c++ - 关于 C++ 中常量的隐式转换

Question

#include <iostream>
int foo(const char* keke) {
  std::cout << keke;
  return 0;
}
int main()
{
  char* keke = new char(10);
  char* const haha = keke;
  return foo(haha);
}

为什么编译上述代码时没有任何错误/警告？

haha的类型是char* const，而foo只接收const char*类型的参数. char* const 可以隐式转换为 const char* 吗？

Answer 1

是的。它叫做qualification conversions (隐式转换之一):

(强调我的)

A prvalue of type pointer to cv-qualified type T can be converted to a prvalue pointer to a more cv-qualified same type T (in other words, constness and volatility can be added).

"More" cv-qualified means that

a pointer to unqualified type can be converted to a pointer to const;
...

这意味着char*可以隐式转换为const char*。

指针本身的

const 限定符在这里无关紧要，参数 keke 本身被声明为按值传递，从 haha(即 const 指针；char* const)。