据我所知,这应该可行,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,模板类构造函数给我一个错误“只有构造函数可以有基/成员初始化列表”。为什么会这样?
最佳答案
我认为您只是在模板结构的定义中有错别字:您编写了 ViewOfFile
而不是 ViewOfFileAs
作为构造函数。
关于c++ - 初始化基类,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33059989/