C ++ int到byte数组

Question

I have this method in my java code which returns byte array for given int:

private static byte[] intToBytes(int paramInt)
{
     byte[] arrayOfByte = new byte[4];
     ByteBuffer localByteBuffer = ByteBuffer.allocate(4);
     localByteBuffer.putInt(paramInt);
     for (int i = 0; i < 4; i++)
         arrayOfByte[(3 - i)] = localByteBuffer.array()[i];
     return arrayOfByte;
}

Can someone give me tip how can i convert that method to C++?

Answer 1

You don't need a whole function for this; a simple cast will suffice:

int x;
static_cast<char*>(static_cast<void*>(&x));

Any object in C++ can be reinterpreted as an array of bytes. If you want to actually make a copy of the bytes into a separate array, you can use std::copy :

int x;
char bytes[sizeof x];
std::copy(static_cast<const char*>(static_cast<const void*>(&x)),
          static_cast<const char*>(static_cast<const void*>(&x)) + sizeof x,
          bytes);

Neither of these methods takes byte ordering into account, but since you can reinterpret the int as an array of bytes, it is trivial to perform any necessary modifications yourself.

Answer 2

Using std::vector<unsigned char> :

#include <vector>
using namespace std;

vector<unsigned char> intToBytes(int paramInt)
{
     vector<unsigned char> arrayOfByte(4);
     for (int i = 0; i < 4; i++)
         arrayOfByte[3 - i] = (paramInt >> (i * 8));
     return arrayOfByte;
}

Answer 3

Another useful way of doing it that I use is unions:

union byteint
{
    byte b[sizeof int];
    int i;
};
byteint bi;
bi.i = 1337;
for(int i = 0; i<4;i++)
    destination[i] = bi.b[i];

This will make it so that the byte array and the integer will "overlap"( share the same memory ). this can be done with all kinds of types, as long as the byte array is the same size as the type( else one of the fields will not be influenced by the other ). And having them as one object is also just convenient when you have to switch between integer manipulation and byte manipulation/copying.

Answer 4

Int to byte and vice versa.

unsigned char bytes[4];
unsigned long n = 1024;

bytes[0] = (n >> 24) & 0xFF;
bytes[1] = (n >> 16) & 0xFF;
bytes[2] = (n >> 8) & 0xFF;
bytes[3] = n & 0xFF;

printf("%x %x %x %x\n", bytes[0], bytes[1], bytes[2], bytes[3]);


int num = 0;
for(int i = 0; i < 4; i++)
 {
 num <<= 8;
 num |= bytes[i];
 }


printf("number %d",num);

Answer 5

You can get individual bytes with anding and shifting operations:

byte1 =  nint & 0x000000ff
byte2 = (nint & 0x0000ff00) >> 8
byte3 = (nint & 0x00ff0000) >> 16
byte4 = (nint & 0xff000000) >> 24

Answer 6

std::vector<unsigned char> intToBytes(int value)
{
    std::vector<unsigned char> result;
    result.push_back(value >> 24);
    result.push_back(value >> 16);
    result.push_back(value >>  8);
    result.push_back(value      );
    return result;
}

Answer 7

An int (or any other data type for that matter) is already stored as bytes in memory. So why not just copy the memory directly?

memcpy(arrayOfByte, &x, sizeof x);

A simple elegant one liner that will also work with any other data type.

---

---

If you need the bytes reversed you can use std::reverse

memcpy(arrayOfByte, &x, sizeof x);
std::reverse(arrayOfByte, arrayOfByte + sizeof x);

or better yet, just copy the bytes in reverse to begin with

BYTE* p = (BYTE*) &x;
std::reverse_copy(p, p + sizeof x, arrayOfByte);

---

If you don't want to make a copy of the data at all, and just have its byte representation

BYTE* bytes = (BYTE*) &x;

Answer 8

return ((byte[0]<<24)|(byte[1]<<16)|(byte[2]<<8)|(byte[3]));

：d

Answer 9

I know this question already has answers but I will give my solution to this problem. I am using template function and integer constraint on it.

Here is my solution:

#include <type_traits>
#include <vector>

template <typename T,
          typename std::enable_if<std::is_arithmetic<T>::value>::type* = nullptr>
std::vector<uint8_t> splitValueToBytes(T const& value)
{
    std::vector<uint8_t> bytes;

    for (size_t i = 0; i < sizeof(value); i++)
    {
        uint8_t byte = value >> (i * 8);
        bytes.insert(bytes.begin(), byte);
    }

    return bytes;
}

Answer 10

I hope mine helps

template <typename t_int>
std::array<uint8_t, sizeof (t_int)> int2array(t_int p_value) {
    static const uint8_t _size_of (static_cast<uint8_t>(sizeof (t_int)));
    typedef std::array<uint8_t, _size_of> buffer;
    static const std::array<uint8_t, 8> _shifters = {8*0, 8*1, 8*2, 8*3, 8*4, 8*5, 8*6, 8*7};

    buffer _res;
    for (uint8_t _i=0; _i < _size_of; ++_i) {
        _res[_i] = static_cast<uint8_t>((p_value >> _shifters[_i]));
    }
    return _res;
}