在流缓冲区中连续扫描字符串的最佳方法 [英] Best approach to continuously scan for a string in a streaming buffer

问题描述

我有这种情况，我的功能连续接收各种长度的数据。数据可以是任何东西。我想找到最好的方法，我在这个数据中寻找特定的字符串。

I have this situation where my function continuously receive data of various length. The data can be anything. I want to find the best way I to hunt for particular string in this data. The solution will require somehow to buffer previous data but I cannot wrap my head around the problem.

这里是一个问题的例子：

Here is an example of the problem:

资料输入 - > [\x00\x00\x01\x23B] [] [LABLABLABLABLA\x01TO] [KEN] [BLA\x01] ...

DATA IN -> [\x00\x00\x01\x23B][][LABLABLABLABLA\x01TO][KEN][BLA\x01]...

如果每个[...]表示一个数据块，[]表示没有项的数据块，扫描字符串TOKEN的最好方法是什么？

if every [...] represents a data chunk and [] represents a data chunk with no items, what is the best way to scan for the string TOKEN?

UPDATE：
我意识到这个问题有点复杂。 []不是分隔符。我只是用它们来描述每个上面的例子的块的结构。此外，TOKEN不是静态字符串。它是可变长度。我认为逐行读取的最好方法是如何读取可变长度的流缓冲区成行。

UPDATE: I realised the question is a bit more complex. the [] are not separators. I just use them to describe the structure of the chunk per above example. Also TOKEN is not a static string per-se. It is variable length. I think the best way to read line by line but than the question is how to read a streaming buffer of variable length into lines.

推荐答案

p>对不起，我投票删除我以前的答案，因为我的理解的问题是不正确的。我没有仔细阅读enouogh，并认为[]是符号分隔符。

Sorry, I voted to delete my previous answer as my understanding of the question was not correct. I didn't read carefully enouogh and thought that the [] are token delimiters.

对于你的问题，我建议基于一个简单的计数器构建一个小状态机：
对于每个字符，你执行类似下面的伪代码：

For your problem I'd recommend building a small state machine based on a simple counter: For every character you do something like the following pseudo code:

if (received_character == token[pos]) {
    ++pos;
    if (pos >= token_length) {
        token_received = 1;
    }
}
else {
    pos = 0; // Startover
}

这需要最少的处理器周期和最小的内存aso你不需要缓冲任何东西，除了刚收到的块。

This takes a minimum of processor cycles and also a minimum of memory aso you don't need to buffer anything except the chunk just received.

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