表达式求值 son bin.doc

#include #include/定义全局变量-int x;/定义栈放操作数和操作符-typedef struct char a100;int b100;int top;stack;/初始化栈-void initstack(stack &s,char )s.top =0;s.a0=#;void initstack(stack &s,int )s.top =0;s.b0=0;/-进栈-void push(stack &s,char y)if (s.top =99)coutstack overflow; elses.top =s.top +1;s.as.top=y;void push(stack &s,int x)if (s.top =99)coutstack overflow;elses.top =s.top +1;s.bs.top=x;/退栈-int shupop(stack &s)if(s.top =-1)coutstack underflow;elsereturn s.b s.top -;char prtpop(stack& s)if(s.top =-1)coutstack underflown;elsereturn s.a s.top -;/运算-int operite(int a,char b,int c)switch (b) case -:return -(a-c); /为什么要加负号？ case +:return (a+c); case *:return (a*c); case /:return (c/a);/为什么是c/a而不是a/c？;/把字符变为int-int change( char c)return x=x*10+(int)c-48;/操作符的定位-int priority(char op)switch(op) case +:return 1;case -:return 2;case *:return 3;case /:return 4; case#:return 0;/操作符的比较-bool compare(char a,char b)if(priority(a)priority(b)return 1;return 0;/应用栈顶元素-char gettop(stack &prt,char=e)return prt.a prt.top ;int gettop(stack &shu,int=7)return shu.bshu.top ;/接收表达式-int accept(stack &shu,stack &prt)char c;int j=0;cout请输入表达式,以#结束,按回车确认下一个运算符- if( priority(theta)priority(gettop(prt,y) ) temp=operite(a,theta,shupop(shu);push(shu,temp); /栈顶运算符下一个运算符- else if(priority(theta)priority(gettop(prt,y) int qian=shupop(shu),hou=shupop(shu); temp=operite(qian,prtpop(prt),hou); push(shu,temp);push(shu,a);push(prt,theta); /连减时结合- else if(priority(theta)=2 & priority(theta)=priority(gettop(prt,y) temp=operite(a,+,shupop(shu); push(shu,temp); /连除时结合- else if(priority(theta)=4 & priority(theta)=priority(gettop(prt,y) temp=operite(a,*,shupop(shu); push(shu,temp); /栈顶运算符=下一个运算符（除去“-”和“/”号）- else temp=operite(a,