C语言 C if 语句 OR AND 上的逻辑运算符
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C Logical operators on if statement OR AND
提问by zikdaljin
Something is wrong with this logic it detects everything and prints bad pkt detected.
这个逻辑有问题，它检测一切并打印bad pkt detected。
if((buff[0] != 0x0a || buff[0] != 0x0e) && (len == 210))
{
printf("badpkt detected from %s\n", xpi);
} else {
if(mysend(ssl_sd ? ssl_sd[i] : NULL, sd[i], buff, len) <= 0) MULTI_SKIP_QUIT
}
The logic is DENY all packets that are 210 in length. Except if the 1st byte is 0x0A or 0x0E.
逻辑是拒绝所有长度为 210 的数据包。除非第一个字节是 0x0A 或 0x0E。
This code is working though:
这段代码虽然有效：
if((buff[0] != 0x0a) && (len == 210))
{
printf("badpkt detected from %s\n", xpi);
} else {
if(mysend(ssl_sd ? ssl_sd[i] : NULL, sd[i], buff, len) <= 0) MULTI_SKIP_QUIT
}
But I need both 0x0aand 0x0eto be the only 210 length packet allowed.
但是我需要两者0x0a并且0x0e是唯一允许的 210 长度数据包。
edit
编辑
What was I thinking, maybe its the lack of sleep.
我在想什么，也许是睡眠不足。
回答by Pol0nium
If you want both 0x0aand 0x0eallowed, you need to use this condition :
如果您既想要0x0a又0x0e允许，则需要使用此条件：
if((buff[0] == 0x0a || buff[0] == 0x0e) && (len == 210))
回答by Jacob Seleznev
This (buff[0] != 0x0a || buff[0] != 0x0e)is always true.
这(buff[0] != 0x0a || buff[0] != 0x0e)总是正确的。
It should be if(buff[0] != 0x0a && buff[0] != 0x0e && (len == 210))
它应该是 if(buff[0] != 0x0a && buff[0] != 0x0e && (len == 210))
回答by nos
You need this logic:
你需要这个逻辑：
if(buff[0] != 0x0a && buff[0] != 0x0e && len == 210)
With this condition:
在这个条件下：
if((buff[0] != 0x0a || buff[0] != 0x0e) && (len == 210))
imagine buff[0]is 0x0e. Then buff[0] != 0x0awill be true, which makes the whole sub expression (buff[0] != 0x0a || buff[0] != 0x0e)be true.
想象buff[0]IS 0x0e。然后buff[0] != 0x0a将为真，这使得整个子表达式(buff[0] != 0x0a || buff[0] != 0x0e)为真。
You could also invert the condition:
您还可以反转条件：
if((buff[0] == 0x0a || buff[0] == 0x0e) //always allow these
|| (len != 210)) //and allow anything thats not of length 210
{
if(mysend(ssl_sd ? ssl_sd[i] : NULL, sd[i], buff, len) <= 0)
MULTI_SKIP_QUIT
} else {
printf("badpkt detected from %s\n", xpi);
}
