当前位置：文江博客话题详情

如何将向量转换为到整数？

发布于 2024-09-29 16:26:50 字数 119 浏览 1 评论 0原文

我有 vector 归档了二进制数据。比方说，我需要从向量（2 个字节）中获取 2 个项目并将其转换为整数。不使用 C 风格如何做到这一点？

原文

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

森末i 2024-10-06 16:26:51

如果你不想关心大/小端，你可以使用：

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]));

If you don't want to care about big/little endian, you can use:

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]));

回复收藏 0 原文

若水般的淡然安静女子 2024-10-06 16:26:50

请使用移位运算符/按位运算。

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

这里提出的所有基于强制转换/联合的解决方案都是 AFAIK 未定义的行为，并且可能在利用严格别名（例如 GCC）。

Please use the shift operator / bit-wise operations.

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

All the solutions proposed here that are based on casting/unions are AFAIK undefined behavior, and may fail on compilers that take advantage of strict aliasing (e.g. GCC).

回复收藏 0 原文

旧时光的容颜 2024-10-06 16:26:50

你可以这样做：

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

You may do:

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

回复收藏 0 原文

谁对谁错谁最难过 2024-10-06 16:26:50

v[0]*0x100+v[1]

回复收藏 0 原文

泪痕残 2024-10-06 16:26:50

那么，另一种方法是包装对 memcpy 的调用：

#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。

本例中没有尺寸检查。我们假设在每次调用 extract 之前 v 实际上有足够的元素。

祝你好运！

Well, one other way to do it is to wrap a call to 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;
}

The extract function template also works with double, long int, float and so on.

There are no size checks in this example. We assume v actually has enough elements before each call to extract.

Good luck!

回复收藏 0 原文