- Basics and Background
- Compiler-Based Mechanisms
- Execution-Based Mechanisms
- Example Information Flow Controls
- Further Reading
- Exercises
This chapter is from the book
This chapter is from the book
This chapter is from the book
15.3 Execution-Based Mechanisms
The goal of an execution-based mechanism is to prevent an information flow that violates policy. Checking the flow requirements of explicit flows achieves this result for statements involving explicit flows. Before the assignment
y = f(x1, ..., xn)
is executed, the execution-based mechanism verifies that
-
lub(x1, ..., xn) ≤ y
If the condition is true, the assignment proceeds. If not, it fails. A naïve approach, then, is to check information flow conditions whenever an explicit flow occurs.
Implicit flows complicate checking.
EXAMPLE: Let x and y be variables. The requirement for certification for a particular statement y op x is that x ≤ y. The conditional statement
if x = 1 then y := a;
causes a flow from x to y. Now, suppose that when x ≠ 1, x = High and y = Low. If flows were verified only when explicit, and x ≠ 1, the implicit flow would not be checked. The statement may be incorrectly certified as complying with the information flow policy.
Fenton explored this problem using a special abstract machine.
15.3.1 Fenton's Data Mark Machine
Fenton [313] created an abstract machine called the Data Mark Machine to study handling of implicit flows at execution time. Each variable in this machine had an associated security class, or tag. Fenton also included a tag for the program counter (PC).
The inclusion of the PC allowed Fenton to treat implicit flows as explicit flows, because branches are merely assignments to the PC. He defined the semantics of the Data Mark Machine. In the following discussion, skip means that the instruction is not executed, push(x, x) means to push the variable x and its security class x onto the program stack, and pop(x, x) means to pop the top value and security class off the program stack and assign them to x and x, respectively.
Fenton defined five instructions. The relationships between execution of the instructions and the classes of the variables are as follows.
-
The increment instruction
x := x + 1
is equivalent to
if PC ≤ x then x := x + 1; else skip -
The conditional instruction
if x = 0 then goto n else x := x – 1
is equivalent to
if x = 0 then { push(PC, PC); PC = lub(PC, x); PC := n; } else { if PC ≤ x then { x := x – 1; } else skip }This branches, and pushes the PC and its security class onto the program stack. (As is customary, the PC is incremented so that when it is popped, the instruction following the if statement is executed.) This captures the PC containing information from x (specifically, that x is 0) while following the goto.
-
The return
return
is equivalent to
pop(PC, PC);This returns control to the statement following the last if statement. Because the flow of control would have arrived at this statement, the PC no longer contains information about x, and the old class can be restored.
-
The branch instruction
if' x = 0 then goto n else x := x – 1
is equivalent to
if x = 0 then { if x ≤ PC then { PC := n; } else skip } else { if PC ≤ x then { x := x – 1; } else skip }This branches without saving the PC on the stack. If the branch occurs, the PC is in a higher security class than the conditional variable x, so adding information from x to the PC does not change the PC's security class.
-
The halt instruction
halt
is equivalent to
if program stack empty then halt execution
The program stack being empty ensures that the user cannot obtain information by looking at the program stack after the program has halted (for example, to determine which if statement was last taken).
EXAMPLE: Consider the following program, in which x initially contains 0 or 1. [2]
1. if x = 0 then goto 4 else x := x – 1 2. if z = 0 then goto 6 else z := z – 1 3. halt 4. z := z + 1 5. return 6. y := y + 1 7. return
This program copies the value of x to y. Suppose that x = 1 initially. The following table shows the contents of memory, the security class of the PC at each step, and the corresponding certification check.
|
x |
y |
z |
PC |
PC |
stack |
certification check |
|
1 |
0 |
0 |
1 |
Low |
— |
|
|
0 |
0 |
0 |
2 |
Low |
— |
Low ≤ x |
|
0 |
0 |
0 |
6 |
x |
(3, Low) |
|
|
0 |
1 |
0 |
7 |
x |
(3, Low) |
PC ≤ y |
|
0 |
1 |
0 |
3 |
Low |
— |
Fenton's machine handles errors by ignoring them. Suppose that, in the program above, y ≤ x. Then at the fifth step, the certification check fails (because PC = x). So, the assignment is skipped, and at the end y = 0 regardless of the value of x. But if the machine reports errors, the error message informing the user of the failure of the certification check means that the program has attempted to execute step 6. It could do so only if it had taken the branch in step 2, meaning that z = 0. If z = 0, then the else branch of statement 1 could not have been taken, meaning that x = 0 initially.
To prevent this type of deduction, Fenton's machine continues executing in the face of errors, but ignores the statement that would cause the violation. This satisfies the requirements. Aborting the program, or creating an exception visible to the user, would also cause information to flow against policy.
The problem with reporting of errors is that a user with lower clearance than the information causing the error can deduce the information from knowing that there has been an error. If the error is logged in such a way that the entries in the log, and the action of logging, are visible only to those who have adequate clearance, then no violation of policy occurs. But if the clearance of the user is sufficiently high, then the user can see the error without a violation of policy. Thus, the error can be logged for the system administrator (or other appropriate user), even if it cannot be displayed to the user who is running the program. Similar comments apply to any exception action, such as abnormal termination.
15.3.2 Variable Classes
The classes of the variables in the examples above are fixed. Fenton's machine alters the class of the PC as the program runs. This suggests a notion of dynamic classes, wherein a variable can change its class. For explicit assignments, the change is straightforward. When the assignment
y := f(x1, ..., xn)
occurs, y's class is changed to lub(x1, . . , xn). Again, implicit flows complicate matters.
EXAMPLE: Consider the following program (which is the same as the program in the example for the Data Mark Machine). [3]
proc copy(x : integer class { x };
var y : integer class { y });
var z : integer class variable { Low };
begin
y := 0;
z := 0;
if x = 0 then z := 1;
if z = 0 then y := 1;
end;
In this program, z is variable and initially Low. It changes when something is assigned to z. Flows are certified whenever anything is assigned to y. Suppose y < x.
If x = 0 initially, the first statement checks that Low ≤ y (trivially true). The second statement sets z to 0 and z to Low. The third statement changes z to 1 and z to lub(Low, x) = x. The fourth statement is skipped (because z = 1). Hence, y is set to 0 on exit.
If x = 1 initially, the first statement checks that Low ≤ y (trivially true). The second statement sets z to 0 and z to Low. The third statement is skipped (because x = 1). The fourth statement assigns 1 to y and checks that lub(Low, z) = Low ≤ y (again, trivially true). Hence, y is set to 1 on exit.
Information has therefore flowed from x to y even though y < x. The program violates the policy but is nevertheless certified.
Fenton's Data Mark Machine would detect the violation (see Exercise 4).
Denning [239] suggests an alternative approach. She raises the class of the targets of assignments in the conditionals and verifies the information flow requirements, even when the branch is not taken. Her method would raise z to x in the third statement (even when the conditional is false). The certification check at the fourth statement then would fail, because lub(Low, z) = x ≤ y is false.
Denning ([242], p. 285) credits Lampson with another mechanism. Lampson suggested changing classes only when explicit flows occur. But all flows force certification checks. For example, when x = 0, the third statement sets z to Low and then verifies x ≤ z (which is true if and only if x = Low).