在不带constexpr的模板非类型参数中进行类型转换 [英] Type conversion at template non-type argument without constexpr

问题描述

请考虑以下代码：

  struct A {
 constexpr operator int（）{return 42; } 
}; 
 
 template< int> 
 void foo（）{} 
 
 void bar（A a）{
 foo< a>（）; 
} 
 
 int main（）{
 foo< A {}>（）; 
 
 const int i = 42; 
 foo< i>（）; //（1）
 
 A a {}; 
 
 static_assert（i == a，）; 
 bar（a）; 
 foo< a>（）; // error here 
} 
  

Clang 3.7 with c ++ 14接受这个，而gcc 5.2.0与c ++ 14不会，产生以下消息：

  / tmp / gcc-explorer-compiler1151027-68-1f801jf / example.cpp：在函数'int main（）'中：
 26：错误：'a'的值在常量表达式中不可用
 foo< a>（）; 
 ^ 
 23：注意：'a'未声明为'constexpr'
 A a {}; 
 ^ 
编译失败
  

gcc修复了gcc编译错误，但没有 constexpr ccc建议的code> a code>，编译器是正确的吗？





对我来说，似乎 a 在常量表达式，如 static_assert ceritifies。此外， i 可以以相同的方式使用（标记为（1））， code> bar（）编译，也让我认为gcc是错误的。

UPD 报告了针对gcc的错误： https://gcc.gnu.org/bugzilla/show_bug .cgi？id = 68588 //eel.is/c++draft/expr.const#4rel =nofollow> [expr.const] /（4.1），而且我没有看到一个适用的项目符号， a href =http://eel.is/c++draft/expr.const#2 =nofollow> [expr.const] / 2 ，这将阻止您的表达式成为常量表达式。事实上，要求是如此宽松，声明 a 为

  A一个; 
  

是仍然给出一个格式良好的程序，即使 a 没有 constexpr 默认构造函数等，因为转换运算符 constexpr 并且没有成员被求值。

正如你所看到的，GCC是矛盾的，因为它允许 static_assert中的 a 条件，而不是模板参数。

Consider the following code:

struct A {
    constexpr operator int() { return 42; }
};

template <int>
void foo() {}

void bar(A a) {
    foo<a>();
}

int main() {
    foo<A{}>();

    const int i = 42;
    foo<i>();  // (1)

    A a{};

    static_assert(i == a, "");
    bar(a);
    foo<a>();  // error here
}

Clang 3.7 with c++14 accepts this, while gcc 5.2.0 with c++14 does not, producing the following message:

/tmp/gcc-explorer-compiler1151027-68-1f801jf/example.cpp: In function 'int main()':
26 : error: the value of 'a' is not usable in a constant expression
foo<a>();
^
23 : note: 'a' was not declared 'constexpr'
A a{};
^
Compilation failed

Changing a to be constexpr as suggested by gcc fixes the gcc compilation error, but without constexpr, which compiler is right?

For me, it seems that a should be "usable in constant expression", as static_assert ceritifies. Moreover, the fact that i can be used the same way (marked (1)), and the fact that bar() compiles, also makes me think that gcc is wrong.

UPD: reported a bug against gcc: https://gcc.gnu.org/bugzilla/show_bug.cgi?id=68588

解决方案

The user-defined conversion is allowed by [expr.const]/(4.1), and I don't see a single applicable bullet point in [expr.const]/2 that would prevent your expression from being a constant one. In fact, the requirements are so loose that declaring a as

A a;

is still giving a well-formed program, even if a didn't have a constexpr default constructor etc., since the conversion operator is constexpr and no members are evaluated.

As you saw yourself, GCC is contradictory in that it allows a in the static_assert condition but not a template-argument.

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