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Mail Archives: djgpp/1997/03/29/20:28:41

From: "Thomas Harte" <T DOT Harte AT btinternet DOT com>
Newsgroups: comp.os.msdos.djgpp
Subject: Allegro 2.2 question
Date: 29 Mar 1997 16:19:58 GMT
Organization: SoftySoft
Lines: 43
Message-ID: <01bc3c5c$e0867fe0$LocalHost@default>
NNTP-Posting-Host: host5-99-62-53.btinternet.com
To: djgpp AT delorie DOT com
DJ-Gateway: from newsgroup comp.os.msdos.djgpp

	Is there a command similar to polygon3d_f, but which reads the necessary
number of co-ordinates from one long int? e.g. if I wrote :-

------------------------------------------------------------
polygon3d_f(BITMAP *bmp, int type, BITMAP *texture,
------------------------------------------------------------

	. . . . as normal, and then finished with . . . .

------------------------------------------------------------
n, vertices);
------------------------------------------------------------

	. . . . where n is the number of points, and then it reads point one as
'vertices[0]', point two as 'vertices[1]' and so on until it reaches
'vertices[n-1]'. I think polygon3d_f would let me do this already, but I am
unsure how. Just supplying vertices as a V3D_f structure doesn't work
(incompatible pointer type apparently), so I am not sure what to do. Can
anyone help?
	Also later on, in the documentation it states

------------------------------------------------------------
void triangle3d(BITMAP *bmp, int type, BITMAP *tex, V3D *v1, *v2, *v3);
------------------------------------------------------------

 as equivalent to

------------------------------------------------------------
polygon3d(bmp, type, tex, 3, v1, v2, v3);
------------------------------------------------------------

	and not

------------------------------------------------------------
polygon3d(bmp, type, tex, 3, v);
------------------------------------------------------------

	where 'v' contains 3 sets of co-ordinates. However, in my program, the
number of sides is never known except in a variable, so I do need a way of
doing this.
	Thanks in advance,

		-Thomas

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