blocks|key|1131794|text|apache+commons-io+FileUtils.readFileToByteArray|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|1131795|entityMap^0|I|T|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

apache commons-io FileUtils.readFileToByteArray

blocks|key|1131871|text|它不是很慢，至少没有办法让它更快。但这是错误的。如果文件足够大，方法read()将不会返回第一次调用的所有字节。此方法返回它设法读取的字节数作为返回值。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1131872|正确的方法是在循环中调用此方法：|1131873|++public+static+void+copy(InputStream+input,
++++++OutputStream+output,
++++++int+bufferSize)
++++++throws+IOException+{
++++byte[]+buf+=+new+byte[bufferSize];
++++int+bytesRead+=+input.read(buf);
++++while+(bytesRead+!=+-1)+{
++++++output.write(buf,+0,+bytesRead);
++++++bytesRead+=+input.read(buf);
++++}
++++output.flush();
++}|code-block|syntax|javascript|1131874|按如下方式调用：|1131875|ByteArrayOutputStream+baos+=+new+ByteArrayOutputStream();
copy(new+FileInputStream(myfile),+baos);
byte[]+bytes+=+baos.toByteArray();|1131876|这样的东西在很多包中都实现了，例如@Andrey+Borisov+(%2B1)提到的FileUtils.readFileToByteArray()。|1131877|编辑|BOLD|1131878|我认为在你的例子中速度慢的原因是你创建了如此巨大的数组。你确定你真的需要它吗？试着重新思考你的设计。我相信您不必将此文件读入数组，并且可以增量地处理数据。|1131879|entityMap^0|Y|6|0|0|0|0|0|14|V|0|0|2|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|12|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|13|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|14|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|15|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|16|8|@$9|17|A|18|B|C]]|D|@]|E|$]]|$1|S|3|T|5|6|7|19|8|@$9|1A|A|1B|B|U]]|D|@]|E|$]]|$1|V|3|W|5|6|7|1C|8|@]|D|@]|E|$]]|$1|X|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]]|Y|$]]

It is not very slow, at least there is not way to make it faster. BUT it is wrong. If file is big enough the method <code>read()</code> will not return all bytes from fist call. This method returns number of bytes it managed to read as return value. 

The right way is to call this method in loop:

<pre><code> public static void copy(InputStream input,
 OutputStream output,
 int bufferSize)
 throws IOException {
 byte[] buf = new byte[bufferSize];
 int bytesRead = input.read(buf);
 while (bytesRead != -1) {
 output.write(buf, 0, bytesRead);
 bytesRead = input.read(buf);
 }
 output.flush();
 }
</code></pre>

call this as following:

<pre><code>ByteArrayOutputStream baos = new ByteArrayOutputStream();
copy(new FileInputStream(myfile), baos);
byte[] bytes = baos.toByteArray();
</code></pre>

Something like this is implemented in a lot of packages, e.g. <code>FileUtils.readFileToByteArray()</code> mentioned by @Andrey Borisov (+1)

EDIT

I think that reason for slowness in your case is the fact that you create so huge array. Are you sure you really need it? Try to re-think your design. I believe that you do not have to read this file into array and can process data incrementally.

I want to read a file into a byte array. So, I am reading it using:

<pre><code> int len1 = (int)(new File(filename).length());
 FileInputStream fis1 = new FileInputStream(filename);
 byte buf1[] = new byte[len1];
 fis1.read(buf1);
</code></pre>

However, it is realy very slow. Can anyone inform me a very fast approach (possibly best one) to read a file into byte array. I can use java library also if needed.

Edit: Is there any benchmark which one is faster (including library approach).

Java - File To Byte Array - Fast One

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我想把一个文件读入一个字节数组。因此，我使用以下命令来阅读它：    int len1 = (int)(new File(filename).length());    FileInputStream fis1 = new FileInputStream(filename);    byte buf1[] = new ...

问Java -文件到字节数组- Fast One
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