C宏乘法发生了什么

6 回答

• 5

  你应该总是在宏定义周围加上括号：

  #define LOCK_BYTES (LOCK_NAME_BYTES + LOCK_TYPE_BYTES + PID_BYTES + TID_BYTES)
  

  否则，代码扩展为：

  cout << 0 * LOCK_NAME_BYTES + LOCK_TYPE_BYTES + PID_BYTES + TID_BYTES
  

  输出 LOCK_TYPE_BYTES + PID_BYTES + TID_BYTES 的值 .

  更好的是，除非你真的需要，否则不要使用宏 . 这些更好地表示为常量变量 .

  回复于 2024-05-03T00:04:02+08:00
• 2

  std::cout << "LOCK_BYTES * 0: " << 0 * LOCK_BYTES << std::endl;
  

  扩展到

  std::cout << "LOCK_BYTES * 0: " << 0 * LOCK_NAME_BYTES + LOCK_TYPE_BYTES + PID_BYTES + TID_BYTES << std::endl;
  

  反过来扩展到

  std::cout << "LOCK_BYTES * 0: " << 0 * 256 + 1 + 4 + 4 << std::endl;
  

  并为优先规则增加了一些parens：

  std::cout << "LOCK_BYTES * 0: " << ((((0 * 256) + 1) + 4) + 4) << std::endl;
  

  评估为

  std::cout << "LOCK_BYTES * 0: " << 9 << std::endl;
  

  将您的代码更改为

  std::cout << "LOCK_BYTES * 0: " << 0 * (LOCK_BYTES) << std::endl;
  

  甚至更好，使用 const unsigned int 值：

  const unsigned int NAME_BYTES = 256;
  const unsigned int VERSION_BYTES = 256;
  const unsigned int SIZE_BYTES = 32;
  const unsigned int USED_LOCK_COUNT_BYTES = 32;
  const unsigned int LOCK_NAME_BYTES = 256;
  const unsigned int LOCK_TYPE_BYTES = 1;
  const unsigned int PID_BYTES = 4;
  const unsigned int TID_BYTES = 256;
  const unsigned int LOCK_BYTES = LOCK_NAME_BYTES + LOCK_TYPE_BYTES + PID_BYTES + TID_BYTES;
  const unsigned int HEADER_BYTES = NAME_BYTES + VERSION_BYTES + SIZE_BYTES + USED_LOCK_COUNT_BYTES;
  

  好哇！突然间，你不再有奇怪的问题了 .

  回复于 2024-05-03T00:04:02+08:00
• 11

  我觉得你算错了一点 . macros 不像变量那样，它们的值在预处理器编译之前插入到它们的位置 . 因此，编译器将看到的是：

  0* LOCK_NAME_BYTES + LOCK_TYPE_BYTES + PID_BYTES + TID_BYTES
  

  这是：

  0 * 256 + 1 + 4 + 4
  

  根据Order of Operations，C的运算符优先级所基于的，乘法首先发生，所以它将等于 9 而不是 0 .

  P.S如果你不是为嵌入式系统或过时的游戏机开发，比如Gameboy Color（注意我的gravatar），我强烈建议你使用 const 关键字而不是 #define 来做这些事情 .

  回复于 2024-05-03T00:04:02+08:00
• 9

  问题是你使用宏 . 您的

  #define LOCK_BYTES LOCK_NAME_BYTES + LOCK_TYPE_BYTES + PID_BYTES + TID_BYTES
  

  不做你的想法 . 它的作用是用 LOCK_NAME_BYTES + LOCK_TYPE_BYTES + PID_BYTES + TID_BYTES 以文本方式替换 LOCK_BYTES 的每一次出现 . 所以

  0 * LOCK_BYTES
  

  扩展到

  0 * LOCK_NAME_BYTES + LOCK_TYPE_BYTES + PID_BYTES + TID_BYTES
  

  这是C. Avoid macros whenever possible. 我们有 const .

  ---

  这对我来说很好用：

  #include <iostream>
  
  const int name_bytes = 256;
  const int version_bytes = 256;
  const int size_bytes = 32;
  const int used_lock_count_bytes = 32;
  const int lock_name_bytes = 256;
  const int lock_type_bytes = 1;
  const int pid_bytes = 4;
  const int tid_bytes = 4;
  const int lock_bytes = lock_name_bytes + lock_type_bytes + pid_bytes + tid_bytes;
  const int header_bytes = name_bytes + version_bytes + size_bytes + used_lock_count_bytes;
  
  int main() {
    std::cout << "lock_bytes: " << lock_bytes << std::endl;
    std::cout << "header_bytes: " << header_bytes << std::endl;
    std::cout << "lock_bytes * 0: " << 0 * lock_bytes << std::endl;
  }
  

  ---

  你有 a good C++ book 要学习吗？你应该 .

  回复于 2024-05-03T00:04:02+08:00
• 6

  std::cout << "LOCK_BYTES * 0: " << 0 * LOCK_BYTES << std::endl;

  扩展到

  std::cout << 0 * LOCK_NAME_BYTES + LOCK_TYPE_BYTES + PID_BYTES + TID_BYTES << std::endl;

  从基本的操作顺序来看，它不等同于 0 * (that whole thing) . 始终将表达式括在括号内的宏定义中以避免此类错误 - 请记住，预处理器会逐字扩展宏（或多或少） .

  回复于 2024-05-03T00:04:02+08:00
• 1

  发布完整性：

  const unsigned NAME_BYTES = 256;
  const unsigned VERSION_BYTES = 256;
  const unsigned SIZE_BYTES = 32;
  const unsigned USED_LOCK_COUNT_BYTES = 32;
  const unsigned LOCK_NAME_BYTES = 256;
  const unsigned LOCK_TYPE_BYTES = 1;
  const unsigned PID_BYTES = 4;
  const unsigned TID_BYTES = 4;
  const unsigned LOCK_BYTES = LOCK_NAME_BYTES + LOCK_TYPE_BYTES + PID_BYTES + TID_BYTES;
  const unsigned HEADER_BYTES = NAME_BYTES + VERSION_BYTES + SIZE_BYTES + USED_LOCK_COUNT_BYTES;
  

  宏扩大了，但是还没有问题 .

  回复于 2024-05-03T00:04:02+08:00