利用堆栈的基本操作来实现中缀表达式的计算

1、#include #include #include #include using namespace std; typedef struct double *data; int top; nstack; typedef struct char *data; int top; cstack; nstack num; cstack op; char suffix1010,s1010; double num1,num2,num3; void init(nstack &s) s.data=new double 100; s.top=-1; void init(cstack &s) s.data=ne

2、w char 100; s.top=-1; bool empty(nstack s) if(s.top=-1) return 1; return 0; bool empty(cstack s) if(s.top=-1) return 1; return 0; bool top(nstack s,int &e) if(s.top=-1) return 0; e=s.datas.top; return 1; bool top(cstack s,char &e) if(s.top=-1) return 0; e=s.datas.top; return 1; bool pop(cstack &s,ch

3、ar &e) if(s.top=-1) return 0; e=s.datas.top-; return 1; bool pop(nstack &s,double &e) if(s.top=-1) return 0; e=s.datas.top-; return 1; void push(cstack &s,char e) s.data+s.top=e; void push(nstack &s,double e) s.data+s.top=e; int opmember(char c) if(c=+|c=-|c=*|c=/) return 1; if(c=() return 2; if(c=)

4、 return 3; return 0; int getpre(char c) if(c=#) return -1; if(c=() return 0; if(c=+|c=-) return 1; if(c=) return 2; if(c=*|c=/) return 2; bool compare(char s1,char s2) if(getpre(s1)=getpre(s2) return 1; return 0; bool isnum(char c) if(c=0&c=9) return 1; return 0; bool islogic(char *s) int len=strlen

5、(s); int l=0,r=0; if(opmember(s0=1|opmember(s0)=3) return 0; if(opmember(slen-1)=1|opmember(slen-1)=2) return 0; for(int i=0;ilen-1;i+) if(si=() l+; if(si=) r+; if(opmember(si)=3&isnum(si+1) return 0; if(opmember(si)=1&opmember(si+1)=1) return 0; if(opmember(si)=1&opmember(si+1)=3) return 0; if(opme

6、mber(si)=2&opmember(si+1)=3) return 0; if(opmember(si)=3&isnum(si+1) return 0; if(isnum(si)&opmember(si+1)=2) return 0; if(si=/&si+1=0) return 0; if(slen-1=) r+; if(l!=r) return 0; return 1; double operate(double a,char op,double b) if(op=+) return a+b; else if(op=-) return a-b; else if(op=*) return

7、 a*b; else if(op=/) return a/b; void transform(char *suffix,char *s) init(op); push(op,#); int i=0,k=0; char ch=s0,c; while(!empty(op) if(isnum(ch) suffixk+=ch; else suffixk+= ; switch(ch) case (:push(op,ch);break; case ): pop(op,c); while(c!=() suffixk+=c; pop(op,c); break; default: while(top(op,c)

8、&(compare(c,ch) suffixk+=c;pop(op,c); if(ch!=#) push(op,ch); break; if(ch!=#) i+;ch=si; suffixk=0; void calculate() init(num); int len=strlen(suffix),i=0; while(ilen) if(suffixi= ) i+;continue; else if(suffixi=#) break; else double cur=0; bool hasnum=0; while(isnum(suffixi) cur*=10; cur+=suffixi+-0; hasnum=1; if(hasnum) push(num,cur); if(suffixi= ) i+;continue; if(opmember(suffixi)=1) pop(num,num1); pop(num,num2); num3=operate(num2,suffixi+,num1); push(num,num3); double ans; pop(num,ans); printf(%.2lfn,ans); int main() while(scanf(%s,s)!=eo