blocks|key|3522070|text|当然，您可以使用内存映射文件。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3522071|以下是示例代码的两个很好的链接：|3522072|3522073|Thinking+in+Java:+Memory-mapped+files|unordered-list-item|offset|length|3522074|Java+Tips:+How+to+create+a+memory-mapped+file|3522075|3522076|3522077|如果您不想走这条路，只需使用普通的InputStream+(例如将其包装在BufferedInputStream中之后的DataInputStream。|style|CODE|3522078|entityMap|0|LINK|mutability|MUTABLE|url|http://www.linuxtopia.org/online_books/programming_books/thinking_in_java/TIJ314_029.htm|1|http://www.java-tips.org/java-se-tips/java.nio/how-to-create-a-memory-mapped-file-19.html^0|0|0|0|0|11|0|0|0|19|1|0|0|0|H|B|11|J|1O|F|0^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|12|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|13|8|@]|9|@]|A|$]]|$1|E|3|F|5|G|7|14|8|@]|9|@$H|15|I|16|1|17]]|A|$]]|$1|J|3|K|5|G|7|18|8|@]|9|@$H|19|I|1A|1|1B]]|A|$]]|$1|L|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|1E|8|@$H|1F|I|1G|P|Q]|$H|1H|I|1I|P|Q]|$H|1J|I|1K|P|Q]]|9|@]|A|$]]|$1|R|3|-4|5|6|7|1L|8|@]|9|@]|A|$]]]|S|$T|$5|U|V|W|A|$X|Y]]|Z|$5|U|V|W|A|$X|10]]]]

Sure, you could use a memory mapped file.

Here are two good links with sample code:

<ul>
<li><a href="http://www.linuxtopia.org/online_books/programming_books/thinking_in_java/TIJ314_029.htm" rel="nofollow">Thinking in Java: Memory-mapped files</a></li>
<li><a href="http://www.java-tips.org/java-se-tips/java.nio/how-to-create-a-memory-mapped-file-19.html" rel="nofollow">Java Tips: How to create a memory-mapped file</a></li>
</ul>

<hr>

If you don't want to go this route, just use an ordinary <code>InputStream</code> (such as a <code>DataInputStream</code> after wrapping it in a <code>BufferedInputStream</code>.

blocks|key|185681|text|看看这篇关于如何在Java中将二进制文件读入字节数组的博客文章：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|185682|http://www.spartanjava.com/2008/read-a-file-into-a-byte-array/|offset|length|185683|从链接复制：|185684|File+file+=+new+File("/somepath/myfile.ext");
FileInputStream+is+=+new+FileInputStream(file);

//+Get+the+size+of+the+file
long+length+=+file.length();

if+(length+>+Integer.MAX_VALUE)+{
++++throw+new+IOException("The+file+is+too+big");
}

//+Create+the+byte+array+to+hold+the+data
byte[]+bytes+=+new+byte[(int)length];

//+Read+in+the+bytes
int+offset+=+0;
int+numRead+=+0;
while+(offset+<+bytes.length
+++++++&&+(numRead=is.read(bytes,+offset,+bytes.length-offset))+>=+0)+{
++++offset+%2B=+numRead;
}

//+Ensure+all+the+bytes+have+been+read+in
if+(offset+<+bytes.length)+{
++++throw+new+IOException("The+file+was+not+completely+read:+"%2Bfile.getName());
}

//+Close+the+input+stream,+all+file+contents+are+in+the+bytes+variable
is.close()++++|code-block|syntax|javascript|185685|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|1Q|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@]|9|@$D|V|E|W|1|X]]|A|$]]|$1|F|3|G|5|6|7|Y|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|Z|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|C]]]]

Take a look at this blog post here on how to read a binary file into a byte array in Java:

<a href="http://www.spartanjava.com/2008/read-a-file-into-a-byte-array/" rel="nofollow">http://www.spartanjava.com/2008/read-a-file-into-a-byte-array/</a>

Copied from link:

<pre><code>File file = new File("/somepath/myfile.ext");
FileInputStream is = new FileInputStream(file);

// Get the size of the file
long length = file.length();

if (length &gt; Integer.MAX_VALUE) {
 throw new IOException("The file is too big");
}

// Create the byte array to hold the data
byte[] bytes = new byte[(int)length];

// Read in the bytes
int offset = 0;
int numRead = 0;
while (offset &lt; bytes.length
 &amp;&amp; (numRead=is.read(bytes, offset, bytes.length-offset)) &gt;= 0) {
 offset += numRead;
}

// Ensure all the bytes have been read in
if (offset &lt; bytes.length) {
 throw new IOException("The file was not completely read: "+file.getName());
}

// Close the input stream, all file contents are in the bytes variable
is.close() 
</code></pre>

blocks|key|3522154|text|在这里，使用Java+SDK的DataInputStream会很有帮助。如果需要的话，DataInputStream提供了readByte()或readChar()这样的函数。一个简单的例子可以是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3522155|DataInputStream+dis+=+new+DataInputStream(new+FileInputStream("file.dat"));+
try+{
+++while(true)+{
++++++byte+b+=+dis.readByte();
++++++//Do+something+with+the+byte
+++}+
}+catch+(EOFException+eofe)+{
//Stream+Ended
}+catch+(IOException+ioe)+{
//Input+exception
}|code-block|syntax|javascript|3522156|希望能有所帮助。当然，您可以将整个流读取到一个字节数组中，也可以遍历它……|3522157|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Using the DataInputStream of the Java SDK can be helpful here. DataInputStream provide such functions as readByte() or readChar(), if that's what needed.
A simple example can be:

<pre><code>DataInputStream dis = new DataInputStream(new FileInputStream("file.dat")); 
try {
 while(true) {
 byte b = dis.readByte();
 //Do something with the byte
 } 
} catch (EOFException eofe) {
//Stream Ended
} catch (IOException ioe) {
//Input exception
}
</code></pre>

Hope it helps. You can, of course, read the entire stream to a byte array and iterate through it as well...

blocks|key|3522206|text|大多数文件不需要内存映射，而是可以通过标准Java+I/O读取，特别是在文件非常小的情况下。读取所述文件的一种合理方式是使用BufferedInputStream。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3522207|InputStream+in+=+new+BufferedInputStream(new+FileInputStream("somefile.ext"));|code-block|syntax|javascript|3522208|缓冲已经在Java中针对大多数计算机进行了优化。如果你有一个更大的文件，比如说100MB，那么你会考虑进一步优化它。|3522209|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Most files will not need memory mapping but can simply be read by the standard Java I/O, especially since your file is so small. A reasonable way to read said files is by using a BufferedInputStream.

<pre><code>InputStream in = new BufferedInputStream(new FileInputStream("somefile.ext"));
</code></pre>

Buffering is already optimized in Java for most computers. If you had a larger file, say 100MB, then you would look at optimizing it further.

blocks|key|3522237|text|从磁盘读取文件将是最慢的部分，所以这可能没有任何区别。当然，这个单独的操作-+JVM仍然需要十年的时间才能启动，所以要加上这段时间。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3522238|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Reading the file from the disk is going to be the slowest part by miles, so it's likely to make no difference whatsoever. Of this individual operation, of course- the JVM still takes a decade to start up, so add that time in.

blocks|key|185764|text|如果您使用内存映射文件或常规缓冲区，您将能够以硬件允许的速度读取数据。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|185765|File+tmp+=+File.createTempFile("deleteme",+"bin");
tmp.deleteOnExit();
int+size+=+1024+*+1024+*+1024;

long+start0+=+System.nanoTime();
FileChannel+fc0+=+new+FileOutputStream(tmp).getChannel();
ByteBuffer+bb+=+ByteBuffer.allocateDirect(32+*+1024).order(ByteOrder.nativeOrder());

for+(int+i+=+0;+i+<+size;+i+%2B=+bb.capacity())+{
++++fc0.write(bb);
++++bb.clear();
}
long+time0+=+System.nanoTime()+-+start0;
System.out.printf("Took+%25.3f+ms+to+write+%25,d+MB+using+ByteBuffer%25n",+time0+/+1e6,+size+/+1024+/+1024);

long+start+=+System.nanoTime();
FileChannel+fc+=+new+FileInputStream(tmp).getChannel();
MappedByteBuffer+buffer+=+fc.map(FileChannel.MapMode.READ_ONLY,+0,+size);
LongBuffer+longBuffer+=+buffer.order(ByteOrder.nativeOrder()).asLongBuffer();
long+total+=+0;+//+used+to+prevent+a+micro-optimisation.
while+(longBuffer.remaining()+>+0)
++++total+%2B=+longBuffer.get();
fc.close();
long+time+=+System.nanoTime()+-+start;
System.out.printf("Took+%25.3f+ms+to+read+%25,d+MB+MemoryMappedFile%25n",+time+/+1e6,+size+/+1024+/+1024);

long+start2+=+System.nanoTime();
FileChannel+fc2+=+new+FileInputStream(tmp).getChannel();
bb.clear();
while+(fc2.read(bb)+>+0)+{
++++while+(bb.remaining()+>+0)
++++++++total+%2B=+bb.get();
++++bb.clear();
}
fc2.close();
long+time2+=+System.nanoTime()+-+start2;
System.out.printf("Took+%25.3f+ms+to+read+%25,d+MB+File+via+NIO%25n",+time2+/+1e6,+size+/+1024+/+1024);|code-block|syntax|javascript|185766|打印|185767|Took+305.243+ms+to+write+1,024+MB+using+ByteBuffer
Took+286.404+ms+to+read+1,024+MB+MemoryMappedFile
Took+155.598+ms+to+read+1,024+MB+File+via+NIO|185768|这是针对一个比你想要的文件大10倍的文件。它这么快是因为数据被缓存在内存中(我有一个SSD驱动器)。如果你有快速的硬件，那么读取数据的速度就会很快。|185769|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

If you use a memory mapped file or regular buffer you will be able to read the data as fast your hardware allows.

<pre><code>File tmp = File.createTempFile("deleteme", "bin");
tmp.deleteOnExit();
int size = 1024 * 1024 * 1024;

long start0 = System.nanoTime();
FileChannel fc0 = new FileOutputStream(tmp).getChannel();
ByteBuffer bb = ByteBuffer.allocateDirect(32 * 1024).order(ByteOrder.nativeOrder());

for (int i = 0; i &lt; size; i += bb.capacity()) {
 fc0.write(bb);
 bb.clear();
}
long time0 = System.nanoTime() - start0;
System.out.printf("Took %.3f ms to write %,d MB using ByteBuffer%n", time0 / 1e6, size / 1024 / 1024);

long start = System.nanoTime();
FileChannel fc = new FileInputStream(tmp).getChannel();
MappedByteBuffer buffer = fc.map(FileChannel.MapMode.READ_ONLY, 0, size);
LongBuffer longBuffer = buffer.order(ByteOrder.nativeOrder()).asLongBuffer();
long total = 0; // used to prevent a micro-optimisation.
while (longBuffer.remaining() &gt; 0)
 total += longBuffer.get();
fc.close();
long time = System.nanoTime() - start;
System.out.printf("Took %.3f ms to read %,d MB MemoryMappedFile%n", time / 1e6, size / 1024 / 1024);

long start2 = System.nanoTime();
FileChannel fc2 = new FileInputStream(tmp).getChannel();
bb.clear();
while (fc2.read(bb) &gt; 0) {
 while (bb.remaining() &gt; 0)
 total += bb.get();
 bb.clear();
}
fc2.close();
long time2 = System.nanoTime() - start2;
System.out.printf("Took %.3f ms to read %,d MB File via NIO%n", time2 / 1e6, size / 1024 / 1024);
</code></pre>

prints

<pre><code>Took 305.243 ms to write 1,024 MB using ByteBuffer
Took 286.404 ms to read 1,024 MB MemoryMappedFile
Took 155.598 ms to read 1,024 MB File via NIO
</code></pre>

This is for a file 10x larger than what you want. Its this fast because the data is being cached in memory (and I have an SSD drive). If you have fast hardware, the data can be read pretty fast.

I have a binary file (about 100&nbsp;MB) that I need to read in quickly. In C++ I could just load the file into a char pointer and march through it by incrementing the pointer. This of course would be very fast.

Is there a comparably fast way to do this in Java?

Reading a binary file in Java vs C++

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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我有一个二进制文件(大约100兆字节)，我需要快速读入。在C++中，我可以将文件加载到一个字符指针中，并通过递增指针来遍历它。这当然是非常快的。在Java中有没有更快的方法来做到这一点？

问在Java与C++中读取二进制文件
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