any ideas (integer encryption)

I'm trying to come up with a way to take a buffer of integers such as the numbers '123456' and translate them singly into their character reprenstation form....


for use with network hand shake of a server/client I am creating.

I've tried quite a few ways of looping through each index of a character array to do this, but to no avail....

Code:

---

int CreateHandShake(char *buf)
{
        char data[10];
        char a22[1];
        char *b22;
        strcpy(data, "123456");
        b22 = data;
        for(int cnt=0; cnt < 4; cnt++)
                        {
                strncpy(a22,b22,1);
                b22 + 1;
                strcpy(buf,a22);
                strcpy(a22, "");
                }
  return 0;
}

---

any ideas? anyone?

Quote:

---

Originally posted by TokersBall_CDXX
I'm trying to come up with a way to take a buffer of integers such as the numbers '123456' and translate them singly into their character reprenstation form....


for use with network hand shake of a server/client I am creating.

I've tried quite a few ways of looping through each index of a character array to do this, but to no avail....

Code:

---

int CreateHandShake(char *buf)
{
        char data[10];
        char a22[1];
        char *b22;
        strcpy(data, "123456");
        b22 = data;
        for(int cnt=0; cnt < 4; cnt++)
                        {
                strncpy(a22,b22,1);
                b22 + 1;
                strcpy(buf,a22);
                strcpy(a22, "");
                }
  return 0;
}

---

any ideas? anyone?

---

not sure what you're trying to do here, but
b22 + 1;
does nothing, you probably mean ++b22;
strncpy(a22,b22,1);
is the same as
*a22=*b22;

to run through each number in the buffer, and translate it into it's character representation.

I was using strncpy() to return 1 character from the buffer..

hmm still stumped..

Code:

---

int CreateHandShake(char *buf, HWND hDlg2)
{
        char data[10];
        char a22[1];
        int c22;
        char *b22;
        strcpy(data, "123156");
        b22 = data;
        for(int cnt=0; cnt < 6; cnt++)
                {
                c22=(int)data[cnt];
                buf[cnt] = (char)c22;
                }
  return 0;
}

---

doesn't translate the character though
buf = 123156
instead of the character representation?

hmmm...
any ideas?

Use atoi() defined in the <cstdlib> header to convert from text to int.

Quote:

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Originally posted by transcendental
Use atoi() defined in the <cstdlib> header to convert from text to int.

---

Code:

---

{
        char data[10];
        char a22[1];
        int c22;
        strcpy(data, "123156");
        for(int cnt=0; cnt < 6; cnt++)
                {
                c22=atoi(data[cnt]);
                buf[cnt] = (char)c22;
                }
  return 0;
}

---

gives me an error because data[cnt] isn't a buffer/string ending with /0....

Error example.c: 423 type error in argument 1 to `atoi'; found `char' expected `pointer to char'
Compilation + link time:3.5 sec, Return code: 1

I've already tried that..hmmm

I've also tried using a pointer, and stepping the location of the memmory address incremented to each character of data.... also doesn't work.

Quote:

---

Originally posted by TokersBall_CDXX
doesn't translate the character though
buf = 123156
instead of the character representation?

hmmm...
any ideas?

---

:confused:

123156 is already the char representation since buf is a char array.

I think you are utterly confused with numbers and their ascii character representations. Whatever numbers you sees on your computer, let it be your web browser, VC IDE or anything, are already in their textual form. It is just that they don't display them inside double quotes. We need to put double quotes when programming is because we want the compiler to know that literal is a string. And when compiling, compilers automatically convert for integer or float , from the textual form we type in our source to their equivalent binary form.

An simple example to explain

Code:

---

int i=20;
cout<< i <<endl;// display 20

---

The 20 displayed on the console is text converted from int by cout.

Another explanation

123156 is text. It takes up 6 bytes.

1234563 is text. It takes up 7 bytes.

Integer type only takes up 4 bytes(32 bit).

If you have the character '1' and you want the integer value 1, you can simply do it like so (assuming all your numbers are 0-9):

Code:

---

char c = '1';
int i = c - '0';  // equivalent to  '\1'
cout << "i is " << i << endl;

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of course, if you're just displaying it there's no need for this, but i gathered from your question that you want to perform mathematical operations on it.