使用模板函数访问 POD 数据类型和字符串的原始字节

Question

我正在尝试制作一个函数模板，让我可以处理 POD 数据类型 和 字符串的原始字节。

我想出了这个有点幼稚的方法（我没有经常使用模板，而且我能够编写它们仅用于非常简单的情况）：

#include <cstdio>

template<typename T>
void PrintBytes(T object)
{
  unsigned char* p = reinterpret_cast<unsigned char*>(&object);
  for (int i = 0; i < sizeof object; i++)
  {
    printf("%02x ", *p++);
  }

  printf("\n");
}

int main()
{
  int a = 10;
  PrintBytes(a);    // works as expected      
  short b = 12;
  PrintBytes(b);    // works as expected    
  const char* foo = "ABC";
  PrintBytes(foo);  // works as expected, but doesn't do what I want
  const char bar[] = "ABC";
  PrintBytes(bar);  // works as expected, but doesn't do what I want
}

输出

0a 00 00 00
0c 00
a4 ac 5b 8c f6 7f 00 00
a4 fb 7f 07 b7 00 00 00

最后两种情况的期望输出

41 42 43
41 42 43

显然，前两种情况有效，而后两种情况无效，因为foo 和bar 已经是指针，而unsigned char* p = reinterpret_cast<unsigned char*>(&object) 应该是unsigned char* p = object 而sizeof object 应该是strlen(object)。

如何使用 C++ 模板来做到这一点？我可以使用 c++17。

Answer 1

你可以使用C++17的if constexpr根据T的类型来执行对应的地址和大小的方法，大概是这样：

#include <cstdio>
#include <cstring>
#include <utility>
#include <type_traits>

template<typename T>
void PrintBytes(T object)
{
  auto [p, size] = [&] {
    if constexpr (std::is_same_v<T, const char*>)
      return std::pair{object, strlen(object)};
    else
      return std::pair{reinterpret_cast<unsigned char*>(&object), sizeof(object)};
  }();
  for (int i = 0; i < size; i++)
  {
    printf("%02x ", *p++);
  }
  printf("\n");
}

Demo.

Answer 2

让我们扩展编译器在使用字符数组或指针调用 PrintBytes 时的作用。

你打电话：

const char* foo = "ABC";
PrintBytes(foo);  // works as expected, but doesn't do what I want

然后编译器必须为给定的参数实例化函数 PrintBytes，一个 const char*，它给出：

void PrintBytes(const char* object)
{
  unsigned char* p = reinterpret_cast<unsigned char*>(&object);
  for (int i = 0; i < sizeof (object); i++)  // sizeof(const char*) is 8, for a 64 biut app.
  {
    printf("%02x ", *p++); // this will print the value of object, NOT what it points to.
  }

  printf("\n");
}

要强制 PrintBytes(const char*) 打印字符串的 cntexnts，而不是其地址，您应该为字符串创建特定类型的覆盖，如下所示：

void PrintBytes(const char* sz)
{
  // you may want to check for sz itself being NULL before proceding
  if (sz)
  {
    do
    {
      printf("%02x ", *sz); // this will print the value of object, NOT its address
    } while (*sz++ != 0);
  }

  printf("\n");
}

Answer 3

我终于想到了这个：

#include <cstdio>
#include <cstring>

void PrintBytes(const unsigned char *p, size_t len)
{
  for (int i = 0; i < len; i++)
  {
    printf("%02x ", *p++);
  }

  printf("\n");
}

template<typename T>
void PrintBytes(T object)
{
  const unsigned char* p = reinterpret_cast<const unsigned char*>(&object);
  PrintBytes(p, sizeof object);
}

template<>
void PrintBytes(const char *object)
{
  auto len = strlen(object);
  const unsigned char* p = reinterpret_cast<const unsigned char*>(object);
  PrintBytes(p, len);
}

template<>
void PrintBytes(char* object)
{
  auto len = strlen(object);
  const char* p = reinterpret_cast<const char*>(object);
  PrintBytes(p);
}


int main()
{
  int a = 10;
  PrintBytes(a);   // works as expected
  
  short b = 12;
  PrintBytes(b);   // works as expected

  float f = 1.2f;  
  PrintBytes(f);   // works as expected

  const char* foo = "ABC";   // works as expected
  PrintBytes(foo);

  char bar[] = "ABC";  // works as expected
  PrintBytes(bar);
}

Answer 4

#include <cstdio>

template<typename T>
void PrintBytes(T& object)
{
  const unsigned char* p = reinterpret_cast<const unsigned char*>(&object);
  for (int i = 0; i < sizeof object; i++)
  {
    printf("%02x ", p[i]);
  }

  printf("\n");
}

int main()
{
  int a = 10;
  PrintBytes(a);    // works as expected      
  short b = 12;
  PrintBytes(b);    // works as expected

  const char* foo = "ABC";
  PrintBytes(foo);
  // Print the content of the pointer foo, which is an address;
  // I think this work as expected as the "object" is just a pointer;
  // if you want to print the content of the data it points to instead,
  // you need another implementation logic of this "PrintBytes",
  // which would be an function overload, like some other answers show.

  const char bar[] = "ABC";
  PrintBytes(bar);  // print "41 42 43 00"
}