在 C++ 中获取地址的内容
如果我没有指向内存中地址的指针，但我确实将地址本身作为 uint_32 ，那么在 C++ 中是否可以获取内存中地址的内容？
谢谢
编辑
我正在写入这些位置，并试图查看我的 write() 函数是否正常工作，所以我想手动从地址进行读取。上面提到的“内容”的类型是uint64_t，这就是我尝试做的，但这给了我一个Segmentation Failure。
uint64_t *contents = reinterpret_cast<uint64_t*>(start_address);
cout<< hex << "Contents: " << *contents << endl;
我在这里做错了什么？
Is it possible in C++ to get the contents of an address in memory if I don't have a pointer to it, but I do have the address itself as a uint_32?
Thanks
EDIT
I am writing to those locations, and was trying to see if my write() function was working properly, so I wanted to manually do a read from the addresses. The 'contents' mentioned above are of type uint64_t and this is what I've tried doing, but this gives me a Segmentation Fault.
uint64_t *contents = reinterpret_cast<uint64_t*>(start_address);
cout<< hex << "Contents: " << *contents << endl;
What am I doing wrong here?
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（5）
不确定可移植性，但您可以使用
reinterpret_cast。例如Unsure of portability, but you can use
reinterpret_cast<>. e.g.是的，但前提是该地址位于您的地址空间中（或者如果您在微控制器上运行）或在内核空间中）。否则会导致段错误。
只需将 uint_32 转换为 int 指针，然后取消引用它：
Yes, but only if the address is in your address space (or if you are running on a microcontroller or in kernel space). Otherwise it will cause a seg fault.
Simply cast the uint_32 to an int pointer, and dereference it:
是的，您可以，如果您将此整数转换为指针：
但是您必须 100% 确定您的整数确实是一个地址，否则可能会发生不好的事情。所以这被认为是不好的做法，通常是一种黑客行为。
Yes, you can, if you cast this integer to a pointer:
But you have to be 100% sure that your integer really is an address, or bad things are likely to happen. So this is considered bad practice and usually a hack.
您只需将其转换为指针并取消引用即可。
You just cast it to pointer and dereference.
C 风格变体：
C-style variant: