Fenton`s Data Mark Machine

written by: Fred Foster; article published: year 2007, month 04;


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Fenton 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.

  1. The increment instruction

    x := x + 1

    is equivalent to

    if PC <= x then x := x + 1; else skip

  2. 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.

  3. 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.

  4. 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.

  5. 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.

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.

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