Question

在 C++11 中,我正在使用一个 ConstExplor 函数作为模板参数的默认值 - 它看起来像 :

template <int value>
struct bar
{
    static constexpr int get()
    {
        return value;
    }
};

template <typename A, int value = A::get()>
struct foo
{
};

int main()
{
    typedef foo<bar<0>> type;

    return 0;
}

G++ 4. 5 和 4. 7 编译此, 但 Clang++ 3. 1 不。来自 cng 的错误信息是 :

clang_test.cpp:10:35: error: non-type template argument is not a constant expression
template <typename A, int value = A::get()>
                                  ^~~~~~~~
clang_test.cpp:17:19: note: while checking a default template argument used here
        typedef foo<bar<3>> type;
                ~~~~~~~~~^~
clang_test.cpp:10:35: note: undefined function  get  cannot be used in a constant expression
template <typename A, int value = A::get()>
                                  ^
clang_test.cpp:4:23: note: declared here
        static constexpr int get()
                             ^
1 error generated.

<% 1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Answer 1

在LLLVM IRC频道的Richard Smith(zygoloid) 与我简短谈论了这个问题,

<litb> hello folks
<litb> zygoloid, what should happen in this case?
<litb> http://stackoverflow.com/questions/10721130/calling-constexpr-in-default-template-argument
<litb> it seems to be clang s behavior is surprising
<litb> zygoloid, i cannot apply the "point of instantiation" rule to constexpr 
  function templates. if i call such a function template, the called definition s 
  POI often is *after* the specialization reference, which means at the point of 
  the call, the constexpr function template specialization is "undefined".
<zygoloid> it s a horrible mess. Clang does not do what the standard intends, but 
  as you note, the actual spec is gloriously unclear
<d0k> :(
<zygoloid> we should instantiate bar<3>::get(), because it is odr-used, but we 
  don t, because we incorrectly believe it s used in an unevaluated context
<zygoloid> conversely, the point of instantiation is too late :/
<zygoloid> PR11851

因此,似乎有时,Clang Comptiates 被称为功能模板或类模板的成员功能,但是它们的即时化为时太晚,无法引起人们的注意,在其他情况下,它甚至没有立即化它们,因为它认为它永远不会需要它们(未经评价的背景)。

Answer 2

我认为海合会 ~~Clang是正确的~~

从 n3290 引用 :

14.3.2 模板非类型参数[tmp.arg.non类型]

A template-argument for a non-type, non-template template-parameter shall be one of:

for a non-type template-parameter of integral or enumeration type, a converted > constant expression (5.19) of the type of the template-parameter; or
...

5.19 3

A literal constant expression is a prvalue core constant expression of literal type, but not pointer type. An integral constant expression is a literal constant expression of integral or unscoped enumeration type. [ Note: Such expressions may be used as array bounds (8.3.4, 5.3.4), as bit-field lengths (9.6), as enumerator initializers if the underlying type is not fixed (7.2), as null pointer constants (4.10), and as alignments (7.6.2). —end note ] A converted constant expression of type T is a literal constant expression, implicitly converted to type T, where the implicit conversion (if any) is permitted in a literal constant expression and the implicit conversion sequence contains only user-defined conversions, lvalue-to-rvalue conversions (4.1), integral promotions (4.5), and integral conversions (4.7) other than narrowing conversions (8.5.4).

[注:这些表达式可用作案件表达式(6.4.2),如果基本类型固定(7.2),可作为数字初始化器,并作为整体或列举非类型模板参数(14.3)——尾注]

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