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c++ - 使用 clang 和 g++ 编译此 "simple"程序时出现链接错误

转载 作者:塔克拉玛干 更新时间:2023-11-03 06:46:31 26 4
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我正在做一个项目,似乎 clang 无法生成有效的字节码(因为链接器无法链接,在链接时找不到模板类中的某些 static constexpr )我可以在类中使用静态 getter 来修复它,但这会导致一些非常难看/重载的代码。

这是使错误(是错误吗?)出现的“最小”代码示例。不幸的是,在这里,g++ 会产生相同的链接错误。

我想问的是这是否是一个编译器错误,或者我只是做错了什么,以及是否有避免此错误的解决方案。(如果我做错了什么,为什么 相同 构造在我的更大项目中起作用??)

注意:更大的项目在 github 上名为 yaggler,但仍处于早期阶段

#include <type_traits>
#include <iostream>

// exemple vector classes
struct vector2
{
constexpr vector2(unsigned int _x = 0, unsigned int _y = 0) : x(_x), y(_y) {} // for clang

unsigned int x;
unsigned int y;
};
struct vector3 : public vector2 // uh ;)
{
constexpr vector3(unsigned int _x = 0, unsigned int _y = 0, unsigned int _z = 0) : vector2(_x, _y), z(_z) {} // for clang
unsigned int z;
};

// simple templated generic vector type
// we could make a more generic one, but this would require something like a tuple.
template<unsigned int... Vals>
struct vector
{
static_assert(!(sizeof...(Vals) + 1), "[...]");
};
template<unsigned int X>
struct vector<X>
{
using vec_type = unsigned int;
static constexpr unsigned int value = X;
};
template<unsigned int X, unsigned int Y>
struct vector<X, Y>
{
using vec_type = vector2;
static constexpr vector2 value = vector2(X, Y);
};
template<unsigned int X, unsigned int Y, unsigned int Z>
struct vector<X, Y, Z>
{
using vec_type = vector3;
static constexpr vector3 value = vector3(X, Y, Z);
};

// a simple wrapper
template<typename V>
struct some_wrapper
{
static constexpr typename V::vec_type value = V::value;
};

// a dummy function that print something to stdout.
void do_something(int32_t id, const vector3 &value)
{
std::cout << id << " " << value.z << std::endl;
}
void do_something(int32_t id, const vector2 &value)
{
std::cout << id << " " << value.y << std::endl;
}
void do_something(int32_t id, int value)
{
std::cout << id << " " << value << std::endl;
}

// the class used to create the error
template< typename... Args>
class exemple
{
private:
// an initialisation that recurse over the Args... template arguments
template<typename Current, typename... Others>
void __rec_init() const
{
do_something(0, Current::value);
__rec_init<Others...>();
}

// end of recursion
template<size_t = 0>
void __rec_init() const {}

// launch the recursion
void tpl_init() const
{
__rec_init<Args...>();
}

public:
exemple()
{
tpl_init();
}
};

int main()
{
// and here, we get a linker error.
exemple<some_wrapper<vector<4, 4, 5>>, some_wrapper<vector<4, 1>>, some_wrapper<vector<9>>>();
}

编辑:仅提及 gcc 和 clang 版本:gcc 4.7.3/4.8.2 和 clang 3.2/3.3

最佳答案

vector 的特化2 和 3 模板参数的类模板有 static constexpr数据成员value没有命名空间范围定义的文字类型(分别为 vector2vector3)。

您将需要它们,因为您使用 odr-use value传递给 do_something 时绑定(bind)到引用参数功能。

§9.4.2/3 [class.static.mfct]

If a non-volatile const static data member is of integral or enumeration type, its declaration in the class definition can specify a brace-or-equal-initializer in which every initializer-clause that is an assignment-
expression
is a constant expression (5.19). A static data member of literal type can be declared in the class definition with the constexpr specifier; if so, its declaration shall specify a brace-or-equal-initializer in which every initializer-clause that is an assignment-expression is a constant expression. [Note: In both these cases, the member may appear in constant expressions. —end note ] The member shall still be defined in a namespace scope if it is odr-used (3.2) in the program and the namespace scope definition shall not contain an initializer.

编辑:更正我自己,实际上是 some_wrapper<T>::value这需要这个定义(尽管如此,出于上述原因)。所以你需要的是在定义 some_wrapper 之后的命名空间范围内的 this :

template<typename V>
constexpr typename V::vec_type some_wrapper<V>::value;

在那之后,你的代码compiles and runs .

关于c++ - 使用 clang 和 g++ 编译此 "simple"程序时出现链接错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/21344031/

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