据我所知，下面的函数不是 constexpr，而是用 clang 和 g++ 编译的代码。我错过了什么？

Question

我从 N4140 的 §5.19/2 中得到这个例子：

constexpr int incr(int &n) {
    return ++n;
}

据我所知，这不是 constexpr 函数。但是这段代码是用 clang 和 g++ 编译的。参见 live example。我在这里错过了什么？

Answer 1

As far as I can tell, this is not a constexpr function.

为什么这么说？ §5.19/2 中的示例显示为：

constexpr int g(int k) {
    constexpr int x = incr(k); // error: incr(k) is not a core constant
                               // expression because lifetime of k
                               // began outside the expression incr(k)
    return x;
}

incr(k) 不是 核心常量表达式 并不意味着 incr 不能是 constexpr 函数。

根据 C++14 的 constexpr 规则，可以在 constexpr 上下文中使用 incr，例如：

constexpr int incr(int& n) {
    return ++n;
}

constexpr int foo() {
    int n = 0;
    incr(n);
    return n;
}

除非函数体完全不可能是 constexpr（例如，无条件调用非 constexpr 函数），否则编译器没有理由在定义点产生错误。

constexpr 函数甚至可以在主体中包含 paths/branches 而不是 constexpr。只要它们从未在 constexpr context 中使用，您就不会收到错误。例如：

constexpr int maybe_constexpr(bool choice, const int& a, const int& b) {
    return choice ? a : b;
}

constexpr int a = 0;
int b = 1;
static_assert(maybe_constexpr(true, a, b) == 0, "!");

live example

Answer 2

在 C++14 中，constexpr 函数的规则放宽了，论文 N3597: Relaxing constraints on constexpr functions。该论文探讨了基本原理和效果，包括以下内容（强调我的）：

As in C++11, the constexpr keyword is used to mark functions which the implementation is required to evaluate during translation, if they are used from a context where a constant expression is required. Any valid C++ code is permitted in constexpr functions, including the creation and modification of local variables, and almost all statements, with the restriction that it must be possible for a constexpr function to be used from within a constant expression. A constant expression may still have side-effects which are local to the evaluation and its result.

和：

A handful of syntactic restrictions on constexpr functions are retained:

asm-declarations are not permitted.

try-blocks and function-try-blocks are not permitted.

Declarations of variables with static and thread storage duration have some restrictions (see below).

我们可以在 N4140 部分 7.1.5 [dcl.constexpr] 中找到它，它说：

The definition of a constexpr function shall satisfy the following constraints:

it shall not be virtual (10.3);

its return type shall be a literal type;

each of its parameter types shall be a literal type;

its function-body shall be = delete, = default, or a compound-statement that does not contain

an asm-definition,

a goto statement,

a try-block, or

a definition of a variable of non-literal type or of static or thread storage duration or for which no initialization is performed.

最后一个示例显示了如何在 constexpr 中使用 incr：

constexpr int h(int k) {
  int x = incr(k); // OK: incr(k) is not required to be a core
                   // constant expression
  return x;
}

constexpr int y = h(1); // OK: initializes y with the value 2
                        // h(1) is a core constant expression because
                        // the lifetime of k begins inside h(1)

覆盖the lifetime of k begins inside h(1)的规则是：

modification of an object (5.17, 5.2.6, 5.3.2) unless it is applied to a non-volatile lvalue of literal type that refers to a non-volatile object whose lifetime began within the evaluation of e;

7.1.5[dcl.constexpr]中的措辞向我们展示了为什么incr是一个有效的constexpr：

For a non-template, non-defaulted constexpr function or a non-template, non-defaulted, non-inheriting constexpr constructor, if no argument values exist such that an invocation of the function or constructor could be an evaluated subexpression of a core constant expression (5.19), the program is ill-formed; no diagnostic required.

如T.C修改后的例子：

constexpr int& as_lvalue(int&& i){ return i; }

constexpr int x = incr(as_lvalue(1)) ;

表明，我们确实可以使用 incr 作为核心常量表达式的子表达式，因此它不是病式的。

据我所知，下面的函数不是 constexpr，而是用 clang 和 g++ 编译的代码。我错过了什么？

As far as I can tell the function below is not constexpr, but the code compiles in clang and g++. What am I missing?

c++

constexpr

c++14