C memmove 相当于 for 循环 - 段错误

Question

我正在实现具有动态调整大小的环形缓冲区。当 tail 位于 head 后面时，调整大小后，必须将缓冲区末尾的数据移至缓冲区的新末尾。为此，我编写了以下代码:

memmove(self->broadcaster.events+self->broadcaster.events_head+self->broadcaster.events_size,
              self->broadcaster.events+self->broadcaster.events_head,
              self->broadcaster.events_size-self->broadcaster.events_head);

其中 self->broadcaster.events_size 是旧大小 (new_size/2)。不幸的是，它会导致段错误。我认为它与此代码等效:

for (i = 0 ; i < self->broadcaster.events_size - self->broadcaster.events_head ; ++i)
        self->broadcaster.events[self->broadcaster.events_size+self->broadcaster.events_head+i]=
            self->broadcaster.events[self->broadcaster.events_head+i];

但是这种幼稚的 for 循环实现可以正常工作，所以我似乎不知道如何正确使用 memmove 。这两段代码有何不同？

Answer 1

它们仅在 sizeof(*self->broadcaster.events) == 1 时等效.

为了清楚起见，我替换了 self->broadcaster与 b和events与 e并在代码中添加了一些空格。

memmove(b.e + b.e_head + b.e_size, b.e + b.e_head, b.e_size - b.e_head);

将仅复制 b.e_size - b.e_head字节和循环:

for (i = 0 ; i < b.e_size - b.e_head ; ++i)
        b.e[b.e_size + b.e_head + i] = b.e[b.e_head + i];

将复制(b.e_size - b.e_head) * sizeof *b.e字节，因为每个 b.e[...] = b.e[...]作业正在通过 sizeof *b.e字节，每个 ++i正在推进b.e[... + i]的地址通过sizeof *b.e字节。

如果您定义宏，您将得到最好的服务:

#define MOVE(dst, src, count)  memmove((dst), (src), (count) * sizeof *(src))

并使用它代替 memmove .

但是您当然可以更改 memmove 的最后一个参数至

(self->broadcaster.events_size-self->broadcaster.events_head)*sizeof*self->broadcaster.events