Question

让我说我有8个unsigned char，我想转换为unsigned long long。

例如，如果所有char都等于0xFF，则无符号long long将等于0xFFFFFFFFFFFFFFFF。

使用C或C ++最有效的方法是什么？

Answer 1

例如：

unsigned char buffer[8] = { 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF, 0xFF };
unsigned long long target;
memcpy(&target, buffer, sizeof target);

请注意，如果buffer的所有元素都不具有相同的值，则结果将取决于字节顺序（little-endian与big-endian）。

这也假设unsigned long long正好是8个字节。这种情况非常普遍，但并不能保证。（它也不能保证一个字节正好是8位;它可以更多。unsigned char定义为1字节。）

Answer 2

您可以直接分配位

，而不是memcpy

  unsigned char buffer[8] = { 0xff, 0xff, 0xff, 0xff, 0xff, 0xff, 0xff, 0xff };
  unsigned long long l = 0;
  for (int i = 0; i < 8; ++i) {
    l = l | ((unsigned long long)buffer[i] << (8 * i));
  }

我相信这对字节序不敏感。

Answer 3

对不起我最初的愚蠢回答，真的应该更仔细地阅读这个问题。希望这个更有帮助。转换的方式取决于字节数组中长long表示的字节顺序是否与您正在处理的体系结构的字节顺序相同。

C ++：

#include <iostream>
#include <cstring>

using namespace std;

// Assuming ca is at least 8 bytes, the size of long long, interpret the
//    first 8 bytes as long long.
// differentEndian == true means the endianness of the machine is
//    different from the representation in ca.
long long getLongLong( unsigned char * ca, bool differentEndian )
{
   long long retVal;

   if (differentEndian)
   {
       for (int i = 0; i < 4; i++)
       {
           unsigned char _tmpCh = ca[i];
           ca[i] = ca[7-i];
           ca[7-i] = _tmpCh;
       }
   }
   retVal = *reinterpret_cast<unsigned long long *>(ca);

   return retVal;
}

int main()
{
   unsigned char cArray[] = {0xff, 0x1, 0x70, 0x2, 0x61, 0x3, 0x52, 0x4};

   unsigned long long ll = getLongLong( cArray, false );
   cout << "Result for same endian: " << hex << ll << " or " << dec << ll << endl;

   ll = getLongLong( cArray, true );
   cout << "Result for different endian: " << hex << ll << " or " << dec << ll << endl;

   return 0;
}

C：

#include <stdio.h>
#include <string.h>

// Assuming ca is at least 8 bytes, the size of long long, interpret the
//    first 8 bytes as long long.
// differentEndian != 0 means the endianness of the machine is
//    different from the representation in ca.
long long getLongLong( unsigned char * ca, int differentEndian )
{
   long long retVal;

   if (differentEndian)
   {
       int i;
       for (i = 0; i < 4; i++)
       {
           unsigned char _tmpCh = ca[i];
           ca[i] = ca[7-i];
           ca[7-i] = _tmpCh;
       }
   }
   memcpy( &retVal, ca, sizeof(long long));

   return retVal;
}

int main()
{
   unsigned char cArray[] = {0xff, 0x1, 0x70, 0x2, 0x61, 0x3, 0x52, 0x4};

   unsigned long long ll = getLongLong( cArray, 0 );
   printf("Result for same endian: %llx or %llu\n", ll, ll);

   ll = getLongLong( cArray, 1 );
   printf("Result for different endian: %llx or %llu\n", ll, ll);

   return 0;
}

两个版本的输出是：

Result for same endian: 4520361027001ff or 311315039429591551
Result for different endian: ff01700261035204 or 1837509111016818739

如何将unsigned char数组转换为unsigned long long？

3 个答案: