c++ - constexpr 函数可以返回局部对象的指针吗？

Question

constexpr 函数定义为 (c++14)

A constexpr function must satisfy the following requirements:

• it must not be virtual
• its return type must be LiteralType each of its parameters must be LiteralType
• there exists at least one set of argument values such that an invocation of the function could be an evaluated subexpression of a core constant expression (for constructors, use in a constant initializer is sufficient) (since C++14). No diagnostic is required for a violation of this bullet.

the function body must be either deleted or defaulted or contain any statements except:

• an asm declaration
• a goto statement
• a statement with a label other than case and default
• a try-block
• a definition of a variable of non-literal type
• a definition of a variable of static or thread storage duration
• a definition of a variable for which no initialization is performed.

现在下面的func1满足要求编译

constexpr int * func1 (int a)
{
  int b = 4;
  return &b;
}
int main()
{
    constexpr int * a = func1(3);
    int arr[*a];  
    std::cout << a << std::endl;
}

现在我的问题是 func1 为什么是 constexpr。它如何在编译时知道局部变量的地址？

我使用的是 gcc 6.4.0

Answer 1

How does it know address of local variable at compile time?

不是。

你引述中的第三个要点是永不满足:

There exists at least one set of argument values such that an invocation of the function could be an evaluated subexpression of a core constant expression (for constructors, use in a constant initializer is sufficient) (since C++14). No diagnostic is required for a violation of this bullet.

编译器只是不提示它，因为它不是必需的，直到你通过尝试在需要正确的 constexpr 函数的东西中使用 func1 来让它提示，例如:

std::array<int, func(3)> d;

这不会编译，您的编译器会告诉您原因。