Question

从我能收集到的，这应该有效，不是吗？

struct ViewOfFile {
    void* p = nullptr;

    ViewOfFile(HANDLE hMap, int64_t OffsetB, SIZE_T SizeB, bool WriteAccess) {
        DWORD ViewAccessMode = FILE_MAP_READ | (WriteAccess ? FILE_MAP_WRITE : NULL);
        LARGE_INTEGER LI = { OffsetB };
        p = MapViewOfFile(hMap, ViewAccessMode, LI.HighPart, LI.LowPart, SizeB);
        if (p == nullptr) throw Exception("ViewOfFile: Failed to create view.");
    }
    ~ViewOfFile() {
        if (p) UnmapViewOfFile(p);
    }
    operator void*() const { return p; }
    operator char*() const { return reinterpret_cast<char*>(p); }
};

template <typename T>
struct ViewOfFileAs : ViewOfFile {
    T* as;
    ViewOfFile(HANDLE hMap, int64_t OffsetB, SIZE_T SizeB, bool WriteAccess) : ViewOfFile(hMap, OffsetB, SizeB, WriteAccess) {
        as = reinterpret_cast<T*>(p);
    }
};

但是，模板类构造函数给出了一个错误“只有一个构造函数可以有基本/成员初始化列表”。为什么会这样？

Answer 1

我认为你的模板结构定义中只有一个拼写错误：你写了ViewOfFile而不是ViewOfFileAs作为构造函数。

Answer 2

ViewOfFile(HANDLE hMap, int64_t OffsetB, SIZE_T SizeB, bool WriteAccess)

应该是

ViewOfFileAs(HANDLE hMap, int64_t OffsetB, SIZE_T SizeB, bool WriteAccess)

初始化基类

2 个答案: