C++ 类型强制推导

Question

我正在使用 Colvin-Gibbons 技巧在 C++03 中实现移动语义，我得到了以下内容:

#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>

template <typename T>
class buffer {
    struct buffer_ref {
        buffer_ref(T* data) : data_(data) {}
        T* data_;
    };

public:
    buffer() : data_(NULL) {}
    //explicit buffer(T* data) : data_(data)        {}
    buffer(size_t size)      : data_(new T[size]) {}
    buffer(buffer_ref other) : data_(other.data_) { other.data_ = NULL; } 
    buffer(buffer    &other) : data_(other.data_) { other.data_ = NULL; }
   ~buffer()                                      { delete [] data_;    }

    operator buffer_ref() { buffer_ref ref(data_); data_ = NULL; return ref; }
    operator T*()         { return data_;                                    }

private:
    T* data_;
};


int main() {
    buffer<float> data(buffer<float>(128));
    printf("ptr: %p\n", (float*)data);

}   

编辑:格式化

我希望能够在方便的时候将我的缓冲区用作指向基类型的指针，因此我向指针类型添加了一个强制转换运算符，它按预期工作。但是，如果我取消注释采用指针的构造函数，那么转换推导会变得困惑并提示不明确的转换(因为它可以转到 buffer->T*->buffer 或 buffer->buffer_ref->buffer)。我希望指针构造函数上的显式修饰符可以解决此问题，但事实并非如此。比我更了解 C++ 转换推导的人可以解释一下编译器的想法吗？

Answer 1

这是 13.3.1.3 [over.match.ctor] 的直接结果:

When objects of class type are direct-initialized (8.5), or copy-initialized from an expression of the same or a derived class type (8.5), overload resolution selects the constructor. For direct-initialization, the candidate functions are all the constructors of the class of the object being initialized. For copy-initialization, the candidate functions are all the converting constructors (12.3.1) of that class. [...]

因为 buffer<float> data(buffer<float>(128));是直接初始化，您已明确请求 explicit考虑构造函数。

如果你写:

buffer<float> data = buffer<float>(128);

那就没有歧义了。