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<article id="post-CASPP-Lab-第一章题解-1" class="h-entry article article-type-post" itemprop="blogPost" itemscope itemtype="https://schema.org/BlogPosting">
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<a href="/2024/09/23/CASPP-Lab-%E7%AC%AC%E4%B8%80%E7%AB%A0%E9%A2%98%E8%A7%A3-1/" class="article-date">
<time class="dt-published" datetime="2024-09-23T12:13:37.000Z" itemprop="datePublished">2024-09-23</time>
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<a class="article-category-link" href="/categories/CSAPP/">CSAPP</a>
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<a class="p-name article-title" href="/2024/09/23/CASPP-Lab-%E7%AC%AC%E4%B8%80%E7%AB%A0%E9%A2%98%E8%A7%A3-1/">CASPP Lab 第一章题解(1)</a>
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<h1 id="Data-Lab"><a href="#Data-Lab" class="headerlink" title="Data Lab"></a>Data Lab</h1><h2 id="bitXor"><a href="#bitXor" class="headerlink" title="bitXor"></a>bitXor</h2><figure class="highlight c"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">//1</span></span><br><span class="line"><span class="comment">/* </span></span><br><span class="line"><span class="comment"> * bitXor - x^y using only ~ and & </span></span><br><span class="line"><span class="comment"> * Example: bitXor(4, 5) = 1</span></span><br><span class="line"><span class="comment"> * Legal ops: ~ &</span></span><br><span class="line"><span class="comment"> * Max ops: 14</span></span><br><span class="line"><span class="comment"> * Rating: 1</span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"><span class="type">int</span> <span class="title function_">bitXor</span><span class="params">(<span class="type">int</span> x, <span class="type">int</span> y)</span> {</span><br><span class="line"> <span class="keyword">return</span> (x&~y)|(~x&y);</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>蛮简单的，直接异或</p>
<blockquote>
<p>有无更简便的写法呢？</p>
</blockquote>
<h2 id="tmin"><a href="#tmin" class="headerlink" title="tmin"></a>tmin</h2><p>关于<code>tmin</code></p>
<p>我的想法就是</p>
<ul>
<li>对0x00逻辑取反，得到111…111</li>
<li>然后右移一位，得到011…111</li>
<li>再加1，即可得到100…000这种形式</li>
</ul>
<p>有个要注意的，就是逻辑取反必须是对无符号数，因此我们首先要写上<code>unsigned</code></p>
<blockquote>
<p>这个地方还不确定，究竟什么时候需要加<code>unsigned</code></p>
</blockquote>
<figure class="highlight c"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">/* </span></span><br><span class="line"><span class="comment"> * tmin - return minimum two's complement integer </span></span><br><span class="line"><span class="comment"> * Legal ops: ! ~ & ^ | + << >></span></span><br><span class="line"><span class="comment"> * Max ops: 4</span></span><br><span class="line"><span class="comment"> * Rating: 1</span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"><span class="type">int</span> <span class="title function_">tmin</span><span class="params">(<span class="type">void</span>)</span> {</span><br><span class="line"> <span class="keyword">return</span> ( ( ~(<span class="type">unsigned</span>)<span class="number">0x00</span> ) >> <span class="number">1</span> ) + <span class="number">1</span>;</span><br><span class="line"></span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<h2 id="isTmax"><a href="#isTmax" class="headerlink" title="isTmax"></a>isTmax</h2><figure class="highlight c"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">//2</span></span><br><span class="line"><span class="comment">/*</span></span><br><span class="line"><span class="comment"> * isTmax - returns 1 if x is the maximum, two's complement number,</span></span><br><span class="line"><span class="comment"> * and 0 otherwise </span></span><br><span class="line"><span class="comment"> * Legal ops: ! ~ & ^ | +</span></span><br><span class="line"><span class="comment"> * Max ops: 10</span></span><br><span class="line"><span class="comment"> * Rating: 1</span></span><br><span class="line"><span class="comment"> */</span></span><br></pre></td></tr></table></figure>
<p>不能用if之类的控制语句，==这类也不能用，因此需要好好思考思考<br>我们可以先得出最大的数，其实最大数也知道，就是01111…111这样的，我们用一个mask应该可以<br>如果和mask进行异或，得到0，说明是相同的，那就取反好了，反之就不是0；如何对于0输出1，对于其他数输出0？</p>
<figure class="highlight c"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br></pre></td><td class="code"><pre><span class="line"></span><br><span class="line"><span class="type">int</span> <span class="title function_">isTmax</span><span class="params">(<span class="type">int</span> x)</span>{</span><br><span class="line"> <span class="type">unsigned</span> mask = ~(<span class="type">unsigned</span>)<span class="number">0x00</span> >> <span class="number">1</span>;<span class="comment">//得到0111..11</span></span><br><span class="line"> <span class="keyword">return</span> <span class="number">1</span> - (<span class="type">bool</span>)( x ^ mask);<span class="comment">//异或的话，如果位相同则返回0， 位不同则返回1 </span></span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<p>但是，bool类型不能用，这是c…</p>
<p>好吧，忘记了一个重要的事情，那就是——没有bool我可以用<code>!</code>啊…</p>
<p>那么<code>!</code>怎么实现呢？（这就是第9题的事情了）</p>
<p>下面是正确代码</p>
<figure class="highlight c"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br></pre></td><td class="code"><pre><span class="line"><span class="type">int</span> <span class="title function_">isTmax</span><span class="params">(<span class="type">int</span> x)</span>{</span><br><span class="line"> <span class="type">unsigned</span> mask = ~(<span class="type">unsigned</span>)<span class="number">0x00</span> >> <span class="number">1</span>;<span class="comment">//得到0111..11</span></span><br><span class="line"> <span class="keyword">return</span> ! ( x ^ mask);<span class="comment">//异或的话，如果位相同则返回0， 位不同则返回1 </span></span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<blockquote>
<p>看了一个讲解，还有一个思路：</p>
<p>目前我们能用的一个确定的数就是0，因此如果需要执行比较判断，我们可以试着从当前数通过一系列操作映射到目标数0<br>现在就有一个问题，执行操作的时候，可能会同时将其他一些数也映射到0，这个过程大概率不是一一对应的，当然也有可能hhh，可以想一想<br>总之，对于那些多出来的数，我们需要设计一些方法将它们排除掉</p>
</blockquote>
<h2 id="allOddBits"><a href="#allOddBits" class="headerlink" title="allOddBits"></a>allOddBits</h2><figure class="highlight c"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">/* </span></span><br><span class="line"><span class="comment"> * allOddBits - return 1 if all odd-numbered bits in word set to 1</span></span><br><span class="line"><span class="comment"> * where bits are numbered from 0 (least significant) to 31 (most significant)</span></span><br><span class="line"><span class="comment"> * Examples allOddBits(0xFFFFFFFD) = 0, allOddBits(0xAAAAAAAA) = 1</span></span><br><span class="line"><span class="comment"> * Legal ops: ! ~ & ^ | + << >></span></span><br><span class="line"><span class="comment"> * Max ops: 12</span></span><br><span class="line"><span class="comment"> * Rating: 2</span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"><span class="comment">//直接告诉你了，一共32位，确实得告诉，不过假如不知道上限的话，怎么去做呢？或者怎么用这些写出循环判断呢？</span></span><br><span class="line"><span class="comment">//所有奇数位为1，mask一下？先做一个奇数位的mask,然后进行&的操作，我们就取出了奇数位，1，3，</span></span><br><span class="line"></span><br><span class="line"><span class="type">int</span> <span class="title function_">allOddBits</span><span class="params">(<span class="type">int</span> x)</span> {</span><br><span class="line"> <span class="keyword">return</span> !(<span class="number">0xAAAAAAAA</span> ^ (x & <span class="number">0xAAAAAAAA</span>));</span><br><span class="line">}</span><br><span class="line"><span class="comment">//通过了，可以可以</span></span><br><span class="line"><span class="comment">//如果不知道上限位数呢？那么这个mask就需要我们自己去获取</span></span><br><span class="line"><span class="comment">//其实很容易，我们可以构建从mask到0的映射，然后去掉杂项即可</span></span><br><span class="line"><span class="comment">//映射为：x + x >> 1 = -1 = 1111...1111</span></span><br><span class="line"><span class="comment">//这个好理解，因为101010...1010这种，加上自己右移一位的结果，恰好就是全1</span></span><br><span class="line"><span class="comment">//那么倒推呢，还有谁可以通过这个映射计算出全1？</span></span><br><span class="line"><span class="comment">//稍微推导一下就知道，没有别的了，因为最低位相加为1，每一位对应相加为1，因此必然是0101的，而首位直接等于1（过程无进位），就推出x必然为1010...1010的形式</span></span><br><span class="line"><span class="type">int</span> <span class="title function_">allOddBits</span><span class="params">(<span class="type">int</span> x)</span>{</span><br><span class="line"> <span class="keyword">return</span> !(<span class="number">-1</span> ^ (x + x >> <span class="number">1</span>));</span><br><span class="line">}</span><br><span class="line"><span class="comment">//其实自己想错了，构建映射然后去掉杂项，只能应用于判断条件，而非求值</span></span><br><span class="line"><span class="comment">//可是这个映射下只有1010...1010可以映射到0，能否映射回去呢？看看这个过程是不是双射，感觉就是吧，一个x对应一个结果，而结果为0的情况下只有x可以满足，就是双射</span></span><br><span class="line"><span class="comment">//尝试构建逆运算</span></span><br><span class="line"><span class="comment">//没搞出来，也许可以直接从Tmax下手，直接搞算数运算，之后想到了再试试吧</span></span><br><span class="line"></span><br><span class="line"><span class="comment">//其实有个问题，即实验要求只能使用两位16进制，即0x00这种，因此0xAAAAAAAA需要通过0xAA来构造，倒是不麻烦</span></span><br></pre></td></tr></table></figure>
<h2 id="negate"><a href="#negate" class="headerlink" title="negate"></a>negate</h2><figure class="highlight c"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">/* negate - return -x </span></span><br><span class="line"><span class="comment"> * Example: negate(1) = -1.</span></span><br><span class="line"><span class="comment"> * Legal ops: ! ~ & ^ | + << >></span></span><br><span class="line"><span class="comment"> * Max ops: 5</span></span><br><span class="line"><span class="comment"> * Rating: 2</span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"><span class="comment">//要求负数，我加个负号？</span></span><br><span class="line"><span class="comment">//好好好，还真能过，有点懵逼，真的</span></span><br><span class="line"><span class="type">int</span> <span class="title function_">negate</span><span class="params">(<span class="type">int</span> x)</span> {</span><br><span class="line"> <span class="keyword">return</span> -x;</span><br><span class="line">}</span><br><span class="line"><span class="comment">//其实不对，没有满足要求</span></span><br><span class="line"><span class="type">int</span> <span class="title function_">negate</span><span class="params">(<span class="type">int</span> x)</span> {</span><br><span class="line"> <span class="keyword">return</span> ~x + <span class="number">1</span>;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
<h2 id="isAsciiDigit"><a href="#isAsciiDigit" class="headerlink" title="isAsciiDigit"></a>isAsciiDigit</h2><figure class="highlight c"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">//3</span></span><br><span class="line"><span class="comment">/* </span></span><br><span class="line"><span class="comment"> * isAsciiDigit - return 1 if 0x30 <= x <= 0x39 (ASCII codes for characters '0' to '9')</span></span><br><span class="line"><span class="comment"> * Example: isAsciiDigit(0x35) = 1.</span></span><br><span class="line"><span class="comment"> * isAsciiDigit(0x3a) = 0.</span></span><br><span class="line"><span class="comment"> * isAsciiDigit(0x05) = 0.</span></span><br><span class="line"><span class="comment"> * Legal ops: ! ~ & ^ | + << >></span></span><br><span class="line"><span class="comment"> * Max ops: 15</span></span><br><span class="line"><span class="comment"> * Rating: 3</span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"><span class="comment">//0x30 = 0011 0000</span></span><br><span class="line"><span class="comment">//0x39 = 0011 1001</span></span><br><span class="line"><span class="comment">//判断最高位是否为第6位且第5第6位是否为1，构建一个mask，0011 0000，然后结果右移4位，要是非0就说明不是，用一个flag记录一下</span></span><br><span class="line"><span class="comment">//然后若4为1则mask一下23，若4不为1则直接输出1</span></span><br><span class="line"><span class="comment">//看看咋写</span></span><br><span class="line"><span class="type">int</span> <span class="title function_">isAsciiDigit</span><span class="params">(<span class="type">int</span> x)</span> {</span><br><span class="line"> <span class="type">int</span> flag_1 = !( (<span class="number">0x30</span> ^ x) >> <span class="number">4</span> );</span><br><span class="line"> <span class="type">int</span> a = (<span class="number">0x08</span> & x) >> <span class="number">3</span>, b = (<span class="number">0x04</span> & x) >> <span class="number">2</span>, c = (<span class="number">0x02</span> & x) >> <span class="number">1</span>;<span class="comment">//取出后4位的前3位</span></span><br><span class="line"> <span class="type">int</span> flag_2 = !( (a & b) | (a & c) );</span><br><span class="line"> <span class="keyword">return</span> flag_1 & flag_2;</span><br><span class="line">}</span><br><span class="line"><span class="comment">//通过了，还蛮容易，不过运算符刚好用了15个，试试能不能再简化简化</span></span><br><span class="line"></span><br><span class="line"><span class="comment">//看到了一个思路，说是根据上下界来判断，可以之后思考一下</span></span><br></pre></td></tr></table></figure>
<h2 id="conditional"><a href="#conditional" class="headerlink" title="conditional"></a>conditional</h2><figure class="highlight c"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">/* </span></span><br><span class="line"><span class="comment"> * conditional - same as x ? y : z </span></span><br><span class="line"><span class="comment"> * Example: conditional(2,4,5) = 4</span></span><br><span class="line"><span class="comment"> * Legal ops: ! ~ & ^ | + << >></span></span><br><span class="line"><span class="comment"> * Max ops: 16</span></span><br><span class="line"><span class="comment"> * Rating: 3</span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"><span class="comment">//判定逻辑就是，x是否非0，如果非0，输出y；如果是0，输出z。</span></span><br><span class="line"><span class="comment">//想到了一个，我们只需要最后输出统一格式：mut + y</span></span><br><span class="line"><span class="comment">//然后中途修改mut即可</span></span><br><span class="line"><span class="comment">//mut满足：当x=0时，mut=z-y；当x!=0时，mut=0</span></span><br><span class="line"><span class="comment">//下面就是这个逻辑的实现，还蛮有意思的，通过0和非0的映射</span></span><br><span class="line"><span class="comment">//其实思路就是，我们直接用0作为选择器来生成mut变量，反正mut变量要么是z-y要么是0</span></span><br><span class="line"><span class="comment">//怎么选择呢？用或呗：00000和11111，就可以实现选择了</span></span><br><span class="line"><span class="comment">//比如，当x=0时，我们将门设为111...111，当x!=0时，我们将门设为000...000；</span></span><br><span class="line"><span class="comment">//然后将门与z-y或一下就行</span></span><br><span class="line"><span class="comment">//怎么从0和非0生成111...111和000...000，这个更简单了，000...000和111...111之间其实就差1，0和1也差1，你懂我意思吧</span></span><br><span class="line"><span class="comment">//看代码：</span></span><br><span class="line"><span class="type">int</span> <span class="title function_">conditional</span><span class="params">(<span class="type">int</span> x, <span class="type">int</span> y, <span class="type">int</span> z)</span> {</span><br><span class="line"> <span class="type">int</span> mut = ( ( ( ! ! (<span class="type">unsigned</span>)x ) + ~<span class="number">1</span> + <span class="number">1</span> ) ) & (z + ~y + <span class="number">1</span>);</span><br><span class="line"> <span class="keyword">return</span> mut + y;</span><br><span class="line">}</span><br><span class="line"><span class="comment">//通过了~</span></span><br></pre></td></tr></table></figure>
<h2 id="isLessOrEqual"><a href="#isLessOrEqual" class="headerlink" title="isLessOrEqual"></a>isLessOrEqual</h2><figure class="highlight c"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">/* </span></span><br><span class="line"><span class="comment"> * isLessOrEqual - if x <= y then return 1, else return 0 </span></span><br><span class="line"><span class="comment"> * Example: isLessOrEqual(4,5) = 1.</span></span><br><span class="line"><span class="comment"> * Legal ops: ! ~ & ^ | + << >></span></span><br><span class="line"><span class="comment"> * Max ops: 24</span></span><br><span class="line"><span class="comment"> * Rating: 3</span></span><br><span class="line"><span class="comment"> */</span></span><br><span class="line"><span class="comment">//相减，按照符号位进行判断输出，可以用上一题的结论</span></span><br><span class="line"><span class="type">int</span> <span class="title function_">isLessOrEqual</span><span class="params">(<span class="type">int</span> x, <span class="type">int</span> y)</span> {</span><br><span class="line"> <span class="keyword">return</span> ( ( ( ! ! ((y + ~x + <span class="number">1</span>) >> <span class="number">31</span>) ) + ~<span class="number">0</span>) ) & <span class="number">1</span>;<span class="comment">//稍微化简一下</span></span><br><span class="line">}</span><br><span class="line"><span class="comment">//事实上通过不了，我们没考虑溢出</span></span><br><span class="line"><span class="comment">//那咋办，考虑一下呗</span></span><br><span class="line"><span class="comment">//思考一下，溢出有两种情况：x正y负，x负y正，前者溢出结果为正，后者溢出结果为负</span></span><br><span class="line"><span class="comment">//那么我们直接对于所有x负y正输出1，x正y负输出0？其余的正常判断</span></span><br><span class="line"><span class="comment">//那么需要两次判断，可以通过&和|实现01的优先条件判断</span></span><br><span class="line"><span class="comment">//比如，flag_np标记x负y正，flag_pn反之</span></span><br><span class="line"><span class="comment">//那么，当flag_np=1时，我们直接输出1；flag_pn=1时，直接输出0</span></span><br><span class="line"><span class="comment">//flag_np=0和flag_pn=0时我们就不管了</span></span><br><span class="line"><span class="comment">//假设原输出为output</span></span><br><span class="line"><span class="comment">//修正后的输出为：(2 + ~flag_pn) & (flag_np | output)</span></span><br><span class="line"><span class="comment">//逻辑是这样的：</span></span><br><span class="line"><span class="comment">//如果flag_pn=1，那么前面就爆0，直接输出0；如果flag_pn=0，进入后面的判断条件</span></span><br><span class="line"><span class="comment">//如果flag_np=1，那么直接输出1，否则输出output</span></span><br><span class="line"><span class="type">int</span> <span class="title function_">isLessOrEqual</span><span class="params">(<span class="type">int</span> x, <span class="type">int</span> y)</span> {</span><br><span class="line"> <span class="type">int</span> xh = ((<span class="type">unsigned</span>)x) >> <span class="number">31</span>;</span><br><span class="line"> <span class="type">int</span> yh = ((<span class="type">unsigned</span>)y) >> <span class="number">31</span>;</span><br><span class="line"> <span class="type">int</span> flag_pn = !xh & yh;</span><br><span class="line"> <span class="type">int</span> flag_np = !yh & xh;</span><br><span class="line"> <span class="keyword">return</span> (<span class="number">2</span> + ~flag_pn) & (flag_np | ( ( ( ( ! ! ((y + ~x + <span class="number">1</span>) >> <span class="number">31</span>) ) + ~<span class="number">0</span>) ) & <span class="number">1</span> ) );</span><br><span class="line">}</span><br><span class="line"><span class="comment">//然后终于通过了...好耶！</span></span><br></pre></td></tr></table></figure>
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