I had a strange problem i was appending an array using delete/new

but after creation of new array i had to put the structs in it (texture loading- whenever i loaded texture that wasnt 256x256 i had to resize it -> ther i called mytexture.~TGLTexture(); //it seems that was the problem i had to write another function that wasnt named ~TGLTexture() with the exact code.

So my question is: does calling a destructor w/o delete, deletes the object? it seems that yes but i didint find anythig on the net about it

yeah but that wasnt the question ;] now i have a kill() procedure that is called in destructor. so i can safely call kill() without caling ~TGLTexture(), and everything works..

void kill()
{
	if (texloaded)
{
	if (pdata != NULL) delete [] pdata;
	glDeleteTextures(1, &texture);
	texloaded = false;
}
}

~TGLTexture()
{
kill();
}

so again the same question does calling ~TGLTexture() deletes the object? and why :P

I had a strange problem i was appending an array using delete/new

but after creation of new array i had to put the structs in it (texture loading- whenever i loaded texture that wasnt 256x256 i had to resize it -> ther i called mytexture.~TGLTexture(); //it seems that was the problem i had to write another function that wasnt named ~TGLTexture() with the exact code.

So my question is: does calling a destructor w/o delete, deletes the object? it seems that yes but i didint find anythig on the net about it

You should never call the destructor directly. Always use delete.

	if (pdata != NULL) delete [] pdata;

If you do that, do this:

if (pdata != NULL)
{
    delete[] pdata;
    pdata = NULL;
}

In other words, make sure that your object is in a valid state in case someone calls kill twice, or kill once and delete once.

    delete[] pdata;
    pdata = NULL;