C ++ 11和C ++ 14中的constexpr（与const关键字没有区别） [英] constexpr in C++11 and C++14 (not a difference to const keyword)

问题描述

在Mike Isaacson的探索C ++ 17及更高版本中的演讲中（ https ：//youtu.be/-ctgSbEfRxU？t = 2907 ）关于写作存在疑问：

In the "Exploring C++17 and beyond" presentation by Mike Isaacson at one point (https://youtu.be/-ctgSbEfRxU?t=2907) there is question about writing:

const constexpr ....

vs单个常量。 Mike表示，在C ++ 11中constexpr隐含const，而在C ++ 14中则不隐含。
是真的吗？
我试图找到证明，但是我做不到。

vs single const. Mike said that in C++11 constexpr implies const and in C++14 it does not. Is it true? I've tried to find prove for that but I couldn't.

我并不是在问const和constexpr之间的区别（因为很多其他问题），但是关于constexpr在两个版本的C ++标准中的区别。

I'm not asking about difference between const and constexpr (as a lot of other questions) but about difference in constexpr in two versions of C++ standard.

推荐答案

请考虑以下这段代码：

struct T { 
  int x;

  constexpr void set(int y) {
    x = y;
  }
};

constexpr T t = { 42 };

int main() {
  t.set(0);
  return 0;
}

它应该在C ++ 14中编译，但在C +中给出错误+11，就像成员函数 set 被声明为 const 一样。这是因为 constexpr 在C ++ 11中隐含了 const ，而在该标准的后续版本中不再如此。

It should compile in C++14, but give an error in C++11 as if the member function set was declared const. This is because constexpr implies const in C++11, while this is not true anymore in subsequent versions of the standard.

constexpr 关键字保证对象不会在运行时动态初始化。对于函数，它保证可以在编译时评估函数。在C ++ 11中，这与被存储在只读存储器中混为一谈，但是很快就将其识别为错误。

The constexpr keyword guarantees that the object will not be dynamically initialized at runtime. In case of functions, it guarantees that the function can be evaluated at compile-time. In C++11, this was confounded with "being stored in read-only memory", but this was soon recognized as an error.

此外，在C ++ 14中， constexpr 函数和成员的功能更加强大。例如，他们可以修改非 const 对象，因此它们不能隐式地成为 const 。

In C++14, moreover, constexpr functions and members are much more capable. They can modify non-const objects, for instance, so they cannot be implicitly const.

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