学步园

//一个简单的自顶向下语法分析<br /> #include <cassert><br /> #include <cstdio><br /> using namespace std;<br /> char str[1024];<br /> long curpos;<br /> int num()<br /> {<br /> int ret;<br /> if (str[curpos] >= '0' && str[curpos] <= '9')<br /> {<br /> ret = str[curpos] - '0';<br /> ++curpos;<br /> return ret;<br /> }<br /> printf("ERROR!not a num.pos=%d/r/n", curpos);<br /> assert(0);<br /> return 0;<br /> }<br /> void match(int c)<br /> {<br /> if (str[curpos] == c)<br /> {<br /> ++curpos;<br /> return;<br /> }<br /> printf("ERROR!not =%c/r/n", c);<br /> assert(0);<br /> }<br /> int A();<br /> int C()<br /> {<br /> int ret;<br /> if (str[curpos] == '(')<br /> {<br /> match('(');<br /> ret = A();<br /> match(')');<br /> }<br /> else<br /> {<br /> ret = num();<br /> }<br /> return ret;<br /> }<br /> int B();<br /> void B2(int& token, int& num)<br /> {<br /> if (str[curpos] == '*')<br /> {<br /> match('*');<br /> token = '*';<br /> num = B();<br /> }<br /> else<br /> {<br /> token = 0;<br /> }<br /> }<br /> int B()<br /> {<br /> int ret = C();<br /> int token, num;<br /> B2(token, num);<br /> if (token == '*')<br /> {<br /> return ret * num;<br /> }<br /> else if (token == '/')<br /> {<br /> return ret / num;<br /> }<br /> else<br /> {<br /> return ret;<br /> }<br /> }<br /> void A2(int& token, int& num)<br /> {<br /> if (str[curpos] == '+')<br /> {<br /> match('+');<br /> token = '+';<br /> num = A();<br /> }<br /> else if (str[curpos] == '-')<br /> {<br /> match('-');<br /> token = '-';<br /> num = A();<br /> }<br /> }<br /> int A()<br /> {<br /> int token, num;<br /> int ret = B();<br /> A2(token, num);<br /> if (token == '+')<br /> {<br /> return ret + num;<br /> }<br /> else if (token == '-')<br /> {<br /> return ret - num;<br /> }<br /> else<br /> {<br /> return ret;<br /> }<br /> }<br /> int main()<br /> {<br /> printf("请输入表达式:/r/n");<br /> for (; <img src="https://admin.xuebuyuan.com/wp-includes/images/smilies/icon_wink.gif" alt=";)" class="wp-smiley" /><br /> {<br /> scanf("%s", &str);<br /> curpos = 0;<br /> printf("结果为:%d/r/n", A());<br /> match(0);<br /> printf("Corrent!/r/n");<br /> }<br /> return 0;<br /> }<br />