c - 将可变大小的数组分配给固定大小的数组 c

Question

我有一个字符缓冲区，定义为char *buffer。当我发送 XBEE 数据包数据时，有固定的大小要求(84 字节)。我的问题是我不知道如何在迭代时将缓冲区转换为 84 字节 block 。这是我得到的:

warning: assignment from incompatible pointer type [-Wincompatible-pointer-types]
     packet = (buffer + i);
            ^

我的代码:

int sendSerialBuffer(char *buffer, int length)
{
  int i = 0;
  int result = 0;
  char (* packet)[84];

  // Iterate through all the bytes in the buffer
  for (i = 0; i < length; i = i + 84)
  {
    packet = (buffer + i);
    result = sendPacket(packet)
  }
  return 0;
}

编辑:我还尝试过声明像 char packet[84]; 和 get:

这样的数据包

error: assignment to expression with array type
     packet = (buffer + i);
            ^

此外，sendPacket 定义为:sendPacket(char packet[84]);

Answer 1

int i = 0;
int result = 0;

// Iterate through all the bytes in the buffer
for (i = 0; i < length; i = i + 84)
{
  char packet[84];
  memcpy(packet, buffer[i],84);
  result = sendPacket(packet);
}

return 0;  

请注意，这不处理缓冲区不是 84 的精确倍数的情况

如果 sendPAcket 期望的是 char * 而不是 char [84]，那么生活会更容易。然后代码将读取

int i = 0;
int result = 0; 

// Iterate through all the bytes in the buffer
for (i = 0; i < length; i = i + 84)
{
  char packet[84];
  memcpy(packet, buffer[i],84);
  result = sendPacket(buffer[i]);
}

return 0;