Does anyone have a code snippet or a class that will take a long long and turn it into a 16 byte Hex string?
I'm looking to turn data like this
long long decimalRepresentation = 1719886131591410351;
and turn it into this
//Base 16 Hex Output: 17DE435307A07300
The %x operator doesn't want to work for me
NSLog(#"Hex: %x",decimalRepresentation);
//console : "Hex: 7a072af"
As you can see that's not even close. Any help is truly appreciated!
%x prints an unsigned integer in hexadecimal representation and sizeof(long long) != sizeof(unsigned). See e.g. "Data Type Size and Alignment" in the 64bit transitioning guide.
Use the ll specifier (thats two lower-case L) to get the desired output:
NSLog(#"%llx", myLongLong);
Related
I want to get the UTF-16 code unit at a given index in ABAP.
Same can be done in JavaScript with charCodeAt().
For example "d".charCodeAt(); will give back 100.
Is there a similar functionality in ABAP?
This can be done with class CL_ABAP_CONV_OUT_CE
DATA(lo_converter) = cl_abap_conv_out_ce=>create( encoding = '4103' ). "Litte Endian
TRY.
CALL METHOD lo_converter->convert
EXPORTING
data = 'a'
n = 1
IMPORTING
buffer = DATA(lv_buffer). "lv_buffer will 0061
CATCH ...
ENDTRY.
Codepage 4102 is for UTF-16 Big endian.
It is possible to encode not just a single character, but a string as well:
EXPORTING
data = 'abc'
n = 3
"n" always stands for the length of the string you want to be encoded. It could be less, than the actual length of the string.
When you say you "want to get the UTF-16 code unit",
either you mean the Unicode code point, e.g. the character d is always U+0064 (official "name" of Unicode character, the two bytes 0x0064 being the hexadecimal representation of decimal 100),
or you mean you want to encode d to UTF-16 little endian (SAP code page 4103) or big endian (SAP code page 4102) which gives respectively 2 bytes 0x4400 or 2 bytes 0x0044.
For the second case, see József answer.
For the first case, you may get it using the method UCCP (UniCode Code Point) or UCCPI (UniCode Code Point Integer) of class CL_ABAP_CONV_OUT_CE:
DATA: l_unicode_point_hex TYPE x LENGTH 2,
l_unicode_point_int TYPE i.
l_unicode_point_hex = cl_abap_conv_out_ce=>UCCP( 'd' ).
ASSERT l_unicode_point_hex = '0064'.
l_unicode_point_int = cl_abap_conv_out_ce=>UCCPI( 'd' ).
ASSERT l_unicode_point_int = 100.
EDIT: Note that the two methods return always the same values whatever the SAP system code page is (4102, 4103 or whatever).
I understand that you have a hex string and perform SHA256 on it twice and then byte-swap the final hex string. The goal of this code is to find a Merkle Root by concatenating two transactions. I would like to understand what's going on in the background a bit more. What exactly are you decoding and encoding?
import hashlib
transaction_hex = "93a05cac6ae03dd55172534c53be0738a50257bb3be69fff2c7595d677ad53666e344634584d07b8d8bc017680f342bc6aad523da31bc2b19e1ec0921078e872"
transaction_bin = transaction_hex.decode('hex')
hash = hashlib.sha256(hashlib.sha256(transaction_bin).digest()).digest()
hash.encode('hex_codec')
'38805219c8ac7e9a96416d706dc1d8f638b12f46b94dfd1362b5d16cf62e68ff'
hash[::-1].encode('hex_codec')
'ff682ef66cd1b56213fd4db9462fb138f6d8c16d706d41969a7eacc819528038'
header_hex is a regular string of lower case ASCII characters and the decode() method with 'hex' argument changes it to a (binary) string (or bytes object in Python 3) with bytes 0x93 0xa0 etc. In C it would be an array of unsigned char of length 64 in this case.
This array/byte string of length 64 is then hashed with SHA256 and its result (another binary string of size 32) is again hashed. So hash is a string of length 32, or a bytes object of that length in Python 3. Then encode('hex_codec') is a synomym for encode('hex') (in Python 2); in Python 3, it replaces it (so maybe this code is meant to work in both versions). It outputs an ASCII (lower hex) string again that replaces each raw byte (which is just a small integer) with a two character string that is its hexadecimal representation. So the final bit reverses the double hash and outputs it as hexadecimal, to a form which I usually call "lowercase hex ASCII".
Could someone explain why I can not use int() to convert an integer number represented in string-scientific notation into a python int?
For example this does not work:
print int('1e1')
But this does:
print int(float('1e1'))
print int(1e1) # Works
Why does int not recognise the string as an integer? Surely its as simple as checking the sign of the exponent?
Behind the scenes a scientific number notation is always represented as a float internally. The reason is the varying number range as an integer only maps to a fixed value range, let's say 2^32 values. The scientific representation is similar to the floating representation with significant and exponent. Further details you can lookup in https://en.wikipedia.org/wiki/Floating_point.
You cannot cast a scientific number representation as string to integer directly.
print int(1e1) # Works
Works because 1e1 as a number is already a float.
>>> type(1e1)
<type 'float'>
Back to your question: We want to get an integer from float or scientific string. Details: https://docs.python.org/2/reference/lexical_analysis.html#integers
>>> int("13.37")
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: invalid literal for int() with base 10: '13.37'
For float or scientific representations you have to use the intermediate step over float.
Very Simple Solution
print(int(float(1e1)))
Steps:-
1- First you convert Scientific value to float.
2- Convert that float value to int .
3- Great you are able to get finally int data type.
Enjoy.
Because in Python (at least in 2.x since I do not use Python 3.x), int() behaves differently on strings and numeric values. If you input a string, then python will try to parse it to base 10 int
int ("077")
>> 77
But if you input a valid numeric value, then python will interpret it according to its base and type and convert it to base 10 int. then python will first interperet 077 as base 8 and convert it to base 10 then int() will jsut display it.
int (077) # Leading 0 defines a base 8 number.
>> 63
077
>> 63
So, int('1e1') will try to parse 1e1 as a base 10 string and will throw ValueError. But 1e1 is a numeric value (mathematical expression):
1e1
>> 10.0
So int will handle it as a numeric value and handle it as though, converting it to float(10.0) and then parse it to int. So Python will first interpret 1e1 since it was a numric value and evaluate 10.0 and int() will convert it to integer.
So calling int() with a string value, you must be sure that string is a valid base 10 integer value.
int(float(1e+001)) will work.
Whereas like what others had mention 1e1 is already a float.
I am looking for a way to convert an array of 16-bit unsigned integer into ASCII char array. I am using char to do the conversion
D=[65 65 65 65];
char(D)
which will show 4 'A'. However, since each number in D is 16-bit, I expect it to convert each number to 2 chars. For example, if I have
D=[16707]
char(D)
I expect it gives me two chars 'A' and 'C'. But char always return 1 character. Is that anyway to force char to convert like the way I stated? Thanks.
For this, you need to write your own function.
You can use char() to convert most significant byte and least significant byte separately.
k = 16707;
first = char(bitand(bitshift(k, -8), 255));
second = char(bitand(k, 255));
Have a look at
http://www.mathworks.com/help/matlab/ref/char.html
It cleatly states that the char function is valid only for 8 bit numbers. you can convert each part of cell of the array with this and contact the results for each two cells.
Use typecast to convert each uint16 to two uint8, and then apply char. Make sure that the input to typecastr is really of type uint16.
If you need to reverse char order, use swapbytes on the uint16 vector.
>> D = [16707 16708];
>> char(typecast(uint16(D),'uint8'))
ans =
CADA
>> char(typecast(swapbytes(uint16(D)),'uint8'))
ans =
ACAD
I am working on a small task which requires some base64 encoding. I am trying to do it in head but getting lost .
I have a 13 digit number in java long format say: 1294705313608 , 1294705313594 , 1294705313573
I do some processing with it, bascially I take this number append it with stuff put it in a byte array and then convert it to base64 using:
String b64String = new sun.misc.BASE64Encoder().encodeBuffer(bArray);
Now , I know that for my original number, the first 3 digits would never change. So 129 is constant in above numbers. I want to find out how many chars corresponding to those digits would not change in the resultant base64 string.
Code to serialize long to the byte array. I ignore the first 2 bytes since they are always 0:
bArray[0] = (byte) (time >>> 40);
bArray[1] = (byte) (time >>> 32);
bArray[2] = (byte) (time >>> 24);
bArray[3] = (byte) (time >>> 16);
bArray[4] = (byte) (time >>> 8);
bArray[5] = (byte) (time >>> 0);
Thanks.
Notes:
I know that base64 would take 6 bits and make one character out of it. So if first 3 digits do not change in long how many chars would not change in base64.
This in NOT a HW assignment, but I am not very familiar with encoding.
1290000000000 is 10010110001011001111111011110010000000000 in binary.
1299999999999 is 10010111010101110000010011100011111111111 in binary.
Both are 41 bits long, and they differ after the first 7 bits. Your shift places bits 41-48 in the first byte, which will always be 00000001. The following byte will always be 00101110, 00101101, or 00101110. So you've got the leading 14 bits in common across all your possible array values, which (at 6 bits per encoded base64 char) means 2 characters in common in the encoded string.
Appears you're on the right track. I think what you want to do is convert a long to a byte array, then convert the byte array to Base64.
How do I convert Long to byte[] and back in java shows you how to convert it to bytes.