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CRC CCITT (0xFFFF) Implementation in Python

I need to implement calculation of crc from string of zeroes and one-s like 10000000 etc
I have found this code but its for CRC CCITT(XModem)
POLYNOMIAL = 0x1021
PRESET = 0
def _initial(c):
crc = 0
c = c << 8
for j in range(8):
if (crc ^ c) & 0x8000:
crc = (crc << 1) ^ POLYNOMIAL
else:
crc = crc << 1
c = c << 1
return crc
_tab = [ _initial(i) for i in range(256) ]
def _update_crc(crc, c):
cc = 0xff & c
tmp = (crc >> 8) ^ cc
crc = (crc << 8) ^ _tab[tmp & 0xff]
crc = crc & 0xffff
return crc
def crc(str):
crc = PRESET
for c in str:
crc = _update_crc(crc, ord(c))
return crc
def crcb(*i):
crc = PRESET
for c in i:
crc = _update_crc(crc, c)
print crc
return crc
Changing PRESET to 0xFFFF as it is initial value for CRC CCITT (0xFFFF) doesnt help.
Does anyone can help to rewrite or adapt this code? For reaching a result? For checking uses https://www.lammertbies.nl/comm/info/crc-calculation.html
Downloading and installation crc16 library not allowed due security reasons
Python2.7

CRC32 hash of python string

Using an existing C example algorithm, I want to generate the correct CRC32 hash for a string in python. However, I am receiving incorrect results. I mask the result of every operation and attempt to copy the original algorithm's logic. The C code was provided by the same website which has a webpage string hash checking tool, so it is likely to be correct.
Below is a complete Python file including C code in its comments which it attempts to mimic. All pertinent information is in the file.
P_32 = 0xEDB88320
init = 0xffffffff
_ran = True
tab32 = []
def mask32(n):
return n & 0xffffffff
def mask8(n):
return n & 0x000000ff
def mask1(n):
return n & 0x00000001
def init32():
for i in range(256):
crc = mask32(i)
for j in range(8):
if (mask1(crc) == 1):
crc = mask32(mask32(crc >> 1) ^ P_32)
else:
crc = mask32(crc >> 1)
tab32.append(crc)
global _ran
_ran = False
def update32(crc, char):
char = mask8(char)
t = crc ^ char
crc = mask32(mask32(crc >> 8) ^ tab32[mask8(t)])
return crc
def run(string):
if _ran:
init32()
crc = init
for c in string:
crc = update32(crc, ord(c))
print(hex(crc)[2:].upper())
check0 = "The CRC32 of this string is 4A1C449B"
check1 = "123456789" # CBF43926
run(check0) # Produces B5E3BB64
run(check1) # Produces 340BC6D9
# Check CRC-32 on http://www.lammertbies.nl/comm/info/crc-calculation.html#intr
"""
/* http://www.lammertbies.nl/download/lib_crc.zip */
#define P_32 0xEDB88320L
static int crc_tab32_init = FALSE;
static unsigned long crc_tab32[256];
/*******************************************************************\
* *
* unsigned long update_crc_32( unsigned long crc, char c ); *
* *
* The function update_crc_32 calculates a new CRC-32 value *
* based on the previous value of the CRC and the next byte *
* of the data to be checked. *
* *
\*******************************************************************/
unsigned long update_crc_32( unsigned long crc, char c ) {
unsigned long tmp, long_c;
long_c = 0x000000ffL & (unsigned long) c;
if ( ! crc_tab32_init ) init_crc32_tab();
tmp = crc ^ long_c;
crc = (crc >> 8) ^ crc_tab32[ tmp & 0xff ];
return crc;
} /* update_crc_32 */
/*******************************************************************\
* *
* static void init_crc32_tab( void ); *
* *
* The function init_crc32_tab() is used to fill the array *
* for calculation of the CRC-32 with values. *
* *
\*******************************************************************/
static void init_crc32_tab( void ) {
int i, j;
unsigned long crc;
for (i=0; i<256; i++) {
crc = (unsigned long) i;
for (j=0; j<8; j++) {
if ( crc & 0x00000001L ) crc = ( crc >> 1 ) ^ P_32;
else crc = crc >> 1;
}
crc_tab32[i] = crc;
}
crc_tab32_init = TRUE;
} /* init_crc32_tab */
"""

There's just one thing that's wrong with the current implementation and the fix is actually just one line of code to the end of your run function which is:
crc = crc ^ init
Which if added to your run function look like this:
def run(string):
if _ran:
init32()
crc = init
for c in string:
crc = update32(crc, ord(c))
crc = crc ^ init
print(hex(crc)[2:].upper())
This will give you the correct results you are expecting.The reason that this is necessary is after you are done updating the CRC32, the finalization of it is XORing it with the 0xFFFFFFFF. Since you only had the init table and update functions and not the finalize, you were one step off from the actual crc.
Another C implimentation that is a little more straightforward is this one it's a little bit easier to see the whole process. The only thing slightly obsure is the init poly ~0x0 is the same (0xFFFFFFFF).

Float to Real48

I am trying to convert a floating number to Real48 (Borland Pascal) in python. I have found this method written i C# (I think)
static byte[] DoubleToReal48(double d)
{
byte[] r = new byte[6];
long bits = BitConverter.DoubleToInt64Bits(d);
bool negative = ((bits >> 63) & 1) != 0;
long exponent = ((bits >> 52) & 0x7FF) - 1023;
long mantissa = bits & 0xFFFFFFFFFFFFFL;
long raw = (negative ? 1 : 0);
raw = (raw << 39) | (mantissa >> 13);
raw = (raw << 8) | ((exponent + 129) & 0xFF);
for (int k = 0; k < 6; k++)
{
r[k] = (byte)(raw & 0xFF);
raw >>= 8;
}
return r;
}
I don't know C#, so I'm having a hard time translating it to python. Can anyone help me? Or perhaps there is a better way?
I have found this but it seems to only convert Real48 to double, not the other way around as I need.

Let me try to translate the C# to python
1.- Convert the bits from a double number to a number (unsigned long long).
long bits = BitConverter.DoubleToInt64Bits(d);
translates to
struct.unpack("Q", struct.pack("d", python_double) )[0]
it uses the struct module from standard library.
2.- Extract each field from the double representation
bool negative = ((bits >> 63) & 1) != 0;
long exponent = ((bits >> 52) & 0x7FF) - 1023;
long mantissa = bits & 0xFFFFFFFFFFFFFL;
That can translated alsmot verbatim, since the same operators exist in python.
negative = ((bits >> 63) & 1) # Integer 1 for negative, 0 for positive
exponent = ((bits >> 52) & 0x7FF) - 1023
mantissa = bits & 0xFFFFFFFFFFFFF
3.- Pack again the number into 48 bits
long raw = (negative ? 1 : 0);
raw = (raw << 39) | (mantissa >> 13);
raw = (raw << 8) | ((exponent + 129) & 0xFF);
Same as number 2,
raw = negative
raw = (raw << 39) | (mantissa >> 13)
raw = (raw << 8) | ((exponent + 129) & 0xFF)
4.- Convert the number to a string of 6 bytes
for (int k = 0; k < 6; k++)
{
r[k] = (byte)(raw & 0xFF);
raw >>= 8;
}
Use struck as in step 1,
data = struck.pack('Q', raw)
data = data[:6] # Drop the 16 MSB
NOTE: I'm dropping the last 2 chars because struct uses the processors' endianness by default (x86 in my case), but that can be controled with the first argument.
TIP: If you want to convert a number to a binary representation you could use '{:064b}'.format(number). This will pad it with zeros to fill 64 chars.

Print number of ways for non-consecutive one's

Given a positive integer N, print all integers between 1 and 2^N such that there is no consecutive 1’s in its Binary representation.
I have below code but it is printing duplicate sometimes. Is it possible to print without duplicates?
#include <stdio.h>
int a[100];
void foo(int i, int size)
{
if (i >= size) {
int i;
for (i=0;i<size;i++)
printf("%d\n", a[i]);
printf("----\n");
return;
}
if (a[i-1] == 1 || a[i-1] == 0)
a[i] = 0;
foo(i+1, size);
if (a[i-1] == 0)
a[i] = 1;
foo(i+1, size);
}
int main(void) {
int i = 0;
int size = 5;
a[i] = 1;
foo(1, size);
return 0;
}
I have this http://ideone.com/cT4Hco python program which uses hash maps to print the elements but I think we can do this without hashmaps also.

Couple of notes:
you shouldn't start the backtracking from index 1. Instead, start from 0 since your numbers would be in the range [0, n-1] in array a
you shouldn't initialize a[0] to 1 since a[0] = 0 is also a valid case.
if (a[i-1] == 1 || a[i-1] == 0) is redundant
Code:
#include <stdio.h>
int a[100];
void foo(int i, int size)
{
if (i >= size) {
int i;
for (i=0;i<size;i++)
printf("%d ", a[i]);
printf("\n----\n");
return;
}
a[i] = 0;
foo(i+1, size);
if ( i == 0 || a[i-1] == 0) {
a[i] = 1;
foo(i+1, size);
}
}
int main(void) {
int i = 0;
int size = 5;
foo(0, size);
return 0;
}
You might also want to filter the solution 0 0 0 ... 0 during the printing since you need only the numbers from 1 to 2^n. If 2^n is included you should also print it. The backtracking considers the numbers 0, ...., 2^n-1

Convert this CRC32 algorithm to Python 3.3

I need to convert this CRC32 algorithm to python (using 3.3), but I am a python noob. I tried the built in binascii.crc32(), but the CRC was incorrect. Apparently, STMicro does the CRC32 a bit different. I found an algorithm that works, now I just need it to be in python.
//****************************************************************************
DWORD Crc32Fast(DWORD Crc, DWORD Data)
{
static const DWORD CrcTable[16] = { // Nibble lookup table for 0x04C11DB7 polynomial
0x00000000,0x04C11DB7,0x09823B6E,0x0D4326D9,0x130476DC,0x17C56B6B,0x1A864DB2,0x1E475005,
0x2608EDB8,0x22C9F00F,0x2F8AD6D6,0x2B4BCB61,0x350C9B64,0x31CD86D3,0x3C8EA00A,0x384FBDBD };
Crc = Crc ^ Data; // Apply all 32-bits
// Process 32-bits, 4 at a time, or 8 rounds
Crc = (Crc << 4) ^ CrcTable[Crc >> 28]; // Assumes 32-bit reg, masking index to 4-bits
Crc = (Crc << 4) ^ CrcTable[Crc >> 28]; // 0x04C11DB7 Polynomial used in STM32
Crc = (Crc << 4) ^ CrcTable[Crc >> 28];
Crc = (Crc << 4) ^ CrcTable[Crc >> 28];
Crc = (Crc << 4) ^ CrcTable[Crc >> 28];
Crc = (Crc << 4) ^ CrcTable[Crc >> 28];
Crc = (Crc << 4) ^ CrcTable[Crc >> 28];
Crc = (Crc << 4) ^ CrcTable[Crc >> 28];
return(Crc);
}
//****************************************************************************
DWORD Crc32FastBlock(DWORD Crc, DWORD Size, DWORD *Buffer) // 32-bit units
{
while(Size--)
Crc = Crc32Fast(Crc, *Buffer++);
return(Crc);
}

If my understanding is not mistaken, this should be the code that you want:
CRC_TABLE = (0x00000000, 0x04C11DB7, 0x09823B6E, 0x0D4326D9,
0x130476DC, 0x17C56B6B, 0x1A864DB2, 0x1E475005,
0x2608EDB8, 0x22C9F00F, 0x2F8AD6D6, 0x2B4BCB61,
0x350C9B64, 0x31CD86D3, 0x3C8EA00A, 0x384FBDBD)
def dword(value):
return value & 0xFFFFFFFF
def crc32_fast(crc, data):
crc, data = dword(crc), dword(data)
crc ^= data
for _ in range(8):
crc = dword(crc << 4) ^ CRC_TABLE[crc >> 28]
return crc
def crc32_fast_block(crc, buffer):
for data in buffer:
crc = crc32_fast(crc, data)
return crc
def crc32_fast_bytes(crc, bytes_data, byteorder='big'):
if len(bytes_data) & 3:
raise ValueError('bytes_data length must be multiple of four')
for index in range(0, len(bytes_data), 4):
data = int.from_bytes(bytes_data[index:index+4], byteorder)
crc = crc32_fast(crc, data)
return crc
The function crc32_fast_block expects an initial crc value and an iterable of numbers to run the algorithm on. crc32_fast_bytes is almost the same but expects a bytes value with a length being a multiple of four.