c++ - C 风格的指针查询到 C++ 类型转换

Question

typedef uint8_t RANGE;

int main()
{    
   const  uint16_t  FREQ_RATED_MIN_VALUE = 54;

   RANGE* min;

   min = (uint8_t*)&FREQ_RATED_MIN_VALUE; // C style tyecast

   cout << min << endl; // prints value

   min = reinterpret_cast<uint8_t*>(&FREQ_RATED_MIN_VALUE); //ERROR to be removed in this statement to be written in C++ style typecast

   return 0;
}

需要帮助在注释语句中进行 C++ 样式类型转换..

Answer 1

你的转换丢弃了 const 资格。使用 C++ 风格的转换来摆脱 constness 需要 const_cast。试试这个:

min = reinterpret_cast<uint8_t*>(const_cast<uint16_t*>(&FREQ_RATED_MIN_VALUE));