Adding a byte to a string

I am playing around with a plugin sdk for a video editor (Wax), and the code is:

Code:

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memset(psImage->psImageData, 0x80, sizeof(ImageSample)*psImage->dwWidth*psImage->dwHeight);

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My understanding of memset is that it replaces the old information with the second argument, for as many times as you specify in the third argument.

The resulting string(i dont think its actually a string, but I don't know what to call it) is information for an image, with four bytes to each pixel (r, g, b, alpha). sizeof(ImageSample) = 4.

I want to be able to have a loop and put my own colour. So I need to be able to append four bytes at a time. The for loop looks like this:

Code:

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for(count = 0; count < psImage->dwWidth*psImage->dwHeight){

}

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But I don't know what to put inside the loop.

What I think I need is a command that appends a byte to a string or in this case binary information. Thanks.

Look up memcpy and memmove. You can easily write code to add to the end of the memory block:

Code:

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#include <cstdlib>
#include <iostream>

using namespace std;

int main()
{
  unsigned char bytes[4];
  char c = '0';

  for ( int i = 0; i < 4; i++ ) {
    memcpy ( bytes + i, &c, 1 );
    c++;
  }
  for ( int i = 0; i < 4; i++ ) {
    cout<< bytes[i] <<endl;
  }
}

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color model hacks? lol

Thanks. I just have a few questions.

Code:

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memcpy ( bytes + i, &c, 1 );

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Will this access the memory i bytes after the beginning?

And what is the & in front of the c for?

My last code crashed, here is the new code:

Code:

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for(count = 0; count < 4*psImage->dwWidth*psImage->dwHeight; count++){
                memcpy (psImage->psImageData + count, &c, 1 );
}

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EDIT: I also have c = '0' above that. I tested the code and it still crashes, im looking for stupid mistakes now.

>Will this access the memory i bytes after the beginning?
It depends on the type of the pointer. If it's a char pointer then yes, otherwise it will access the memory i * sizeof ( type ) after the beginning.

>And what is the & in front of the c for?
The second argument requires a pointer. The & operator in this context is the address-of operator, and returns the address of the variable.

Quote:

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It depends on the type of the pointer. If it's a char pointer then yes, otherwise it will access the memory i * sizeof ( type ) after the beginning.

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I read that a whille ago and didn't quite get it, but now that I read it again the whole thing seems to make sense. Thanks, I am going to try to figure it out again.

The code works. Thanks so much.