C ++将32位整数复制到字节数组中

Question

Is it possible to copy a 32 bit value into an array of 8 bit characters with a single assignment?

Say I have a byte array (uint8*) with the contents:

01 12 23 45 56 67 89 90

Is it possible to copy into this array (through casts or something) with a single assignment? For example, copy something like 0x555555, so that we end up with:

55 55 55 55 56 67 78 90

Answer 1

*( (unsigned int *)address_of_byte_buffer) = 0x55555555

请注意64位代码下int的大小......您需要在两种体系结构下找到一致32位的数据类型，例如uint32_t 。

Answer 2

You can use reinterpret_cast although you really need to wear steel-toe-capped boots whilst using it.

#include <vector>
#include <algorithm>
#include <iterator>
#include <iostream>

int main()
{
    using std::vector;
    using std::copy;
    using std::back_inserter;
    using std::ostream_iterator;
    using std::cout;

    int a = 0x55555555;

    char* a_begin = reinterpret_cast<char*>(&a);
    char* a_end = a_begin + 4;

    vector<char> chars;

    copy(a_begin, a_end, back_inserter(chars));

    copy(chars.begin(), chars.end(), ostream_iterator<int>(cout, ", "));

    return 1;
}

Output:

85, 85, 85, 85, 

Answer 3

You can use reinterpret_cast to cast anything into anything.

http://msdn.microsoft.com/en-us/library/e0w9f63b(v=vs.80).aspx

Answer 4

You can use something like this:
unsigned long* fake_long = reinterpret_cast<unsigned long*> (char_array); *fake_long = 0x55555555;

But such a solution can work on a big-endian machine. To make it work in a little-endian machine (probably you want), you should convert endianness of your long variable.