The International Obfuscated C Code Contest

A 27^th IOCCC Winner

Best of show - abuse of libc

Nicholas Carlini nicholas@carlini.com
https://nicholas.carlini.com

The code for this entry can be found in prog.c

Judges' comments:

To use:

make
./prog

Try:

# Play with a friend, P1 and P2, squares are numbered 1..9
./prog

Selected Judges Remarks:

This program consists of a single print statement wrapped in a while loop. You would not think that this would amount to much, but you would be very, very wrong. A clue to what is happening and how this works is encoded in the ASCII art of the program source.

Author’s comments:

HOW ABOUT A NICE GAME OF ~~CHESS~~TIC-TAC-TOE?

USAGE

cc -o prog prog.c
./prog

Alternates between P1 and P2. Enter a digit [1-9] to move:

1 | 2 | 3
---------
4 | 5 | 6
---------
7 | 8 | 9

The game ends if:

A player completes three in a row; that player wins
All squares are taken; neither player wins
A player makes an illegal move; their opponent wins

LIMITATIONS

Bad things may happen if the entered move is outside of the range [1..9]

While the game will stop if one player tries to overwrite another player’s move, it will happily allow a player to move on top of themself. However, given that this is an effective no-op and guarantees a loss to a competent opponent, if you do this you probably deserve to lose.

Your terminal may beep at you every turn. We’ll call this a feature that … the program … is hard at work processing your move! That is! Please be patient.

After macro expansion the format string is roughly 100K. This makes some compilers unhappy, and others slow. (For example, passing GCC the flags -Wall -Wextra -pedantic will result in a warning about oversized strings.)

Under some terminals the program may exit early. The following should always work:

echo "1 2 3 4 5 6 7" | ./prog

OBFUSCATION

The entirety of the program consists of a single iterated call to printf.

int main() {
    while(*d) printf(fmt, arg);
}

Here, fmt is a single string, and arg is a series of arguments to printf.

While its primary purpose is to serve as The One True Debugger, printf also happens to be Turing complete. (See “Control-Flow Bending: On the Effectiveness of Control-Flow Integrity” where we introduced this in an actual, published, academic paper. The things you can get away with sometimes.)

We abuse this fact to implement the logic of tic-tac-toe entirely within this one printf call (and a call to scanf() to read user input).

Here’s (briefly) how it works.

Preliminaries

This program uses three printf format specifiers. - %d takes an integer argument and prints it - %s takes a string argument and prints it - %n takes a pointer and writes (!!) the number of bytes printed so far.

Okay, everyone probably knows this. Let’s get a bit more advanced.

Format specifiers can take extra “arguments”. - "%hhn": store the number of bytes written mod 256 to the char pointer - "%2$d": print argument 2 to printf (and not the sequentially next argument) - "%8d": pad the printed integer to 8 characters - "%3$.*4$d": print argument 3 to printf with as many zeros as in argument 4.

For example, the following expression

printf("%1$.*2$d%3$hhn", 5, 10, &x)

will have the same effect as if we had written

x = 10;

because it will print out 0000000005 (5 padded to size 10) and then write the number of bytes written to x.

Printf Oriented Programming

Alright, now we can get to the real fun.

We perform arbitrary computation with printf treating memory as a binary array—one bit per pair of bytes: - The zero bit is represented by the sequence 00 00 - The one-bit is represented by the sequence xx 00 where xx is any non-zero byte.

We can use format strings to compute the OR/NOT of arbitrary “bits”.

We’ll start with the simplest, OR:

printf("%1$s%2$s%3$hhn", a, b, c)

will compute

*c = strlen(a) + strlen(b)

but given that strlen(x) is 1 for a 1-bit and 0 for a 0-bit, we have

*c = a | b

Computing the NOT of a single value is also easy:

printf("%1$255d%1$s%hhn", a, b)

will compute

*b = (strlen(a)+255)%256 = strlen(a)-1

and again, because strlen(x) is either 1 or 0 we have

*c = !b

From here we can compute any binary circuit. Doing something efficient, though, still takes work.

Tic-Tac-Toe

The game itself is represented as a board of 18 bits, 9 bits per player, along with a turn counter that alternates between player 1 and player 2.

To detect who has won, we implement the following logic. Let A, B, and C be pointers to three squares in a row to test, and D be where to save if there is a win or not.

"%A$s%B$s%C$s%1$253d%11$hhn" // r11 = !(*A & *B & *C)
ZERO
"%11$s%1$255d%11hhn" // r11 = !r11
ZERO
"%11$s%D$s%D$hhn" // *D = *D | r11

That is, we set *D to 1 if there is a three-in-a-row. We repeat this for all possible three-in-a-row configurations, for both players.

The ZERO macro ensures that the number of bytes written out is 0 mod 256 with the following expression

"%1$hhn%1$s" (repeated 256 times)

where argument 1 is a pointer to a temporary variable followed by a null byte.

This works because if the current count is 0 mod 256, then “%1$hhn” will write zero to argument 1 and then “%1$s” will never emit any text. If, on the other hand, the count is not 0 mod 256, a length-1 string will be written to argument 1, and then “%1$s” will increment the count by one. By repeating this 256 times we’re eventually going to reach 0 mod 256.

Checking if there has been an invalid move is achieved similarly.

In order to decide what to print out, we have to cast the “in-memory” array of bits to Xs and Os to print out. This is actually rather straightforward. Given in 1$ the pointer to player 1’s square, and 2$ the pointer to player 2’s, and in 3$ the pointer to the board string, we can compute

"%1$s" (repeated 47 times) "%2$s" (repeated 56 times) %1$32d%3$hhn"

which will, in effect, compute

*r3 = (*r1) * 47 + (*r2) * 56 + 32

which will output ‘ ’ if neither are true, ‘X’ if r1 is, or ‘O’ if r2 is.

Further Obfuscations

In order to be able to finally display the board, while still only using one printf statement, we finish the statement with

"\n\033[2J\n%26$s"

which is the escape sequence to clear the screen, and then prints argument 26. Argument 26 is a pointer to a char* in memory, that initially is undefined, but within the printf statement we will construct this string to look like a tic-tac-toe board.

After the board, we need to print one of the following strings:

P1>_
P2>_
P1 WINS
P2 WINS
P1 TIES
P2 TIES

Depending on if it’s P1 or P2’s turn to move, the game is over and someone won, or the game is over and it is a draw.

This turns out not to be as hard as it might look. Using the same trick as before, we set byte four to be

*byte4 = is_win * 'W' + is_tie * 'T'

The byte 'I' and 'S' can always be the same, and we do the same for 'E'/'N'.

We do this same on-the-fly creation of the scanf() format string, but for a different reason. We first want to run printf() to show the first board, and then alternate between runs to scanf() and printf() reading and then displaying moves. importantly, we do not want a final scanf when the game ends. It should just exit.

One option would be to implement the logic as

printf()
while (*ok) {
    scanf();
    printf();
}

but this would DOUBLE the number of calls to printf we require. So instead we implement it like this

while (*ok) {
    scanf();
    printf();
}

(In reality we actually pass scanf() as an argument to avoid the extra statement, but it has the same effect.)

Notice there is now no initial printf(). In order make sure the program doesn’t block before the first printf(), but we initialize the scanf() format to the null string so that it returns right away without blocking. The first time the printf() call runs, it writes out “%hhd” to create the scanf() format string.