C/C++ 中缀表达式转换成后缀表达式并求值

Posted 2023-01-30 胡宝全

#include <stdio.h> 
#include <math.h> 
#include <stdlib.h> 

#define STACK_INIT_SIZE 100 
#define STACKINCREMENT 10 

typedef struct 
    char *base; 
    char *top; 
    int stacksize; 
SqStack1; 

typedef struct 
    double *base; 
    double *top; 
    int stacksize; 
SqStack2; 

SqStack1 OPTR; 
SqStack2 OPND; 

double EvaluateExpression(); 
void CHAR_InitStack(SqStack1 &S); 
void DOUBLE_InitStack(SqStack2 &S); 
char CHAR_GetTop(SqStack1 S); 
double DOUBLE_GetTop(SqStack2 S); 
int CHAR_Push(SqStack1 &S , char e); 
int DOUBLE_Push(SqStack2 &S , double e); 
char CHAR_Pop(SqStack1 &S , char &e); 
double DOUBLE_Pop(SqStack2 &S , double &e); 
char Precede(char a,char b); 
double Operate ( double d ,char e,double f); 
int In(char c); 


int main(void) 
 
    double result; 
    printf("*************************\\n"); 
    printf("*      表达式计算       *\\n"); 
    printf("*************************\\n"); 
    printf("请输入一组表达式(以#号结束):\\n"); 
    result = EvaluateExpression(); 
    printf("\\n\\n计算结果是(结果保留两位小数):%.2f\\n\\n\\n",result); 
 

void CHAR_InitStack(SqStack1 &S) 
 
    S.base=(char *)malloc(STACK_INIT_SIZE * sizeof(char)); 
    if(!S.base)exit(OVERFLOW); 
    S.top = S.base; 
    S.stacksize = STACK_INIT_SIZE; 
 

void DOUBLE_InitStack(SqStack2 &S) 
 
    S.base=(double *)malloc(STACK_INIT_SIZE * sizeof(double)); 
    if(!S.base)exit(OVERFLOW); 
    S.top = S.base; 
    S.stacksize = STACK_INIT_SIZE; 
 

char CHAR_GetTop(SqStack1 S) 
 
    char e; 
    if(S.top == S.base) return -1; 
    e = *(S.top-1); 
    return e; 
 

double DOUBLE_GetTop(SqStack2 S) 
 
    double e; 
    if(S.top == S.base) return -1; 
    e = *(S.top-1); 
    return e; 
 

int CHAR_Push(SqStack1 &S , char e) 
 
    if(S.top - S.base >= S.stacksize) 
     
        S.base = (char *)realloc(S.base , (S.stacksize + STACKINCREMENT) * sizeof(char)); 
        if(!S.base) exit(OVERFLOW); 
        S.top = S.base + S.stacksize; 
        S.stacksize += STACKINCREMENT; 
     
    *S.top++ = e; 
    return 1; 
 

int DOUBLE_Push(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) exit(OVERFLOW); 
        S.top = S.base + S.stacksize; 
        S.stacksize += STACKINCREMENT; 
     
    *S.top++ = e; 
    return 1; 
 

char CHAR_Pop(SqStack1 &S , char &e) 
 
    if(S.top == S.base)return -1; 
    e = * --S.top; 
    return e; 
 

double DOUBLE_Pop(SqStack2 &S , double &e) 
 
    if(S.top == S.base)return -1; 
    e = * --S.top; 
    return e; 
 

int In(char c)  // 判断是否为运算符 
 
    if ( c=='(' || c=='+' || c=='-' || c == '*' || c=='/' || c==')' || c=='#' || c=='^') 
        return 1; 
    else 
        return 0; 
 

char Precede(char a,char b) //判断优先级 
 
    char op; 
    switch(a) 
     
        case '#': 
            if (b=='#')  
                op='='; 
            else op='<'; 
            break; 
        case '+': 
            if (b=='*' || b=='/' || b=='('|| b=='^')  
                op='<'; 
            else op='>'; 
            break; 
        case '-': 
            if (b=='*' || b=='/' || b=='(' || b=='^') 
                op='<'; 
            else op='>'; 
            break; 
        case '*': 
            if (b=='(' || b=='^') 
                op='<'; 
            else op='>'; 
            break; 
        case '/': 
            if (b=='(' || b=='^') 
                op='<'; 
            else op='>'; 
            break; 
        case'^': 
            if(b=='(') 
                op='<'; 
            else op='>'; 
            break; 
        case '(': 
            if (b==')') 
                op='='; 
            else op='<'; 
            break; 
        case ')': 
            op='>'; 
            break; 
     
    return op; 
 

double Operate( double d ,char e,double f) //计算 
 
    double g; 
    switch(e) 
     
        case '+': 
            g=d+f; 
            break; 
        case '-': 
            g=d-f; 
            break; 
        case '*': 
            g=d*f; 
            break; 
        case '/': 
            g=d/f; 
            break; 
        case '^': 
            g=pow(d,f); 
            break; 
     
    return g; 
 

double EvaluateExpression() 
 
    char c=0,theta,x; 
    double a,b,number, n=0; 
    CHAR_InitStack (OPTR); 
    CHAR_Push (OPTR,'#'); 
    DOUBLE_InitStack (OPND); 
    c = getchar(); 
    printf("\\n转化成后缀表达式是:"); 
    while (c!='#'|| CHAR_GetTop(OPTR)!='#') 
     
        if(!In(c)) 
         
            number=0; 
            while(!In(c)) 
             
                if(c=='.') 
                    break; 
                number = number*10 +(c-48); 
                c = getchar(); 
             
            if(c=='.') 
             
                n=1; 
                while(!In(c = getchar())) 
                 
                    number = number +(c-48)*(double)pow(0.1 , n); 
                    n++; 
                 
             
            DOUBLE_Push (OPND,number); 
                printf("%.2f ",number); 
         
        else 
        switch (Precede(CHAR_GetTop(OPTR),c)) 
         
            case '<': 
                CHAR_Push(OPTR , c); 
                c=getchar(); 
                break; 
            case '=': 
                CHAR_Pop(OPTR , x); 
                c=getchar(); 
                break; 
            case '>': 
                CHAR_Pop(OPTR , theta); 
                printf("%c ",theta); 
                DOUBLE_Pop(OPND , a); 
                DOUBLE_Pop(OPND , b); 
                DOUBLE_Push(OPND,Operate(b,theta,a)); 
                break; 
         
     
    return (DOUBLE_GetTop(OPND)); 
//int EvaluateExpression