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A Java OutofMemoryError occurs if when the program attempts to use more heap space than is available. Among other causes, this error may result from the following:

• a A memory leak (see MSC04-J. Do not leak memory)
• an An infinite loop
• limited Limited amounts of default heap memory available
• incorrect Incorrect implementation of common data structures (hash tables, vectors, and so on)
• unbound Unbounded deserialization
• writing Writing a large number of objects to an ObjectOutputStream (see SER10-J. Avoid memory and resource leaks during serialization)
• creating Creating a large number of threads.
• uncompressing Uncompressing a file (see IDS04-J. Limit the size of files passed to Safely extract files from ZipInputStream)

Some of these causes are platform-dependent , and difficult to anticipate. Others are fairly easy to anticipate, such as reading data from a file, are fairly easy to anticipate. As a result, programs must not accept untrusted input in a manner that can cause the program to exhaust memory.

Noncompliant Code Example (readLine())

This noncompliant code example reads lines of text from a file , and adds each one to a vector , until a line with the word "quit" is encountered.:

class ShowHeapErrorReadNames {
  private Vector<String> names = new Vector<String>();
  private final InputStreamReader input;
  private final BufferedReader reader;

  public ShowHeapErrorReadNames(String filename) throws IOException {
    this.input = new FileReader(filename);
    this.reader = new BufferedReader(input);
  }

  public void addNames() throws IOException {
    try {
      String newName;
      while (((newName = reader.readLine()) != null) &&
             !(newName.equalsIgnoreCase("quit") == false)) {)) {
        names.addElement(newName);
        System.out.println("adding " + newName);
      }
    } finally {
      input.close();
    }
  }

  public static void main(String[] args) throws IOException {
    if (args.length != 1) {
      System.out.println("Arguments: [filename]");
      return;
    }
    ShowHeapErrorReadNames demo = new ShowHeapErrorReadNames(args[0]);
    demo.addNames();
  }
}

Wiki MarkupThe code places no upper bounds on the memory space required to execute the program. Consequently, the program can easily exhaust the available heap space in two ways. First, an attacker can supply arbitrarily many lines in the file, causing the vector to grow until memory is exhausted. Second, an attacker can simply supply an arbitrarily long line, causing the {{readLine()}} method to exhaust memory. According to the Java API \[documentation [API 2006|AA. Bibliography#API 06]\2014], {{the BufferedReader.readLine()}} method documentation method

Reads a line of text. A line is considered to be terminated by any one of a line feed ('\n'), a carriage return ('\r'), or a carriage return followed immediately by a linefeed. Wiki Markup\[{{readLine()}}\] Reads a line of text. A line is considered to be terminated by any one of a line feed ('{{\n}}'), a carriage return ('{{\r}}'), or a carriage return followed immediately by a linefeed.

Any code that uses this method is susceptible to a resource exhaustion attack because the user can enter a string of any length.

Compliant Solution (

...

Limited File Size)

This compliant solution imposes limits, both a limit on the length size of each line, and on the total number of items to add to the vector.the file being read. The limit is set with the Files.size() method, which was introduced in Java SE 7. If the file is within the limit, we can assume the standard readLine() method will not exhaust memory, nor will memory be exhausted by the while loop.

class ReadNames

class ShowHeapError {
  // ... otherOther methods and variables

  public static public String readLimitedLine(Reader reader, int limitfinal int fileSizeLimit = 1000000;

  public ReadNames(String filename) throws IOException {
    StringBuilderlong sbsize = new StringBuilder()Files.size( Paths.get( filename));
    forif (intsize i> = 0; i < limit; i++) {
      int c = reader.read(fileSizeLimit) {
      throw new IOException("File too large");
    } else if (csize == -10L) {
      throw new return null;
      }IOException("File size cannot be determined, possibly too large");
    }
  if (((char) c == '\n') || ((char) c == '\r')) { this.input = new FileReader(filename);
    this.reader =   breaknew BufferedReader(input);
  }
}

Compliant Solution (Limited Length Input)

This compliant solution imposes limits both on the length of each line and on the total number of items to add to the vector. (It does not depend on any Java SE 7 or later features.)

class ReadNames {
  // ... Other methods and variables

  public static String readLimitedLine(Reader reader, int limit) 
                                       throws IOException    }
      sb.append((char) c);
    }
    return sb.toString();
  }

  static public final int lineLengthLimit = 1024;
  static public final int lineCountLimit = 1000000;

  public void addNames() throws IOException {
    try {
    StringBuilder sb String newName= new StringBuilder();
      for (int i = 0; i < lineCountLimitlimit; i++) {
      int c newName = readLimitedLine(reader, lineLengthLimitreader.read();
        if (newNamec == null || newName.equalsIgnoreCase("quit"))-1) {
        return  break;
 ((sb.length() > 0) ? sb.toString() : null);
       }

        names.addElement(newName);
        System.out.println("adding " + newName);
if (((char) c == '\n') || ((char) c == '\r')) {
        }break;
    } finally {}
      inputsb.close(append((char) c);
    }
    return sb.toString();
  }

}

The readLimitedLine() method defined above takes a numeric limit, indicating the total number of characters that may exist on one line. If a line contains more characters, the line is truncated, and they are returned on the next invocation. This prevents an attacker from exhausting memory by supplying input with no line breaks.

Compliant Solution (Java 1.7, limited file size)

This compliant solution impose a limit on the size of the file being read. This is accomplished with the Files.size() method which is new to Java 1.7. If the file is within the limit, we can assume the standard readLine() method will not exhaust memory, nor will memory be exhausted by the while loop.

class ShowHeapError {
  static public final int fileSizeLimit = 1000000;

  public ShowHeapError(String filename) throws IOException {
    if (Files.size(Paths.get(filename)) > fileSizeLimit) {  public static final int lineLengthLimit = 1024;
  public static final int lineCountLimit = 1000000;

  public void addNames() throws IOException {
    try {
      String newName;
      for (int i = 0; i < lineCountLimit; i++) {
        newName = readLimitedLine(reader, lineLengthLimit);
        if (newName == null || newName.equalsIgnoreCase("quit")) {
          break;
        }
        names.addElement(newName);
      throw new IOExceptionSystem.out.println("Fileadding " too+ large"newName);
      }
    } finally {
   this.input = new FileReaderinput.close(filename);
    this.reader = new BufferedReader(input);}
  }

  // ...other methods
}

Noncompliant Code Example

In a server-class machine using a parallel garbage collector, the default initial and maximum heap sizes are as follows for Java SE 6 \[[Sun 2006|AA. Bibliography#Sun 06]\]:

The readLimitedLine() method takes a numeric limit, indicating the total number of characters that may exist on one line. If a line contains more characters, the line is truncated, and the characters are returned on the next invocation. This prevents an attacker from exhausting memory by supplying input with no line breaks.

Noncompliant Code Example

In a server-class machine using a parallel garbage collector, the default initial and maximum heap sizes are as follows for Java SE 6 [Sun 2006]:

• Initial initial heap size: larger of 1/64th 64 of the machine's physical memory on the machine or some reasonable minimum.
• maximum Maximum heap size: smaller of 1/4th 4 of the physical memory or 1GB.

This noncompliant code example requires more memory on the heap than is available by default.:

/** Assuming the heap size as 512 MB 
 * (calculated as 1/4th4 of 2 GB2GB RAM = 512 MB512MB)
 *  Considering long values being entered (64 bits each, 
 * the max number of elements
 *  would be 512 MB/64bits512MB/64 bits = 
 * 67108864)
 */
public class ShowHeapErrorReadNames {
  Vector<Long> // Accepts unknown number of records
  Vector<Long> names = new Vector<Long>(); // Accepts unknown number of records
  long newID = 0L;
  int count = 67108865;
  int i = 0;
  InputStreamReader input = new InputStreamReader(System.in);
  Scanner reader = new Scanner(input);

  public void addNames() {
    try {
      do {
        // Adding unknown number of records to a list
        // The user can enter more IDs than the heap can support and as a result,
        // as a result, exhaust the heap. Assume that the record ID
        // is a 64 -bit long value

        System.out.print("Enter recordID (To quit, enter -1): ");
        newID = reader.nextLong();

        names.addElement(newID);
        i++;
      } while (i < count || newID != -1);
    } finally {
      input.close();
    }
  }

  public static void main(String[] args) {
    ShowHeapErrorReadNames demo = new ShowHeapErrorReadNames();
    demo.addNames();
  }
}

Compliant Solution

A simple compliant solution is to reduce the number of names to read.

  // ...
  int count = 10000000;
  // ...

Compliant Solution

The {{OutOfMemoryError}} can be avoided by ensuring the absence of infinite loops, memory leaks, and unnecessary object retention. When memory requirements are known ahead of time, the heap size can be tailored to fit the requirements using the following runtime parameters \[[Java 2006|AA. Bibliography#Java 06]\]:

java -Xms<initial heap size> -Xmx<maximum heap size>

For example,

java -Xms128m -Xmx512m ShowHeapError

Here the initial heap size is set to 128 MB and the maximum heap size to 512 MB.

This setting can be changed either using the Java Control Panel or from the command line. It cannot be adjusted through the application itself.

Risk Assessment

Assuming that infinite heap space is available can result in denial of service.

Related Vulnerabilities

The Apache Geronimo bug described by GERONIMO-4224 results in an OutOfMemoryError exception thrown by the WebAccessLogViewer if the access log file size is too large, such as more than 200 MB.

Related Guidelines

Bibliography

to read:

  // ...
  int count = 10000000;
  // ...

Compliant Solution

The OutOfMemoryError can be avoided by ensuring the absence of infinite loops, memory leaks, and unnecessary object retention. When memory requirements are known ahead of time, the heap size can be tailored to fit the requirements using the following runtime parameters [Java 2006]:

java -Xms<initial heap size> -Xmx<maximum heap size>

For example,

java -Xms128m -Xmx512m ReadNames

Here the initial heap size is set to 128MB and the maximum heap size to 512MB.

These settings can be changed either using the Java Control Panel or from the command line. They cannot be adjusted through the application itself.

Risk Assessment

Assuming infinite heap space can result in denial of service.

Related Vulnerabilities

The Apache Geronimo bug described by GERONIMO-4224 results in an OutOfMemoryError exception thrown by the WebAccessLogViewer when the access log file size is too large.

Automated Detection

Related Guidelines

Bibliography

...

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