memset question

[Geespot]

Hi

Im following through some OpenGL tutorials and ive come up to a line

memset(TextureImage,0,sizeof(void *)*1);

now im not much of a c++ programmer, i only know the basics ( but i do know pointers )
and what that line above does is, clears out the memory that TextureImage is located. I know what it does but i dont know HOW it does it.

The thing that confuses me is the sizeof(void *)*1)
I just cant work out what that is

Can someone please explain this to me?

Thanks

[parksie]

Check your documentation for memset for an explanation.

The sizeof(void*) * 1 basically resolves to the size of a pointer. So, the memset() sets the first 4 bytes after the memory location pointed to by TextureImage to 0.

It's not that useful though, that line. I'd expect it to be used to clear a much larger buffer.

[Geespot]

Thanks,

so

memset(TextureImage,0,sizeof(void *)*1);
and
memset(TextureImage,0,4);

are basically the same?

[parksie]

Yes. But the size of a pointer is implementation dependent. What is that line of code actually trying to do?

[Geespot]

Heres the lines

int LoadGLTextures()
{
AUX_RGBImageRec *TextureImage[1]; // Create Storage Space For The Texture
memset(TextureImage,0,sizeof(void *)*1); // Set The Pointer To NULL
...
}

What is does, is first that function creates some storage space for a texture, then using the memset, it clears the imagerec to make sure its empty.

its only that memset line that has gotten me confused
I was gonna just ignore it and continue, but i really do like to know what each and evey line does.

[parksie]

Why bother using an array for that first one? There's only space for one, so you might as well use:

int LoadGLTextures() {
    AUX_RGPImageRec *TextureImage = NULL;

    ...
}

If you had 5 textures, then it would be more useful:

int LoadGLTextures() {
    AUX_RGPImageRec *TextureImage[5]; // creates an array of five pointers

    memset(TextureImage, 0, sizeof(AUX_RGPImageRec*) * 5);

    ...
}

[Geespot]

dunno, im just following a tutorial lol

memset(TextureImage, 0, sizeof(AUX_RGPImageRec*) * 5);

that makes

sizeof(void*)*1);

more understanding now

thanks

[parksie]

No. In your case, the sizes of the two pointers will be the same. In my example, it clears the five *pointers*, not anything they may point to.