How to convert a Binary String to a base 10 integer in Java
You need to specify the radix. There’s an overload of Integer#parseInt() which allows you to. int foo = Integer.parseInt(“1001”, 2);
You need to specify the radix. There’s an overload of Integer#parseInt() which allows you to. int foo = Integer.parseInt(“1001”, 2);
For solving the left-side trailing zero problem: my_hexdata = “1a” scale = 16 ## equals to hexadecimal num_of_bits = 8 bin(int(my_hexdata, scale))[2:].zfill(num_of_bits) It will give 00011010 instead of the trimmed version.
I think this is a suboptimal solution, but you could do String.format(“%16s”, Integer.toBinaryString(1)).replace(‘ ‘, ‘0’)
Read the binary file content like this: with open(fileName, mode=”rb”) as file: # b is important -> binary fileContent = file.read() then “unpack” binary data using struct.unpack: The start bytes: struct.unpack(“iiiii”, fileContent[:20]) The body: ignore the heading bytes and the trailing byte (= 24); The remaining part forms the body, to know the number of … Read more
I recommend you to do it exactly as you have shown, since it is the most straight forward one. Initialize to -1 which will work always, independent of the actual sign representation, while ~ will sometimes have surprising behavior because you will have to have the right operand type. Only then you will get the … Read more
Something like this? >>> st = “hello world” >>> ‘ ‘.join(format(ord(x), ‘b’) for x in st) ‘1101000 1100101 1101100 1101100 1101111 100000 1110111 1101111 1110010 1101100 1100100’ #using `bytearray` >>> ‘ ‘.join(format(x, ‘b’) for x in bytearray(st, ‘utf-8’)) ‘1101000 1100101 1101100 1101100 1101111 100000 1110111 1101111 1110010 1101100 1100100’
For arbitrary-length integers, bin(n).count(“1”) is the fastest I could find in pure Python. I tried adapting Óscar’s and Adam’s solutions to process the integer in 64-bit and 32-bit chunks, respectively. Both were at least ten times slower than bin(n).count(“1”) (the 32-bit version took about half again as much time). On the other hand, gmpy popcount() … Read more
When shifting left, there is no difference between arithmetic and logical shift. When shifting right, the type of shift depends on the type of the value being shifted. (As background for those readers unfamiliar with the difference, a “logical” right shift by 1 bit shifts all the bits to the right and fills in the … Read more
If you want to find out only whether or not the files are identical, you can use the Windows fc command in binary mode: fc.exe /b file1 file2 For details, see the reference for fc
all numbers are stored in binary. if you want a textual representation of a given number in binary, use bin(i) >>> bin(10) ‘0b1010’ >>> 0b1010 10