c++ - 为什么指针类型在 C++ 中不受限制？

Question

我认为指针的最大值是所用指针大小的最大可能值，但显然不是。

#include <iostream>
#include <limits>

int main() {
  std::cout << std::numeric_limits<char *>::is_bounded << std::endl;
  std::cout << (void *)std::numeric_limits<char *>::min() << std::endl;
  std::cout << (void *)std::numeric_limits<char *>::max() << std::endl;
}

我用 g++ 编译并得到以下输出:

0
0
0

标准说如果 is_bounded 不为假，则 min() 和 max() 是有意义的。它还指出，如果该类型可表示的值集是有限的，则 is_bounded 应该为真。

指针不是有限的吗？为什么我的程序中 is_bounded false 和 max() 为零？

Answer 1

指针不是算术类型。

来自 18.3.2.1 类模板 numeric_limits

The numeric_limits class template provides a program with information about various properties of the implementation’s representation of the arithmetic types. Specializations shall be provided for each arithmetic type, both floating point and integer, including bool. The member is_specialized shall be true for all such specializations of numeric_limits. For all members declared static constexpr in the numeric_limits template, specializations shall define these values in such a way that they are usable as constant expressions. Non-arithmetic standard types, such as complex<T> (26.4.2), shall not have specializations.

因此。您应用非算术类型的测试会导致无意义的默认值。