数据结构表达式运算浮点多位小数字运算源代码fromsm

1、#include #include #include #include#define TURE1#define FALSE0#define OK1#define ERROR0#define INFEASIBLE-1#define OVERLLOW-2#define STACK_INIT_SIZE 100#define STACKINCREMENT 10typedef double Status;typedef structchar *base; char *top;int stacksize;Sqstack1;typedef structdouble *base; double *top;in

2、t stacksize;Sqstack2;int InitStack1(Sqstack1 &s)/初始化 s.base=(char*)malloc(STACK_INIT_SIZE&sizeof(char);if( !s.base ) exit(OVERLLOW); s.top=s.base;s.stacksize=STACK_INIT_SIZE;return OK;int InitStack2(Sqstack2 &s)s.base=(double*)malloc(STACK_INIT_SIZE&sizeof(double);if( !s.base ) exit(OVERLLOW); s.top

3、=s.base;s.stacksize=STACK_INIT_SIZE;return OK;char GetTop1(Sqstack1 s)if( s.top=s.base ) return ERROR;char e=*(s.top-1);return e;double GetTop2(Sqstack2 s)if( s.top=s.base ) return ERROR;double e=*(s.top-1);return e;int Push1(Sqstack1 &s,char e)if(s.top-s.base=s.stacksize)s.base=(char*)realloc(s.bas

4、e,(s.stacksize+STACKINCREMENT)&sizeof(char);if( !s.base ) exit(OVERLLOW);s.top=s.base+s.stacksize;s.stacksize+=STACKINCREMENT;*(s.top)=e;s.top+;return OK;int Push2(Sqstack2 &s,double e)if(s.top-s.base=s.stacksize)s.base=(double*)realloc(s.base,(s.stacksize+STACKINCREMENT)*sizeof(double);if( !s.base

5、) exit(OVERLLOW);/存储分配失败s.top=s.base+s.stacksize;/s.top的值已变s.stacksize+=STACKINCREMENT;*(s.top)=e;s.top+;return OK;char Pop1(Sqstack1 &s)char e;if(s.top=s.base) return ERROR;e=*(s.top-1); s.top-;return e;double Pop2(Sqstack2 &s)double e;if(s.top=s.base) return ERROR;e=*(s.top-1); s.top-;return e;int

6、 In(char x)char a7=+,-,*,/,(,),#; int i; for (i=0;i);return(); case *:if(x2=() return(); case (:if(x2=) return(=);return();case #:if (x2=#) return(=); return(=0&ani=0&ani=9) number=number+(ani-0)*data; /此处为小数data=data/10; i+; /if Push2(s2,number); else /当前字符是运算符 switch(Precede(GetTop1(s1) ,ani) case: char theta=Pop1(s1); double a=Pop2(s2); double b=Pop2(s2); Push