创建易受攻击的应用程序以进行内存损坏攻击

Question

我正在尝试创建易受攻击的应用程序来练习stack overflow和SEH漏洞，但是我得到了这段代码。我的主要问题或困扰是，如何在代码中添加SEH漏洞并弹出计算器？我的原始代码很容易发生堆栈溢出，但动作最多。可以触发jackpot()函数来完成它，该如何解决？

int game(int user_pick) {
    int rand_pick;
    if ((user_pick > 0 && user_pick <= 32000)) {
        printf("Playing the game of chance..\n");
        rand_pick = (rand() % 32000) + 1;
        printf("You picked: %d\n", user_pick);
        printf("Random Value: %d\n", rand_pick);

        if (user_pick == rand_pick)
            jackpot();
        else
            printf("Sorry, you didn't win this time..\n");
    }
    else {
        printf("You must pick a value from 1 - 32000\n");
        printf("Use help or -h for help\n");
        return 0;
    }
}

int jackpot() {
    printf("You just won!\n");
    printf("Congratulations!\n");
    return 0;
}

void foo(char* input) {
    int(*function_ptr) (int user_pick);
    char buffer[20];
    srand(time(NULL));
    function_ptr = game;
    strcpy(buffer, input);
    if ((!strcmp(buffer, "help")) || (!strcmp(buffer, "-h"))){
        printf("Help Text:\n\n");
        printf("This is a game of chance.\n");
        printf("To play, simply guess a number 1 through 32000\n");
        printf("If you guess the number I am thinking of you win.\n");
    }
    else
        function_ptr(atoi(buffer));
}

int main(int argc, char* argv[]) {
    if (argc < 2) {
        printf("Usage: %s <a number 1 - 32000>\n", argv[0]);
        printf("use %s help or %s -h for more help.\n", argv[0], argv[0]);
        exit(0);
    }
    foo(argv[1]);
    return 0;
}