C++ ENCRYPTION (how to encrypt a text)

[jupi13]

i got a problem wtih my encryption...
would you be kind to help me with this...i'm new to this c++ programing...

i have this expansion encryption method..where you will add a char in between the original text..example >>>>. josh..this could be >>>> jxoxsxhx...like that...

my program goes like this

main()
{
char string[10],count;
int x,y;

cout<<"ENTER TEXT:";
cin>>string;
count=strlen(st ring);

for(x=0;x<count ;x++)

cout<<string[x]<<string[2];

//if i'll enter "JOSH" the output would be "JSOSSSHS"
//if i'll enter "jupi" the output would be "jpupppip"

getche();
}



//what i need to do is to store the output in another array.
example. array[] will be equal to the cout in the program above which is
string[x] and string[2]...>>>>> array[x]="string[x]" and "string[2]"

how will i do that?.....
how will i do it?
please help everyone..

[Brosert]

you could try Something like

// Obviously you have to make sure that the original string is less than 10 chars....

char string2[sizeof(string*2)];
int count2;

for(int x = 0; x<count; x++)
{
  count2 = 2*x;
  string2[count2]=string[x];
  string2[coount2 + 1] = string[2];
}

Although I'm sure there are far more elegant ways to do this

[gpraghuram]

This dosent look encryption as your original characters remains as it is with newly added characters.
Try using Blow-Fish for encrypting the string and the blow-fish is available on the net.

Raghu

[jupi13]

you could try Something like

// Obviously you have to make sure that the original string is less than 10 chars....

char string2[sizeof(string*2)];
int count2;

for(int x = 0; x<count; x++)
{
  count2 = 2*x;
  string2[count2]=string[x];
  string2[coount2 + 1] = string[2];
}

Although I'm sure there are far more elegant ways to do this

what will i output in this code?
string2?
how will i do it?
please help?

if you don't mind..can you give the whole code?
im just a newbie to C++;;


thanks..

[Brosert]

If you want to output something at the end of it, you can still use cout....

eg

cout << string2;

This code shoudl fit nicely into your current code...
Where you declare string and count, also declare string and count2...

where your forloop is, try the forloop in the original response.

[jupi13]

ehhyyy....thank s lot...it worked..

hehehe
...thanks man...

ahm,,if you don't mind,can you help me with this also?

if i'll enter "brosert" the output would be "boet" and the missing letters ("rsr")
would be stored in another array....

[Brosert]

This time...

Set up 3 arrays (you could probably do it with 2, but that would be more complex....), loop through, if we are on an even iteration store to one array, otherwise store to the other.
I'll assume you have three equal sized arrays called string1, string2, string3...
string1 recieves input (as before) in a null terminated string

//Create a pointer to the location in the string that we're up to -> this means we
// will still be able to access the beginning of any string we need.
char *strin1ptr = string1;
char *string2ptr = string2;
char *string3ptr = string3; 

bool is2 = true;
//loop while we have more characters in string1
while(string1ptr)
{
  // if we are on an even character
  if(is2)
  {
    // set this location to contain the character in string1
    *string2ptr = *string1ptr;
     //bump string2ptr along
     string2ptr++;
  }
  else
    //likewise for string3
    *string3ptr = string1ptr;
     string3ptr++;
  }
  //bump string1ptr along too
  string1ptr++;
}

cout << string1 <<endl;
count << string2 <<endl;
count << string3 <<endl;

[jupi13]

can you do it without using a pointer?

we're not yet using pointers...

we are only allowed to use array..and loop..

can you do it?
please...

[Brosert]

in that case, use a for loop....imnstea d of the while loop

eg

int count = sizeof(string1);
for(int x=0; x<count; x++)
{
  /Check whether x is a multiple of 2
  if(x%2 == 0)
  {
    string2[x/2] = string1[x];
  }
  else
  {
    string3[x/2] = string1[x];
}

[jupi13]

\brosert....... .....

//the program you give me only works in a 6-letter word and a 7-letter word...


//here it is


main()
{


char string[10],count,string2[20];
int count2;

cout<<"ENTER TEXT:";
cin>>string;
count=strlen(st ring);

for(int x=0;x<count;x++ )
{
count2 = 2*x;
string2[count2]=string[x];
string2[count2 + 1] = string[2];

}
cout<<string2;

getche();

//what's wrong with it.....my problem is that i want to expand the text...it should
work on text with 1-10 characters.....
if "i" is entered it would output.. "ii"...
if "you is entered it would output "yuouuu"... like that..
if "abcd" is enterd it would output "acbcccdcec "...

please help me...
please...

[Brosert]

Part of the problem is that it always takes the THIRD character....
If your string is shorter than 3 characters, it takes a (probably) NULL character, which terminates the string, and would give you a single character output....

where you get string try making the following modification :

int charAt =2;

count=strlen(string);
//If we have a short input, take the last letter instead of the third
if(count <charAt) count=charAt;

then at the point where you actually assign the value, use
string[charAt]
instead of
string[2]

[jupi13]

in that case, use a for loop....imnstea d of the while loop

eg

int count = sizeof(string1);
for(int x=0; x<count; x++)
{
  /Check whether x is a multiple of 2
  if(x%2 == 0)
  {
    string2[x/2] = string1[x];
  }
  else
  {
    string3[x/2] = string1[x];
}

i also have a question brosert....why is it that the index of string2 and string3
is "x/2"? why is it not x?

[Brosert]

because only half of the characters go in each string....
so string2 and string3 will be half the size of string1.

In c (and I think most languages) if we divide a number in a place where we can only use an integer, it will autometically round the number down....
so 0/2 = 0 (obviously)
but 1/2 = 0 (because we can only use the integer part of the answer).

this means, that if we always take the current value/2, we will always be entering into the correct box of each string (might need to think about that for a second....)

it means if string1 has 7 characters,
the ones in position 0,2,4,6 will become 0,1,2,3 in string2
the ones in position 1,3,5 will become 0,1,2 in string3

If you need a better explanation, I can try again....

[jupi13]

brosert.......

this one also...it only works with letters with 4 characters above..


int count = sizeof(string1) ;
for(int x=0; x<count; x++)
{
/Check whether x is a multiple of 2
if(x%2 == 0)
{
string2[x/2] = string1[x];
}
else
{
string3[x/2] = string1[x];
}


waht should i do?

i can't figure it out..
please help..

[Brosert]

Not real sure....
try changing
count = sizeof(string1) to = strlen(string1) (as you had in your first program....

Meanwhile I'll have a deeper look/think