c++ - 初始化列表的分配

Question

下面的代码是我的问题的一个最小示例。我创建了一个包含固定大小数组的简单模板类，并重载了赋值运算符以接受定义方法 size() 的任何类。和begin() (例如，initializer_list s)。我不明白为什么 g++ 无法解析我对此运算符的调用(我使用的是 gcc 4.6):

***.cpp: In function ‘int main()’:
***.cpp:46:22: error: no match for ‘operator=’ in ‘a = {42, -1.0e+0, 3.14158999999999988261834005243144929409027099609375e+0}’
***.cpp:46:22: note: candidates are:
***.cpp:23:8: note: template<class U> A<T, N>::self& A::operator=(const U&) [with U = U, T = double, unsigned int N = 3u, A<T, N>::self = A<double, 3u>]
***.cpp:8:7: note: A<double, 3u>& A<double, 3u>::operator=(const A<double, 3u>&)
***.cpp:8:7: note:   no known conversion for argument 1 from ‘<brace-enclosed initialiser list>’ to ‘const A<double, 3u>&’
***.cpp:8:7: note: A<double, 3u>& A<double, 3u>::operator=(A<double, 3u>&&)
***.cpp:8:7: note:   no known conversion for argument 1 from ‘<brace-enclosed initialiser list>’ to ‘A<double, 3u>&&’

第一个候选已正确列出，但没有关联的错误消息。这是代码:

#include <iostream>
#include <algorithm>
#include <initializer_list>

// ------------------------------------------------------------------------

template <typename T, unsigned N>
class A
{
public:

    typedef A<T,N> self;

    // Default ctor
    A() {}

    // Copy ctor
    template <typename U>
    A( const U& other ) { operator=(other); }

    // Assignemnt
    template <typename U>
    self& operator= ( const U& other )
    {
        if ( other.size() == N )
            std::copy_n( other.begin(), N, m_data );
            return *this;
    }

    // Display contents
    void print() const
    {
        for ( unsigned i = 0; i < N; ++i )
            std::cout << m_data[i] << " ";
        std::cout << std::endl;
    }

private:
    T m_data[N];
};

// ------------------------------------------------------------------------

int main()
{
    A<double,3> a;
    a = {42,-1.0,3.14159};
    a.print();
}

有谁知道为什么这可能不明确，或者我做错了什么？

---

注意:理想情况下，我什至想将 main 的前两行替换为 A<double,3> a = {42,-1.0,3.14159}; 。但我不知道如何，目前出现以下错误:

***: In function ‘int main()’:
***:45:34: error: could not convert ‘{42, -1.0e+0, 3.14158999999999988261834005243144929409027099609375e+0}’ from ‘<brace-enclosed initialiser list>’ to ‘A<double, 3u>’

Answer 1

与auto不同，其中花括号初始化列表被推导为 initializer_list ，模板参数推导认为它是非推导上下文，除非存在类型为 initializer_list<T> 的对应参数。 ，在这种情况下 T可以推断。

来自§14.8.2.1/1 [temp.deduct.call](已添加强调)

Template argument deduction is done by comparing each function template parameter type (call it P) with the type of the corresponding argument of the call (call it A) as described below. If removing references and cv-qualifiers from P gives std::initializer_list<P0> for some P0 and the argument is an initializer list (8.5.4), then deduction is performed instead for each element of the initializer list, taking P0 as a function template parameter type and the initializer element as its argument. Otherwise, an initializer list argument causes the parameter to be considered a non-deduced context (14.8.2.5).

这就是你的 operator= 的论点不会被推断为 initializer_list<double> 。为了使代码正常工作，您必须定义 operator=这需要 initializer_list论证。

template <typename U>
self& operator= ( const std::initializer_list<T>& other )
{
    if ( other.size() == N )
        std::copy_n( other.begin(), N, m_data );
    return *this;
}