runetype.c - sbase - suckless unix tools
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       ---
       runetype.c (1508B)
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            1 /* MIT/X Consortium Copyright (c) 2012 Connor Lane Smith <cls@lubutu.com>
            2  *                            (c) 2015 Laslo Hunhold <dev@frign.de>
            3  *
            4  * Permission is hereby granted, free of charge, to any person obtaining a
            5  * copy of this software and associated documentation files (the "Software"),
            6  * to deal in the Software without restriction, including without limitation
            7  * the rights to use, copy, modify, merge, publish, distribute, sublicense,
            8  * and/or sell copies of the Software, and to permit persons to whom the
            9  * Software is furnished to do so, subject to the following conditions:
           10  *
           11  * The above copyright notice and this permission notice shall be included in
           12  * all copies or substantial portions of the Software.
           13  *
           14  * THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
           15  * IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
           16  * FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT.  IN NO EVENT SHALL
           17  * THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
           18  * LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING
           19  * FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER
           20  * DEALINGS IN THE SOFTWARE.
           21  */
           22 #include "../utf.h"
           23 
           24 int
           25 rune1cmp(const void *v1, const void *v2)
           26 {
           27         Rune r1 = *(Rune *)v1, r2 = *(Rune *)v2;
           28 
           29         return r1 - r2;
           30 }
           31 
           32 int
           33 rune2cmp(const void *v1, const void *v2)
           34 {
           35         Rune r = *(Rune *)v1, *p = (Rune *)v2;
           36 
           37         if(r >= p[0] && r <= p[1])
           38                 return 0;
           39         else
           40                 return r - p[0];
           41 }