是否有可能在传递给函数的输入缓冲区中写入超出预期的值？ （没有堆栈溢出）

Question

我正在做一些数据包翻译补丁。 我的dll注入中文游戏，钩住recv，侦听数据包并转换以中文接收的字符串。 我一直在编码，编码和编码……直到我发现我应该如何用buf来写超过数据包长度的东西？

int __stdcall Hooked_recv(SOCKET s, char *buf, int len, int flags)
{
    h_recv.PreHook();

    int ret_val = recv(s, buf, len, flags);
//ret_val is the number of bytes received. Ok, I can increase it, but...
//what to do with buf? Sure I can write there as much as no access violation appears.
//but I need a safe way.
//I guess if I do buf = new char[NEW_SIZE] then caller will fail to read buf because of pointer changed?
//what could I do to make received packet longer?
//I no want to reverse exe and increase buffer in hex editor. at least for now.

    h_recv.PostHook();

    return ret_val;
}

Answer 1

尽可能多地填充缓冲区。 如果有剩余，请将其保存以供下一次对挂钩的接收函数的调用（首先输入，如果已满，则重复保存新的剩余）。 您将需要使用缓冲区，这是不可避免的。