Question

我在PHP中有以下代码，我正在从c ++代码转换此代码。

这是C ++代码，如果你想分析那么。

这是PHP代码。

// just read the size to avoid reading 2 messages at once
$this->m_MsgSize = socket_recv($Socket, $this->m_MsgBuf, 2, 0);

// for now we expect 2 bytes at once, it should not be splitted
$DataSize = (int) (ord($this->m_MsgBuf[0]) | ord($this->m_MsgBuf[1]) >> 8);
if(($this->m_MsgSize != 2) || ($DataSize > ($this->NetworkMaxSize - 2)))
{
    $this->ResetBuffer();
    return false;
}

// read the real data
$this->m_MsgSize += socket_recv($Socket, $this->m_MsgBuf + 2, $DataSize, 0);

$NewDataSize = (int) (ord($this->m_MsgBuf[0]) | ord($this->m_MsgBuf[1]) >> 8);

echo $NewDataSize."\n";

// we got something unexpected/incomplete
if(($this->m_MsgSize <= 2) || ($NewDataSize != ($this->m_MsgSize - 2)))
{
    $this->ResetBuffer();
    return false;
}

// ok, ...reading starts after the size
$this->m_ReadPos = 2;

问题出在这一行

$this->m_MsgSize += socket_recv($Socket, $this->m_MsgBuf + 2, $DataSize, 0);

从c ++代码转换为此行，链接位于此问题的顶部。

m_MsgSize += recv(socket, (char*)m_MsgBuf+2, datasize, 0);

当我写($m_MsgBuf + 2)时，php返回php only variables should be passed by reference，当我在$m_MsgBuf += 2;之前或之后写$m_MsgBuf = $m_MsgBuf + 2;或socket_recv时..下一个变量名为{ {1}}不会返回任何值。

问题是..如何在PHP中向缓冲区变量添加+ 2缓冲区？

PHP将+2添加到socket_recv返回的缓冲区中

0 个答案: