memset question

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

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.

Thanks,

so

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

are basically the same?

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

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.

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

Code:

---

int LoadGLTextures() {
    AUX_RGPImageRec *TextureImage = NULL;

    ...
}

---

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

Code:

---

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

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

    ...
}

---

dunno, im just following a tutorial lol

VB Code:

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

that makes

VB Code:

sizeof(void*)*1);

more understanding now

thanks

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.