integer to hex (another question)

[mahmoodn]

consider this array:

  char key[16] = {0x00, 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07, 0x08, 0x09, 0x0A, 0x0B, 0x0C, 0x0D, 0x0E, 0x0F};

I want to put it in a loop such that all 16 element change in each iteration. Some thing like this:

for ( int i = 0; i < 100; i++ ) {
   [B]j = hex(i)[/B] [B][I]// what is this[/I][/B]
   char key[16] = {j, j+1, j+2, j+3, j+4, j+5, j+6, j+7, j+8, j+9, j+10, j+11, j+12, j+13, j+14, j+15};
}

How can I do that? what the conversion function? I saw many threads that they wanted to put a hex value on the output stream. I want a function. Does c++ provide such function?

Thanks,

[Oralloy]

Do you mean something like this?

#include <ios>

...

int value = 0x256;
cout << hex << value;

[mahmoodn]

I know that. I don't want to put the variable on the output stream. I want to convert an integer variable to a hex variable and store that.

[Oralloy]

Ummmm, do you mean as a string, or as another integer type?

// convert integer to hex string

#include <ios>
#include <strstream>

...

int value = 0x0256;
ostrstream hexOut;
hexOut << hex << value;

[mahmoodn]

> do you mean as a string, or as another integer type?

To something that I can use it the array

[Oralloy]

Then don't worry about making a conversion.

At the "machine level", integers are typically represented as an ordered sequence of bits. How that gets displayed on your monitor is quite another thing, entirely.

As for code, you can write any of these, they are equivalent:

int x = 0x0256;  // hex

int x = 01126;   // octal

int x = 598;     // decimal

Does that make sense?

[mahmoodn]

so if I have this:
char key[16]={0xA,...}

and want to put it in a loop
for ( i=0; i<100; i++ )
char key[16]={i,i+1,....}

but the result is:
key={0,1,2,..., 15}

however I want this:
key={0,1,2,..., 0xA,0xB,...,0xF }

[Oralloy]

It doesn't matter. Run this through your compiler and see what comes out....

#include <ios>
#include <iostream>

int main(void)
{
  int v0 = 0x0256;
  int v1 = 598;

  if (v0 == v1)
    cout << "wow, they're equal!" << endl;
  else
    cout << "drat, Oralloy made a mistake." << endl;

  cout << "v0(default) = " << v0 << endl;
  cout << "v1(default) = " << v1 << endl;

  cout << "v0(dec) = " << dec << v0 << endl;
  cout << "v0(hex) = " << hex << v0 << endl;

  cout << "v1(dec) = " << dec << v1 << endl;
  cout << "v1(hex) = " << hex << v1 << endl;

  return 0;
}

[mahmoodn]

I know that ostream works. it is fine. However I need another thing.

Again I have to say, this is my code
for ( i=0; i<100; i++ )
char key[16]={i,i+1,....}

I want the key to be
{0x1,0x2,0x3,0x 4,....,0xA,0xB, 0xC,0xD,0xE,0xF }

I don't want to use ostream.

[Oralloy]

@mahmoodn,

What I'm saying is that the representation of an integer is irrespective of how it is expressed as a constant in your code.

On (most) modern computers, the representation of an integer is as 32 binary bits, it's neither decimal or hexadecimal.

Did you run my little bit of code? Especially the "if" statement? The point of that was to demonstrate that "(0x0256 == 598)" is always true.

In other words, writing

char key[16] = {0x00, 0x01, 0x02, 0x03, 0x04, 0x05, 0x06, 0x07, 0x08, 0x09, 0x0A, 0x0B, 0x0C, 0x0D, 0x0E, 0x0F};

is functionally identical to

char key[16] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15};

Again, they're the same thing once compiled. They only look different in your source code file.

Does that make any sort of sense?

[mahmoodn]

Thanks.
I thought you are saying about ostream. I ran the code but didn't attention to the content of 'if'.

I now understand. :)

[Oralloy]

So, please help me - I know I have difficulty expressing myself on occasion, and would like your feedback.

In my third post, starting with "Then don't worry about making a conversion", how could I have made the answer clearer?

[mahmoodn]

Don't worry. It was my fault... :D