Question

我向vector<unsigned char>提交了二进制数据。我需要从矢量（2个字节）中取出2个项目并将其转换为整数。如何才能用C风格做到这一点？

Answer 1

您可以这样做：

vector<unsigned char> somevector;
// Suppose it is initialized and big enough to hold a uint16_t

int i = *reinterpret_cast<const uint16_t*>(&somevector[0]);
// But you must be sure of the byte order

// or
int i2 = (static_cast<int>(somevector[0]) << 8) | somevector[1];
// But you must be sure of the byte order as well

Answer 2

请使用移位运算符/逐位运算。

int t = (v[0] << 8) | v[1];

此处提出的所有基于转换/联合的解决方案都是AFAIK未定义的行为，并且可能会在利用strict aliasing（例如GCC）的编译器上失败。

Answer 3

v [0] *为0x100 + V [1]

Answer 4

好吧，另一种方法是将调用包装到memcpy：

#include <vector>
using namespace std;

template <typename T>
T extract(const vector<unsigned char> &v, int pos)
{
  T value;
  memcpy(&value, &v[pos], sizeof(T));
  return value;
}

int main()
{
  vector<unsigned char> v;
  //Simulate that we have read a binary file.
  //Add some binary data to v.
  v.push_back(2);
  v.push_back(1);
  //00000001 00000010 == 258

  int a = extract<__int16>(v,0); //a==258
  int b = extract<short>(v,0); //b==258

  //add 2 more to simulate extraction of a 4 byte int.
  v.push_back(0);
  v.push_back(0);
  int c = extract<int>(v,0); //c == 258

  //Get the last two elements.
  int d = extract<short>(v,2); // d==0

  return 0;
}

extract 函数模板也适用于double，long int，float等。

此示例中没有大小检查。我们假设v在每次调用 extract 之前实际上都有足够的元素。

祝你好运！

Answer 5

你是什么意思“不是C风格”？使用按位运算（shift和ors）来实现这一点并不意味着它是“C风格！”

出了什么问题：int t = v[0]; t = (t << 8) | v[1];？

Answer 6

如果您不想关心大/小端，可以使用：

vector<unsigned char> somevector;
// Suppose it is initialized and big enough to hold a uint16_t

int i = ntohs(*reinterpret_cast<const uint16_t*>(&somevector[0]));

如何将vector <unsigned char =“”>转换为int？</unsigned>

6 个答案: