76 lines
2.3 KiB
C++
76 lines
2.3 KiB
C++
/*
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Adept MobileRobots Robotics Interface for Applications (ARIA)
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Copyright (C) 2004, 2005 ActivMedia Robotics LLC
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Copyright (C) 2006, 2007, 2008, 2009, 2010 MobileRobots Inc.
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Copyright (C) 2011, 2012, 2013 Adept Technology
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This program is free software; you can redistribute it and/or modify
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it under the terms of the GNU General Public License as published by
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the Free Software Foundation; either version 2 of the License, or
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(at your option) any later version.
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This program is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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GNU General Public License for more details.
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You should have received a copy of the GNU General Public License
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along with this program; if not, write to the Free Software
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Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA
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If you wish to redistribute ARIA under different terms, contact
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Adept MobileRobots for information about a commercial version of ARIA at
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robots@mobilerobots.com or
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Adept MobileRobots, 10 Columbia Drive, Amherst, NH 03031; +1-603-881-7960
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*/
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#include "Aria.h"
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ArPose fn1(void)
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{
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static ArPose pose;
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pose.setX(pose.getX() + 1);
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pose.setY(pose.getY() + 1);
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pose.setTh(pose.getTh() - 1);
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return pose;
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}
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void fn2(ArPose pose)
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{
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pose.log();
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}
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int main(void)
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{
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printf("Entering 100 iterations of incrementing pose X and Y and decrementing Theta...");
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for (int i = 0; i < 100; i++)
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fn2(fn1());
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printf("\nTesting ArPose::operator+(const ArPose&) and ArPose::operator-(const ArPose&)...\n");
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ArPose p1(10, 10, 90);
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ArPose p2(10, 10, 45);
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ArPose p3(0, 0, 0);
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ArPose p4(-20, 0, 360);
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ArPose p5(-20, -20, -180);
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printf("(10,10,90) + (10,10,90) => ");
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(p1 + p1).log();
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printf("(10,10,90) - (10,10,90) => ");
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(p1 - p1).log();
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printf("(10,10,90) + (10,10,45) => ");
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(p1 + p2).log();
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printf("(10,10,90) + (0,0,0) => ");
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(p1 + p3).log();
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printf("(10,10,90) - (0,0,0) => ");
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(p1 - p3).log();
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printf("(0,0,0) + (-20,0,360) => ");
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(p3 + p4).log();
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printf("(0,0,0) - (-20,0,360) => ");
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(p3 - p4).log();
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printf("(10,10,90) + (-20,0,360) => ");
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(p1 + p4).log();
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printf("(-20,0,360) + (-20,0,360) => ");
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(p4 + p4).log();
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printf("(-20,-20,-180) - (10,10,45) => ");
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(p5 - p2).log();
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}
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