1 /* $OpenBSD: divufr.c,v 1.5 2001/03/29 03:58:18 mickey Exp $ */ 2 3 /* 4 * Copyright 1996 1995 by Open Software Foundation, Inc. 5 * All Rights Reserved 6 * 7 * Permission to use, copy, modify, and distribute this software and 8 * its documentation for any purpose and without fee is hereby granted, 9 * provided that the above copyright notice appears in all copies and 10 * that both the copyright notice and this permission notice appear in 11 * supporting documentation. 12 * 13 * OSF DISCLAIMS ALL WARRANTIES WITH REGARD TO THIS SOFTWARE 14 * INCLUDING ALL IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS 15 * FOR A PARTICULAR PURPOSE. 16 * 17 * IN NO EVENT SHALL OSF BE LIABLE FOR ANY SPECIAL, INDIRECT, OR 18 * CONSEQUENTIAL DAMAGES OR ANY DAMAGES WHATSOEVER RESULTING FROM 19 * LOSS OF USE, DATA OR PROFITS, WHETHER IN ACTION OF CONTRACT, 20 * NEGLIGENCE, OR OTHER TORTIOUS ACTION, ARISING OUT OF OR IN CONNECTION 21 * WITH THE USE OR PERFORMANCE OF THIS SOFTWARE. 22 * 23 */ 24 /* 25 * pmk1.1 26 */ 27 /* 28 * (c) Copyright 1986 HEWLETT-PACKARD COMPANY 29 * 30 * To anyone who acknowledges that this file is provided "AS IS" 31 * without any express or implied warranty: 32 * permission to use, copy, modify, and distribute this file 33 * for any purpose is hereby granted without fee, provided that 34 * the above copyright notice and this notice appears in all 35 * copies, and that the name of Hewlett-Packard Company not be 36 * used in advertising or publicity pertaining to distribution 37 * of the software without specific, written prior permission. 38 * Hewlett-Packard Company makes no representations about the 39 * suitability of this software for any purpose. 40 */ 41 42 43 #include "md.h" 44 45 void 46 divufr(opnd1,opnd2,result) 47 48 unsigned int opnd1, opnd2; 49 struct mdsfu_register *result; 50 { 51 /* check divisor for zero */ 52 if (opnd2 == 0) { 53 overflow = TRUE; 54 return; 55 } 56 57 /* check for overflow */ 58 if (opnd1 >= opnd2) { 59 overflow = TRUE; 60 return; 61 } 62 overflow = FALSE; 63 64 /* do the divide */ 65 divu(opnd1,0,opnd2,result); 66 return; 67 } 68