You can use grep anyway to search through the file – it does not really care if the input file is really text or not. From ‘man grep’: -a, –text Process a binary file as if it were text; this is equivalent to the –binary-files=text option. –binary-files=TYPE If the first few bytes of a file … Read more
Of course. It’s more commonly called machine code. It’s basically assembly language without the mnemonic devices. Someone who knows assembly very well could program in machine code with additional effort, referring to opcode listings (e.g. x86) as needed. Would I do it? No. Even assembly is only useful in rare circumstances, and there’s no reason (beside demonstrating your skills) … Read more
If you want to transform a number into another number (not number to string of characters), and you can do with a small range (0 to 1023 for implementations with 32-bit integers), you don’t need to add char* to the solution HalosGhost suggested to compact the code into a single line
Assuming you are referring to Offset binary: https://en.wikipedia.org/wiki/Offset_binary, of which the most famous example would be Excess-3: https://en.wikipedia.org/wiki/Excess-3, then the solution would be: a) 77 + 127 mod 256 = 204 mod 256 = 204 = 11001100 b) -42 + 127 mod 256 = 85 mod 256 = 85 = 01010101 etc…
How can I perform a conversion of a binary string to the corresponding hex value in Python? I have 0000 0100 1000 1101 and I want to get 048D I’m using Python 2.6.
The easiest way is probably to create an std::bitset representing the value, then stream that to cout.
Just to explain the parts of the formatting string: {} places a variable into a string 0 takes the variable at argument position 0 : adds formatting options for this variable (otherwise it would represent decimal 6) 08 formats the number to eight digits zero-padded on the left b converts the number to its binary … Read more
That would be -8 to +7
You would only need 5 bits because you are counting to 26 (if we take only upper or lowercase letters). 5 bits will count up to 31, so you’ve actually got more space than you need. You can’t use 4 because that only counts to 15. If you want both upper and lowercase then 6 … Read more
According to K&R 2nd edition the results are implementation-dependent for right shifts of signed values. Wikipedia says that C/C++ ‘usually’ implements an arithmetic shift on signed values. Basically you need to either test your compiler or not rely on it. My VS2008 help for the current MS C++ compiler says that their compiler does an arithmetic shift.