C++ Primer Plus Chapter 7 Exercise 9

c plus plusChapter 7 ends with a short exercise on function pointers. The simplest way to complete this, which I provided in my source, is to create the two functions add() and calculate(), then call them in main() with a loop. The text mentions if you are feeling adventurous to include other functions in addition to add. For simplicities sake, I did the minimum requirement here. Finally,  we can put this chapter to rest. See my source below:

9. Design a function calculate() that takes two type double values and a pointer to a function that takes two double arguments and returns a double. The calculate() function should also be type double, and it should return the value that the pointed-to function calculates, using the double arguments to calculate(). For example, suppose you have this definition for the add() function: double add(double x, double y){return x + y}. Then, the function call in double q = calculate(2.5, 10.4, add); would cause calculate() to pass the values 2.5 and 10.4 to the add() function and then return the add() return value (12.9). Use these functions and at least one additional function in the add() mold in a program. The program should use a loop that allows the user to enter pairs of numbers. For each pair, use calculate() to invoke add() and at least one other function. If you are feeling adventurous, try creating an array of pointers to add()-style functions and use a loop to successively apply calculate() to a series of functions by using these pointers. Hint: Here’s how to declare such an array of three pointers: double (*pf[3])(double, double); You can initialize such an array by using the usual array initialization syntax and function names as addresses.

#include <iostream>

using namespace std;

double calculate(double x, double y, double (*pf)(double a, double b));
double add(double x, double y);

int main()
{
double x, y;
cout << "Enter two numbers (q to exit): ";
while(cin >> x >> y)
{
if(cin.fail())
break;
calculate(x, y, add);
cout << "The sum is: " << calculate(x, y, add) << "\n\n";
}
return 0;
}

double calculate(double x, double y, double (*pf)(double a, double b))
{
return pf(x, y);
}

double add(double x, double y)
{
return x + y;
}

C++ Primer Plus Chapter 7 Exercise 8

c plus plus Exercise 8 has us re-writing a skeleton program to fill in the missing functions. A combination of structs, pointers, and functions will complete this exercise. Comments describing where and what additions have been made are embedded in the code. See my solution below:

8. This exercise provides practice in writing functions dealing with arrays and structures. The following is a program skeleton. Complete it by providing the described functions:


#include <iostream>

using namespace std;
const int SLEN = 30;
struct student {
char fullname[SLEN];
char hobby[SLEN];
int ooplevel;
};

int getinfo(student pa[], int n);
void display1(student st);
void display2(const student * ps);
void display3(const student pa[], int n);

int main(){
cout << "Enter class size: ";
int class_size;
cin >> class_size;
while (cin.get() != '\n')
continue;
student * ptr_stu = new student[class_size];
int entered = getinfo(ptr_stu, class_size);

for (int i = 0; i < entered; i++) {
display1(ptr_stu[i]);
display2(&ptr_stu[i]);
}
display3(ptr_stu, entered);
delete [] ptr_stu;
cout << "Done\n";

return 0;
}

// getinfo() has two arguments: a pointer to the first element of
// an array of student structures and an int representing the
// number of elements of the array. The function solicits and
// stores data about students. It terminates input upon filling
// the array or upon encountering a blank line for the student
// name. The function returns the actual number of array elements
// filled.
int getinfo(student pa[], int n)
{
int i;
for(i = 0; i < n; i++)
{
cout << "Student Number: " << (i+1) << "\n";
cout << "Student's Name: ";
cin.getline(pa[i].fullname, SLEN);

cout << "Enter students hobby: ";
cin.getline(pa[i].hobby, SLEN);

cout << "Enter students OOP level: ";
cin >> pa[i].ooplevel;
cin.get();
}
return i;
}

// display1() takes a student structure as an argument
// and displays its contents
void display1(student st)
{
cout << "\n=============Display 1================\n";
cout << "Student name: " << st.fullname << "\n";
cout << "Hobby: " <<  st.hobby << "\n";
cout << "OOP Level: " << st.ooplevel << "\n";
cout << "=============End Display 1==============\n";
}

// display2() takes the address of student structure as an
// argument and displays the structure's contents
void display2(const student * ps)
{
cout << "\n============Display 2================\n";
cout << "Students Name: " << ps->fullname << "\n";
cout << "Hobby: " << ps->hobby << "\n";
cout << "OOP Level" << ps->ooplevel << "\n";
cout << "==========End Display 2================\n";
}

// display3() takes the address of the first element of an array
// of student structures and the number of array elements as
// arguments and displays the contents of the structures
void display3(const student pa[], int n)
{
for(int i = 0; i < n; i++)
{
cout << "\n============Display 3================\n";
cout << "Student: " << (i+1) << "\n";
cout << "Fullname: " << pa[i].fullname << "\n";
cout << "OOP Level: " << pa[i].ooplevel << "\n";
cout << "=============End Display 3============\n";
}
}

Linux Password Cracker

python

python

It has been about a month since I have posted, that does not mean I have stopped coding. Lately I’ve been back on my “security” kick. Although for me it’s more of an obsession rather than just a kick. When it comes to security, a programming language like Python can make many common task a breeze to accomplish. Here I have a basic Linux password cracker that can crack the current SHA-512 shadowed hashes from a user supplied dictionary and detect whether a hash is the lesser used MD5 or SHA-256 format. Enjoy.

import crypt

def testPass(cryptPass):
hashType = cryptPass.split("$")[1]
if hashType == '1':
print "[+] Hash Type is MD5"
elif hashType == '5':
print "[+] Hash Type is SHA-256"
elif hashType == '6':
print "[+] Hash Type is SHA-512"
else:
"[+] Hash Type is Unknown"

salt = cryptPass.split("$")[2]
dictFile = open('dictionary.txt', 'r')
for word in dictFile.readlines():
word = word.strip('\n')
pepper = "$" + hashType + "$" + salt
cryptWord = crypt.crypt(word, pepper)
if cryptWord == cryptPass:
print '[+] Found Password: ' + word + '\n'
return
print '[-] Password Not Found.\n'
return

def main():
passFile = open('passwords.txt')
for line in passFile.readlines():
if ':' in line:
user = line.split(':')[0]
cryptPass = line.split(':')[1].strip(' ')
print '[*] Cracking Password For: ' + user
testPass(cryptPass)

if __name__ == '__main__':
main()

Renaming Network Interfaces with Udev Rules

linuxRecently I have noticed something while troubleshooting some network traffic issues. My wireless interface decided to rename itself to an obscure string presumably after an update to either Network Manager or udev. Specifically wlan0 was being renamed to wlp0s22f2u1 when it had previously been wlan0 for as long as I can remember. Digging around, I realized that I no longer had the file: /etc/udev/rules.d/70-persistent-net.rules. This file contains rules that udev will use to name network interfaces, but it was missing. Fortunately it is easy to create or alter if you want to rename network interfaces to stable names.

Let's first grab the hardware address of the problem interface:
ifconfig -a

You will see some output like so:


wlan0: flags=4163<UP,BROADCAST,RUNNING,MULTICAST> mtu 1492
        inet 192.168.2.2  netmask 255.255.255.0  broadcast 192.168.2.255
        inet6 fe80::2c0:caff:fe40:b198  prefixlen 64  scopeid 0x20<link>
        ether 00:01:02:03:04:05  txqueuelen 1000  (Ethernet)
        RX packets 4407  bytes 2748119 (2.6 MiB)
        RX errors 0  dropped 0  overruns 0  frame 0
        TX packets 6714  bytes 1423408 (1.3 MiB)
        TX errors 0  dropped 0 overruns 0  carrier 0  collisions 0

Amongst all the noise, we can see from our output of ifconfig that our hardware address is 00:01:02:03:04:05(ether or hwaddr line*). Now as root, create the file 70-persistent-net.rules with your favorite editor in /etc/udev/rules.d:

# vi /etc/udev/rules.d/70-persistent-net.rules

Now enter or copy/paste in this information. Remember, each rule needs to be on it's own line.

# wlan0
SUBSYSTEM=="net", ACTION=="add", DRIVERS=="?*", ATTR{address}=="00:01:02:03:04:05", ATTR{dev_id}=="0x0", ATTR{type}=="1", KERNEL=="wlan*", NAME="wlan0"

Now you can either reboot or attempt a udev restart. A reboot will ensure rules are reloaded, but you can try some of these commands on various distros:

# udevadm control --reload-rules
or
# udevadm trigger

Closing thoughts. The system does work perfectly fine without renaming interfaces to stable names. And quite possibly a future udev upgrade may wipe this rule out. But for anyone like me who likes clean interface names or anyone needing to rename interfaces for reasons out of context here, this is one way to do this on many Linux distros.

Bubble Sort Algorithm

The Bubble Sort algorithm is one the simplest sorting algorithm to implement. It works by continually looping through pairs in a list and comparing each pair. At each pair, it will perform a swap if the pair happens to be in the wrong order. The order will either be ascending or descending depending on what the programmer wants. This happens until no more swaps are needed. This algorithm apparently gets the name “bubble-sort” because you can imagine the smaller elements bubble to the top of the list. Checkout my bubble sort implementation below. When ran, it will give a good visual reference for the algorithm. It sorts in ascending order, complexity O(n^2). This algorithm approaches O(n) if the elements are already almost in order. Note that is the not the most efficient algorithm for large data sets.

#include <iostream>

using namespace std;

int main()
{
int array[7];
int i, j;

for(i = 0; i <= 6; i++){
cout << "Enter unsorted numbers: " << endl;
cin >> array[i];
}
for(i = 0; i <= 5; i++){ //loop through pair of integers
for(j = i+1; j <= 6; j++){
int temp;
if(array[i] > array[j])
{
temp = array[i]; //swapping mechanisms
array[i] = array[j];
array[j] = temp;
}
}
}
for(i = 0; i <= 6; i++){
cout << array[i]; //sorted output
}
return 0;
}

Fizz Buzz C++

c plus plusThis is a “classic” fizz-buzz solution I created. For those of you who don’t already know what fizz-buzz is, it follows this basic context: “Write a program that prints the numbers from 1 to 100. But for multiples of three print “Fizz” instead of the number and for the multiples of five print “Buzz”. For numbers which are multiples of both three and five print “FizzBuzz””. This question has been asked by recruiters as a method for determining programming competency.  While the fizz-buzz question is not really difficult itself, it does force you to think about one of the most important concepts to programmers, loop structures.


#include <iostream>

using namespace std;

int main(){
for (int i = 0; i <= 100; i++) {
if ((i % 15) == 0) //for multiples of both 3 and 5
cout << "FizzBuzz" << endl;
else if ((i % 3) == 0) //multiples of 3
cout << "Fizz" << endl;
else if ((i % 5) == 0) //multiples of 5
cout << "Buzz" << endl;
else
cout << i << endl; //all other numbers
}
return 0;
}

C++ Primer Plus Chapter 7 Exercise 7

c plus plus

This exercise makes use heavier use of pointers then what we have been doing. However, much of this program can be pulled from listing 7.7 and around that part of the chapter. See source below.

Redo Listing 7.7, modifying the three array-handling functions to each use two pointer parameters to represent a range. The fill_array() function, instead of returning the actual number of items read, should return a pointer to the location after the last location filled; the other functions can use this pointer as the second argument to identify the end of the data.

#include <iostream>

using namespace std;

const int Max = 5;

// function prototypes
double * fill_array(double *first, double *last);
void show_array(const double *first, const double *last);
void revalue(double *first, double *last, double factor);

int main()
{
double properties[Max];
double * last;

last = fill_array(properties, properties+Max);
show_array(properties, last);

cout << "Enter revaluation factor: ";
double factor;
cin >> factor;

revalue(properties, last, factor);
show_array(properties, last);
cout << "Done.\n";

return 0;
}

double *fill_array(double *first, double *last)
{
double temp;
double *pt;
int i=0;
for (pt = first; pt != last; pt++, i++)
{
cout << "Enter value #" << (i + 1) << ": ";
cin >> temp;
if (!cin) // bad input
{
cin.clear();
while (cin.get() != '\n')
continue;
cout << "Bad input; input process terminated.\n";
break;
}
else if (temp < 0) // signal to terminate
break;
*pt = temp;
}
return pt;
}

void show_array(const double *first, const double *last)
{
const double *pt;
int i=0;
for (pt = first; pt != last; pt++, i++)
{
cout << "Property #" << (i + 1) << ": $ \n";
cout << *pt;
}
}

void revalue(double *first, double *last, double factor)
{
double *pt;
for (pt = first; pt != last; pt++)
*pt *= factor;
}