在条件或开关情况下，在C中初始化矩阵（2D阵列）

作者: v-star*위위
发布时间: 2024-10-14 12:59:42 (3天前)
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一生浮华 | 2019-08-31 10-32

<div class =“post-text”itemprop =“text”> <P> 复合文字可以使用自C99。你可以写如下： </p> <pre> <code> void rotate(Point *p, float theta){ int choice = ((int)theta/90)%4; switch(choice){ case 0: matMult(p, (Mat3){{1.0,0.0,0.0}, {0.0,1.0,0.0}, {0.0,0.0,1.0}}); break; case 1: matMult(p, (Mat3){{0.0,-1.0,0.0}, {1.0,0.0,0.0}, {0.0,0.0,1.0}}); break; case 2: matMult(p, (Mat3){{-1.0,0.0,0.0}, {0.0,-1.0,0.0}, {0.0,0.0,1.0}}); break; case 3: matMult(p, (Mat3){{0.0,1.0,0.0}, {-1.0,0.0,0.0}, {0.0,0.0,1.0}}); break; default: matMult(p, (Mat3){{1.0,0.0,0.0}, {0.0,1.0,0.0}, {0.0,0.0,1.0}}); break; } } </code> </pre> </DIV>

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NetworkAttachedStorage | 2019-08-31 10-32

<div class =“post-text”itemprop =“text”> <P> 无法在C中分配矩阵，只能初始化矩阵或分配其各个组件。 </p> <P> 有几种方法可以在这里进行，最简单的方法也许不是一个 <code> switch </code> 但是一阵 <code> Mat3 </code> </p> <pre> <code> static Mat3 const matrix[] = { {{1.0,0.0,0.0}, {0.0,1.0,0.0}, {0.0,0.0,1.0}}, {{0.0,-1.0,0.0}, {1.0,0.0,0.0}, {0.0,0.0,1.0}}, {{-1.0,0.0,0.0}, {0.0,-1.0,0.0}, {0.0,0.0,1.0}}, {{0.0,1.0,0.0}, {-1.0,0.0,0.0}, {0.0,0.0,1.0}}, }; unsigned choice = ((unsigned)(theta/90))%4u; matMult(p, matrix[choice]); </code> </pre> <UL> <LI> 我补充道 <code> const </code> 因为它看起来是你的第二个参数函数是只读的。 </LI> <LI> <code> static </code> 它在真实之前只分配和初始化一次执行代码。 </LI> <LI> <code> unsigned </code> 为了 <code> choice </code> 值保证算术 <code> % </code> 真的会产生价值 <code> 0 </code> 至 <code> 3 </code> ，而不是消极的东西。 </LI> </UL> </DIV>

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