Page History

A Java heap space error occurs if infinite OutofMemoryError occurs when the program attempts to use more heap space is assumed and the program can crash. This error can be generated due to the following possible reasons:

1. A memory leak

2. An infinite loop

3. The program requires more memory than is present by default in the heap

 Non Compliant Code Example

A heap error will be generated if the heap is continued to be accessed even if there is no memory left in the heap.

than is available. Among other causes, this error may result from the following:

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

Some of these causes are platform-dependent and difficult to anticipate. Others, 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 ReadNames {
  private Vector<String> names = new Vector<String>();
  private final InputStreamReader input;
  private final BufferedReader reader;

  public ReadNames(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"))) {
        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;
    }
    ReadNames demo = new ReadNames(args[0]);
    demo.addNames();
  }
}

The 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 2014], the BufferedReader.readLine() 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.

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 a limit on the size of 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 {
  // ... Other methods and variables

  public static final int fileSizeLimit = 1000000;

  public ReadNames(String filename) throws IOException {
    long size = Files.size( Paths.get( filename));
    if (size > fileSizeLimit) {
      throw new IOException("File too large");
    } else if (size == 0L) {
      throw new IOException("File size cannot be determined, possibly too large");
    }
    this.input = new FileReader(filename);
    this.

import java.io.*;
import java.util.*;

public class ShowHeapError {

    Vector<String> names = new Vector<String>();
    String newName=null;
    InputStreamReader input = new InputStreamReader(System.in);
    BufferedReader reader = new 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 voidstatic String addNames(){
    	doreadLimitedLine(Reader reader, int limit) 
                                       throws IOException {
    StringBuilder sb = new StringBuilder();
    for (int i = 0; i < limit; i++) {
    		//adding unknown number of records to a list  int c = reader.read();
      if (c == -1) {
    		System.out.print(" To quit, enter \"quit\"\nEnter record: ");
          	try {
          		newName = reader.readLine(    return ((sb.length() > 0) ? sb.toString() : null);
      }
      if (((char) c == '\n') || ((char) c == '\r')) {
        break;
      }
      sb.append((char) c);
    }
    return sb.toString();
  }

  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  		if(!(newName == null || newName.equalsIgnoreCase("quit")) {
          break;
        }
        names.addElement(newName);
        System.out.println("adding " + newName);
      }
    } finally {
      input.close();
    }
  }

}

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 heap size: larger of 1/64 of the machine's physical memory or some reasonable minimum.
• Maximum heap size: smaller of 1/4 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 1/4 of 2GB RAM = 512MB)
 * Considering long values being entered (64 bits each, 
 * the max number of elements would be 512MB/64 bits = 
 * 67108864)
 */
public class ReadNames {
  // Accepts unknown number of records
  Vector<Long> names = new Vector<Long>(); 
  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, exhaust the heap. Assume that the record ID			//names are continued to be added without bothering about the size on the heap
          			names.addElement(newName);
          		}
   			} catch (IOException e) {
   			}
        // is a 64-bit long value
        System.out.println(newName.print("Enter recordID (To quit, enter -1): ");
        newID = reader.nextLong();

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

  }

  public static void main(String[] args) {
    ReadNames     ShowHeapError 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

This exception The OutOfMemoryError can be avoided by either making sure that there are no ensuring the absence of infinite loops or , memory leaks. If the programmer knows that the application would require a lot of memory then, he can increase the heap size in Java using the following run time parameters, 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 For example:,

java -Xms128m -Xmx512m ShowHeapErrorReadNames

Here we have set the initial  initial heap size as 128Mb is set to 128MB and the maximum heap size as 512Mbto 512MB.

This setting These settings can be done changed either in using the Java Control Panel or on from the command line. This setting They cannot be controlled in adjusted through the application itself.

Risk Assessment

Assuming infinite heap space can result in denial of service.

 Automated Detection

TODO

Related Vulnerabilities

Search for vulnerabilities resulting from the violation of this rule on the CERT website

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

...

Image Added Image Added Image Added

...