly(uint16_t a, uint16_t b)
{
__m128i ma, mb, mp, mt;
ma.m128i_u64[0] = a;
mb.m128i_u64[0] = b;
mp = _mm_clmulepi64_si128(ma, mb, 0x00); /* mp = a*b */
mt = _mm_srli_epi64(mp, 16); /* mt = mp>>16 */
mt = _mm_clmulepi64_si128(mt, invpoly, 0x00); /* mt = mt*ipoly */
mt = _mm_srli_epi64(mt, 15); /* mt = mt>>15 = (a*b)/poly */
mt = _mm_clmulepi64_si128(mt, poly, 0x00); /* mt = mt*poly */
return mp.m128i_u16[0] ^ mt.m128i_u16[0]; /* ret mp^mt */
}
/* external code to generate invpoly */
#define POLY (0x11021u)
static __m128i invpoly; /* 2^31 / poly */
void GenMPoly(void) /* generate __m12i8 invpoly */
{
uint32_t N = 0x10000u; /* numerator = x^16 */
uint32_t Q = 0; /* quotient = 0 */
for(size_t i = 0; i <= 15; i++){ /* 31 - 16 = 15 */
Q <<= 1;
if(N&0x10000u){
Q |= 1;
N ^= POLY;
}
N <<= 1;
}
invpoly.m128i_u16[0] = Q;
}
Explanation: consider the data as separate strings of ever increasing length, padded with zeroes at the end. For the first few bytes of your example, the logic would calculate
CRC = CRC16({39})
CRC ^= CRC16({38 00})
CRC ^= CRC16({37 00 00})
CRC ^= CRC16({36 00 00 00})
...
To speed up this calculation, rather than actually pad with n zero bytes, you can do a carryless multiply of a CRC by 2^{n·8} modulo POLY, where POLY is the 17 bit polynomial used for CRC16:
CRC = CRC16({39})
CRC ^= (CRC16({38}) · (2^08 % POLY)) % POLY
CRC ^= (CRC16({37}) · (2^10 % POLY)) % POLY
CRC ^= (CRC16({36}) · (2^18 % POLY)) % POLY
...
A carryless multiply modulo POLY is equivalent to what CRC16 does, so this translates into pseudo code (all values in hex, 2^8 = 100)
CRC = 0
PDM = 100 ;padding multiplier
PDM = (100 · PDM) % POLY ;main loop (2 lines per byte)
CRC ^= ( 39 · PDM) % POLY
PDM = (100 · PDM) % POLY
CRC ^= ( 38 · PDM) % POLY
PDM = (100 · PDM) % POLY
CRC ^= ( 37 · PDM) % POLY
PDM = (100 · PDM) % POLY
CRC ^= ( 36 · PDM) % POLY
...
Implementing (A · B) % POLY is based on binary math:
(A · B) % POLY = (A · B) ^ (((A · B) / POLY) · POLY)
Where multiply is carryless (XOR instead of add) and divide is borrowless (XOR instead of subtract). Since the divide is borrowless, and most significant term of POLY is x^16, the quotient
Q = (A · B) / POLY
only depends on the upper 16 bits of (A · B). Dividing by POLY uses multiplication by the 16 bit constant IPOLY = (2^31)/POLY followed by a right shift:
Q = (A · B) / POLY = (((A · B) >> 16) · IPOLY) >> 15
The process uses a 16 bit by 16 bit carryless multiply, producing a 31 bit product.
POLY = 0x11021u ; CRC polynomial (17 bit)
IPOLY = 0x08898u ; 2^31 / POLY
; generated by external software
MpyModPoly(A, B)
{
MP = A · B ; MP = A · B
MT = MP >> 16 ; MT = MP >> 16
MT = MT · IPOLY ; MT = MT · IPOLY
MT = MT >> 15 ; MT = (A · B) / POLY
MT = MT · POLY ; MT = ((A · B) / POLY) * POLY
return MP xor MT ; (A·B) ^ (((A · B) / POLY) · POLY)
}
A hardware based carryless multiply would look something like this 4 bit · 4 bit example.
p[] = [a3 a2 a1 a0] · [b3 b2 b1 b0]
p[] is a 7 bit product generated with 7 parallel circuits.
The time for multiply would be worst case propagation time for p3.
p6 = a3&b3
p5 = a3&b2 ^ a2&b3
p4 = a3&b1 ^ a2&b2 ^ a1&b3
p3 = a3&b0 ^ a2&b1 ^ a1&b2 ^ a0&b3
p2 = a2&b0 ^ a1&b1 ^ a0&b2
p1 = a1&b0 ^ a0&b1
p0 = a0&b0
If the xor gates available only have 2 bit inputs, the logic can
be split up. For example:
p3 = (a3&b0 ^ a2&b1) ^ (a1&b2 ^ a0&b3)
I don't know if your VHDL toolset includes a library for carryless multiply. For a 16 bit by 16 bit multiply resulting in a 31 bit product (p30 to p00), p15 has 16 outputs from the 16 ands (in parallel), which could be xor'ed using a tree like structure, 8 xors in parallel feeding into 4 xors in parallel feeding into 2 xor's in parallel into a single xor. So the propagation time would be 1 and and 4 xor propagation times.
Here is an example in C that you can adapt. Since you mentioned VHDL, this is a bit-wise implementation suitable for casting into gates and flip-flops. However, if cycles are more precious to you than memory and gates, then there is also a byte-wise table-driven version that would run in 1/8 the number of cycles.
What this does is the inverse of what is done in a normal CRC calculation. It then applies the same size input in zeros with a normal CRC to get what the normal CRC would have been on that input. Running the zeros through takes the same number of cycles as the inverse CRC, i.e. O(n) where n is the size of the input. If that latency is too large, that can be reduced to O(log n) cycles, with some investment in gates.
#include <stddef.h>
// Update crc with the CRC-16/XMODEM of n zero bytes. (This can be done in
// O(log n) time or cycles instead of O(n), with a little more effort.)
static unsigned crc16x_zeros_bit(unsigned crc, size_t n) {
for (size_t i = 0; i < n; i++)
for (int k = 0; k < 8; k++)
crc = crc & 0x8000 ? (crc << 1) ^ 0x1021 : crc << 1;
return crc & 0xffff;
}
// Update crc with the CRC-16/XMODEM of the len bytes at mem in reverse. If mem
// is NULL, then return the initial value for the CRC. When done,
// crc16x_zeros_bit() must be used to apply the total length of zero bytes, in
// order to get what the CRC would have been if it were calculated on the bytes
// fed in the opposite order.
static unsigned crc16x_inverse_bit(unsigned crc, void const *mem, size_t len) {
unsigned char const *data = mem;
if (data == NULL)
return 0;
crc &= 0xffff;
for (size_t i = 0; i < len; i++) {
for (int k = 0; k < 8; k++)
crc = crc & 1 ? (crc >> 1) ^ 0x8810 : crc >> 1;
crc ^= (unsigned)data[i] << 8;
}
return crc;
}
#include <stdio.h>
int main(void) {
// Do framed example.
unsigned crc = crc16x_inverse_bit(0, NULL, 0);
crc = crc16x_inverse_bit(crc, (void const *)"9876543", 7);
crc = crc16x_inverse_bit(crc, (void const *)"21", 2);
crc = crc16x_zeros_bit(crc, 9);
printf("%04x\n", crc);
// Do another one.
crc = crc16x_inverse_bit(0, NULL, 0);
crc = crc16x_inverse_bit(crc, (void const *)"9876543", 7);
crc = crc16x_inverse_bit(crc, (void const *)"21!\x83\x04\xfc\x1c", 7);
crc = crc16x_inverse_bit(crc, (void const *)"\x9a" "fvC", 4);
crc = crc16x_zeros_bit(crc, 18);
printf("%04x\n", crc);
return 0;
}
Here is the O(log n) version of crc16x_zeros_bit():
// Return a(x) multiplied by b(x) modulo p(x), where p(x) is the CRC
// polynomial. For speed, a cannot be zero.
static inline unsigned multmodp(unsigned a, unsigned b) {
unsigned p = 0;
for (;;) {
if (a & 1) {
p ^= b;
if (a == 1)
break;
}
a >>= 1;
b = b & 0x8000 ? (b << 1) ^ 0x1021 : b << 1;
}
return p & 0xffff;
}
// Return x^(8n) modulo p(x).
static unsigned x2nmodp(size_t n) {
unsigned p = 1; // x^0 == 1
unsigned q = 0x10; // x^2^2
while (n) {
q = multmodp(q, q); // x^2^k mod p(x), k = 3,4,...
if (n & 1)
p = multmodp(q, p);
n >>= 1;
}
return p;
}
// Update crc with the CRC-16/XMODEM of n zero bytes.
static unsigned crc16x_zeros_bit(unsigned crc, size_t n) {
return multmodp(x2nmodp(n), crc);
}
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));
}
For example, I calculated CRC checksum of a file of size 1024 KB and file includes 22 KB of padding of zeros at the end of the file.
If given checksum of 1024 KB and
size of the padding of zeros of given file
Is it possible to calculate the checksum of the file without the passing. That is in above case getting the checksum of 1002 KB of the file. Assuming we don't have to recalculate the checksum again and reuse the checksum already calculated for the entire file with padding.
After a normal CRC is calculated, a CRC can be "reverse cycled" backwards past the trailing zeroes, but rather than actually reverse cycling the CRC, a carryless multiply can be used:
new crc = (crc · (pow(2,-1-reverse_distance)%poly))%poly
The -1 represents the cyclic period for a CRC. For CRC32, the period is 2^32-1 = 0xffffffff .
By generating a table for pow(2,-1-(i*8))%poly) for i = 1 to n, time complexity can be reduced to O(1), doing a table lookup followed by a carryless multiply mod polynomial (32 iterations).
Example code for a 32 byte message with 14 data bytes, 18 zero bytes, with the new crc to be located at msg[{14,15,16,17}]. After the new bytes are stored in the message, a normal CRC calculation on the shortened message will be zero. The example code doesn't use a table, and the time complexity is O(log2(n)) for the pow(2,-1-(n*8))%poly) calculation.
#include <stdio.h>
typedef unsigned char uint8_t;
typedef unsigned int uint32_t;
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++)
crc = (crc<<1)^((0-(crc>>31))&0x04c11db7);
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++){
pd = (pd<<1)^((0-(pd>>31))&0x04c11db7u);
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;
}
/* message 14 data, 18 zeroes */
/* parities = crc cycled backwards 18 bytes */
int main()
{
uint32_t pmr;
uint32_t crc;
uint32_t par;
uint8_t msg[32] = {0x01,0x02,0x03,0x04,0x05,0x06,0x07,0x08,
0x09,0x0a,0x0b,0x0c,0x0d,0x0e,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,
0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00};
GenTbl(); /* generate crc table */
pmr = PowModCrc(-1-(18*8)); /* pmr = pow(2,-1-18*8)%crc */
crc = GenCrc(msg, 32); /* generate crc including 18 zeroes */
par = MpyModCrc(crc, pmr); /* par = (crc*pmr)%crc = new crc */
crc = GenCrc(msg, 14); /* generate crc for shortened msg */
printf("%08x %08x\n", par, crc); /* crc == par */
msg[14] = (uint8_t)(par>>24); /* store parities in msg */
msg[15] = (uint8_t)(par>>16);
msg[16] = (uint8_t)(par>> 8);
msg[17] = (uint8_t)(par>> 0);
crc = GenCrc(msg, 18); /* crc == 0 */
printf("%08x\n", crc);
return 0;
}
Sure. Look at this answer for code that undoes the trailing zeros, crc32_remove_zeros().
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.