Java code that deals with input, for example Scanner
, often buffers its underlying InputStream
.
It is possible to create multiple wrappers around an InputStream
that buffer the input from that InputStream
. Such programs behave significantly differently depending on whether the InputStream
does or does not allow for read ahead. An adversary can exploit this difference in behavior by, for example, redirecting the InputStream
, which could lead to exploitable behavior.
Although the Java standard does not specifically mention this behavior, code compiled with javac
and run with the java
command exhibits this behavior on Java 1.5.0.
Do not create multiple wrappers that buffer their input on an InputStream
; create and use only one, either by passing it as an argument to the methods that need it or centralizing its use in a single place.
Noncompliant Code Example
This noncompliant code example creates multiple BufferedInputStreams
on System.in
, because each time getChar()
is called it makes a new BufferedInputStream
. Note that it does not matter that the old BufferedInputStream
will have "expired" by the time the new one is created. Note that while this code uses a BufferedInputStream
to illustrate that any buffered wrapper is unsafe, it is also exploitable if a Scanner
is used instead.
import java.io.BufferedInputStream; import java.io.EOFException; public final class InputLibrary { public static char getChar() throws EOFException { BufferedInputStream in = new BufferedInputStream(System.in); int input = in.read(); if (input == -1) { throw new EOFException(); } return (char)input; //okay because InputStream guarantees read() fits in a byte if it is not -1 } public static void main(String[] args) { try { System.out.print("Enter first initial: "); char first = getChar(); System.out.println("Your first initial is " + first); System.out.print("Enter last initial: "); char last = getChar(); System.out.println("Your last initial is " + last); } catch(EOFException e) { System.out.println("ERROR"); } } }
Implementation Details
This program was compiled with the command javac InputLibrary.java
on a system with Java 1.5.0. When run from the command line with java InputLibrary
, the program will successfully take two characters as input and print them out. However, when run with java InputLibrary < input
, where input
is a file that contains the exact same input, the program prints "ERROR" because the second call to getChar()
finds no characters to read.
Compliant Solution
Create and use only a single BufferedInputStream
on System.in
. This code example stores the BufferedInputStream
as a class variable so all methods can access it. However, if a program were to use this library in conjunction with other input from a user that also needs some buffered wrapper on System.in
, the library would need to be modified so that all code uses the same buffered wrapper instead of creating separate ones.
import java.io.BufferedInputStream; import java.io.EOFException; public final class InputLibrary { private static BufferedInputStream in = new BufferedInputStream(System.in); public static char getChar() throws EOFException { int input = in.read(); if (input == -1) { throw new EOFException(); } in.skip(1); //This statement now necessary to go to the next line //the Noncompliant code example deceptively worked without it return (char)input; //okay because InputStream guarantees read() fits in a byte if it is not -1 } public static void main(String[] args) { try { System.out.print("Enter first initial: "); char first = getChar(); System.out.println("Your first initial is " + first); System.out.print("Enter last initial: "); char last = getChar(); System.out.println("Your last initial is " + last); } catch(EOFException e) { System.out.println("ERROR"); } } }
Implementation Details
This program was compiled with the command javac InputLibrary.java
on a system with Java 1.5.0. When run from the command line with java InputLibrary
, the program will successfully take two characters as input and print them out. Also, when run with java InputLibrary < input
, where input
is a file that contains the exact same input, the program will successfully take two characters as input and print them out.
Risk Assessment
Creating multiple buffered wrappers on an InputStream
can cause unexpected program behavior when the InputStream
is re-directed.
Rule |
Severity |
Likelihood |
Remediation Cost |
Priority |
Level |
---|---|---|---|---|---|
FIO39-J |
low |
unlikely |
medium |
P2 |
L3 |
Automated Detection
TODO
Related Vulnerabilities
Search for vulnerabilities resulting from the violation of this rule on the CERT website.
References
[[API 06]] method read
[[API 06]] class BufferedInputStream