I can’t well understand your question, but here I show you how to concatenate bits. byte bits2 = 0b_000000_10; byte bits6 = 0b_00_101100; ushort bits14 = 0b_00_10110010001110; uint concatenated = ((uint)bits2 << 20) + ((uint)bits6 << 14) + (uint)bits14; uint test = 0b_00_10_101100_10110010001110; if (concatenated == test‬) { Console.WriteLine(“Ok”); } 4 solved How to perform … Read more
If you like to create a piece of code having that many different variables with a different name, I suggest you start writing your own code generator. The code below should be a good start. #include <sstream> #include <iostream> #include <string> #include <vector> namespace { template <typename Function> void bitExecutor(int nrOfBits, std::vector<bool> &bits, const Function … Read more
To set bit 0 to one, use int_var | 1 To reset bit 0 to zero, use int_var & ~1 The operation in your example doesn’t seem consistent: 10000011 10000010 & ~1 00000001 xor 00001011 00001011 nop 00000000 xor 01110011 01110010 & ~1 00000001 xor 11101100 11101101 & ~1 00000001 xor 11110101 11110100 & ~1 … Read more
The type of 0.5 in C++ is double. If you want a float you need to say 0.5f . This is part of the definition of C and C++. Implied in your question is some desire for automatic type selection based on the number of digits in the value, e.g. we could use float for … Read more
The idea is “recursively defining the problem down to 2-bit values”: 00, 01, 10, 11. They’re not recursive macros, but does represent the technique of recursive decomposition of the problem. The arrangement of macros as a cascade attacks the problem of generating the table of 2^8 values by solving the problem for 2-bits (generating 4 … Read more
Not very hard, all you need is a loop and a cast to make sure your numbers are unsigned for (int i = 0; i < 16; ++i) printf(“%02X”, (unsigned char)data[i]); 4 solved Extracting hex numbers from char array [closed]
The splitter() function here does the job you ask for. It takes quite a lot of arguments, unfortunately. There’s the value to be split (value), the size of the chunk at the least significant end of the value (p1), the size of the middle chunk (p2), and then pointers to the high, medium and low … Read more
What does this code do? How would have you named such a function? I would have named it so /// <summary> /// Convert an Int32 value into its binary string representation. /// </summary> /// <param name=”value”>The Int32 to convert.</param> /// <returns>The binary string representation for an Int32 value.</returns> public String ConvertInt32ToBinaryString(Int32 value) { String pout … Read more
These days, the number of bits per character is not so simple but there are most definitely 8 bits per byte. The number of bits/bytes per character will vary depending on your system and the character set. Original ASCII was 7bits per character. That was extended to 8bits per character and stayed there for a … Read more
There are versions of g++ (and versions of Visual C++ as well for that matter) for which on some platforms sizeof(bool) is not equal to 1. So no, you can’t assume that it will be the same on g++ and Visual C++. You can’t even assume that it’s the same on different versions of the … Read more
Answer 1) With the 4 bits we can write 16 different numbers. As we have 4 different position of bits let’s say ABCD where A,B,C,D are representing 1 bit. Each position A,B,C,D has two possible input 0 or 1 so each position is having 2 possible inputs. So for 4 positions total different outputs = … Read more