意想不到的不恒定std::initializer_list

I was toying a little bit with the indices trick to see where I could go to with and came across a strange error... First, the plain not-so-old indices:

我有点玩弄指数技巧，想看看我能做什么，却发现了一个奇怪的错误……首先，不那么古老的指数:

template<std::size_t...>
struct indices {};

template<std::size_t N, std::size_t... Indices>
struct make_indices:
    make_indices<N-1, N-1, Indices...>
{};

template<std::size_t... Indices>
struct make_indices<0, Indices...>:
    indices<Indices...>
{};

I created a compile-time array class derived from a std::initializer_list and had it indexable (assume that N3471 is support by your compiler. It will be in the next standard anyway). Here it is:

我创建了一个从std:::initializer_list派生的编译时数组类，并使它具有可索引性(假设N3471是由编译器支持的)。无论如何，它将是下一个标准)。这里是:

template<typename T>
struct array:
    public std::initializer_list<T>
{
    constexpr array(std::initializer_list<T> values):
        std::initializer_list<T>(values)
    {}

    constexpr auto operator[](std::size_t n)
        -> T
    {
        return this->begin()[n];
    }
};

So, I tried to create a function that returns a copy of an array after having added 1 to each of its members:

因此，我尝试创建一个函数，在向每个成员添加1之后返回一个数组的副本:

template<typename T, std::size_t... I>
auto constexpr add_one(const array<T>& a, indices<I...>)
    -> const array<T>
{
    return { (a[I]+1)... };
}

And to finish with the code, here is my main:

最后，我想说的是:

int main()
{
    constexpr array<int> a = { 1, 2, 3 };
    constexpr auto b = add_one(a, make_indices<a.size()>());

    return 0;
}

I did not think that code would compile anyway, but I am quite surprised by the error message (Here is the ideone code):

我不认为代码会被编译，但是我对错误消息感到非常惊讶(这里是ideone代码):

In function 'int main()':
error: 'const smath::array<int>{std::initializer_list<int>{((const int*)(& const int [3]{2, 3, 4})), 3u}}' is not a constant expression

So, could someone explain to me what exactly is not constant enough for the compiler in the above code?

那么，谁能向我解释一下上面代码中什么是不足够的常量?

EDIT: Follow-ups for that question

编辑:关于这个问题的后续报道。

• Is it legal to declare a constexpr std::initializer_list object?
• 声明constexpr std::initializer_list对象合法吗?
• Confusion about constant expression
• 常数表达式感到困惑

1 个解决方案

#1

From: the man himself http://www.stroustrup.com/sac10-constexpr.pdf

出处:该男子本人http://www.stroustrup.com/sac10-constexpr.pdf

Specifically: its return type, and the types of its parameters (if any), are literal types (see x2.2). For concreteness, literal types include bool, int, or double; its body is a compound statement of the form { return expr; } where expr is such that if arbitrary constant expressions of appropriate types are substituted for the parameters in expr, then the resulting expression is a constant expression as defined in introductory paragraph of x2. The expression expr is called a potential constant expression.

具体地说:它的返回类型和它的参数类型(如果有的话)是文字类型(参见x2.2)。具体来说，文字类型包括bool、int或double;它的主体是格式{return expr的复合语句;在expr中，如果将适当类型的任意常数表达式替换为expr中的参数，那么得到的表达式就是在x2的引言段中定义的常量表达式。表达式expr称为势常数表达式。