将浮点型算式的中缀表达式转换成后缀表达式并算出式子结果

最近因为需要了解如何将在Win应用程序控制台输入的算式表达式转化成其后缀表达式的算法，所以在网上搜索了一下，看到许多人的程序都只是对应于运算数在0~9的范围内的整型运算式，所以自己就写了一个可以计算浮点型算式的程序，一下是运行时的截图：

式子中的a,b,c是可供用户自行输入的变量。

首先，我先对输入的运算符进行了简单的合法性判断，我的判断代 码如下：

//函数的传入参数为一个字符型数组，该数组存放的是用户输入的表达式，还有一个引用传递的数组下标值  

bool IsValid(char exp[],int& position){bool flag=false;for(position;position<100;position++){if(exp[position]=='\0'){if(!(exp[0]=='(')&&!isdigit(exp[0])&&!(exp[0]=='a')&&!(exp[0]=='b')&&!(exp[0]=='c')){//判断表达式是否以+，-，*，/等运算符开头，如是，则表达式不合格cout<<"the expression can not end of operator"<<endl;return false;}if(!(exp[position-1]==')')&&!isdigit(exp[position-1])&&!(exp[position-1]=='a')&&!(exp[position-1]=='c')&&!(exp[position-1]=='b')){cout<<"the expression can not end of operator"<<endl;//判断表达式是否以+，-，*，/等运算符结尾，如是，则表达式不合格return false;}break;}else{if (!isdigit(exp[position])){if (exp[position]=='('){operatorstack.push(exp[position]);}else{if(exp[position]==')'){if (operatorstack.empty()){cout<<"the expression is unvalid!"<<endl;//若表达式有多余的"(",")"，则不合格return false;}else{operatorstack.pop();}}else{if (exp[position]=='a'||exp[position]=='b'||exp[position]=='c'){continue;}if(exp[position-1]=='+'||exp[position-1]=='-'||exp[position-1]=='*'||exp[position-1]=='^'||exp[position-1]=='/'||exp[position-1]=='%'||exp[position-1]=='&'||exp[position-1]=='|'||exp[position-1]=='>'||exp[position-1]=='<'){cout<<"the expression can not contain two continuous operators"<<endl;//若表达式中有两个以上的+，-，*，/等运算符出现，则不合格return false;}continue;}}}}}cout<<"the expression is valid!"<<endl;return true;}

上面我给出的例子的判断合法性输出为：

初步判断完算式的合法性之后，就进行后缀表达式的输出，我的程序思想是取决于我式子存放的数据类型的，我是将算式存放在一个char型的数组内，所以在输出后缀表达式的时候我就利用了这个存放格式的特点，我的函数In2Pos(char exp[],int& position ,char* tmp)有三个参数，第一个为存放算式的数组，第二个为引用参数，作为数组开始检查的其实位置，第三个参数为存放后缀表达式的char数组：

1，首先，判断是否为算式末尾的回车换行符'\0'，若是，则说明已经到了算式的结尾，将操作符栈的操作符都存放到后缀表达式的数组中，退出程序，否则进行下面的判断，

2，判断当前字符是否为数字，或小数点，或用户赋值的变量a,b,,c,若是则存放到后缀表达式的数组中，若不是这进行第3步判断

3，栈为空时，遇到运算符，直接入栈
4，遇到左括号：将其入栈
5，遇到右括号：执行出栈操作，并将出栈的元素存放到后缀表达式的数组中，直到弹出栈的是左括号，左括号不存放到后缀表达式      的数组中。
6，遇到其他运算符：加减乘除等，弹出所有优先级大于或者等于该运算符的栈顶元素，并将出栈的元素存放到后缀表达式的数组        中，然后将该运算符入栈

7，最后回到第1步

一下是我的代码实现：

void In2Pos(char exp[],int& position ,char* tmp){ int i=0,j;for (j=0;j<100;j++){tmp[j]='e';}stack<char> ope;for (position;position<100;position++){if (exp[position]=='\0'){while (!ope.empty()){char exchange;exchange=ope.top();ope.pop();tmp[i++]=exchange;}return ;}else{if (isdigit(exp[position])||exp[position]=='.'||exp[position]=='a'||exp[position]=='b'||exp[position]=='c'){tmp[i++]=exp[position];}else{if (ope.empty()){ope.push(exp[position]);}else{if (judge(ope.top(),exp[position])){while (!ope.empty()&&judge(ope.top(),exp[position])){ if (exp[position]==')'){while(ope.top()!='('){tmp[i++]=ope.top();ope.pop();}ope.pop(); break;}else{tmp[i++]=ope.top();ope.pop();}}  if (exp[position]!=')')  {  ope.push(exp[position]);  }}else{ ope.push(exp[position]);}}}}}return ;}

以下是后缀表达式的输出：

接下来就是计算算式的值了，我的思想是：

1，首先，判断是否为算式末尾的回车换行符'\0'，若是，则说明已经到了算式的结尾，则将结果sum返回，如不是则接下去判断，

2，先判断字符是否为数字，若是，则不断地从该字符往后检测是否为数字知道遇到下一个运算符，期间将数字的整数部分和小数部分计算出来，然后相加起来，压入数字栈中，如果不是数字则接下去判断，

3，如果是用户赋值的变量，则将对应的数值压入数字栈中，

4，若是运算符，运算符栈为空时，遇到运算符，直接入栈
5，遇到左括号：将其入栈
6，遇到右括号：执行出栈操作，并将出栈的元素输出赋值给operator1，然后弹出数字栈的两个数值进行运算，在将运算结果压入数字栈，然后循环，直到弹出栈的是左括号，左括号直接弹出。
7，遇到其他运算符：加减乘除：弹出所有优先级大于或者等于该运算符的栈顶元素，并将出栈的元素输出赋值给operator1，然后弹出数字栈的两个数值进行运算，在将运算结果压入数字栈，然后将该运算符入栈

8，最后到了结尾，返回1

计算程序的代码为：

float Evaluate(char exp[],int& position){int j;float sum=0,tmp1=0,tmp2=0,tmp3=0;float firstnum,lastnum;char operator1;stack<float> num; <span style="white-space:pre"></span>//数字栈stack<char> ope;<span style="white-space:pre"></span>//运算符栈while(true){if (exp[position]=='\0'){while (!ope.empty()){operator1=ope.top();ope.pop();lastnum=num.top();num.pop();firstnum=num.top();num.pop();tmp3=count(firstnum,lastnum,operator1);num.push(tmp3);}sum=num.top();return sum;}else{if (isdigit(exp[position])||exp[position]=='.'){while (isdigit(exp[position])||exp[position]=='.'){if (exp[position]=='.'){position++;int point=0;float pnum=0;while(isdigit(exp[position])){point++;pnum=(float)(exp[position]-48);for (j=0;j<point;j++){pnum=pnum/10;}tmp2+=pnum;position++;}}else{tmp1=tmp1*10+(int)(exp[position]-48);position++;}}tmp1=tmp1+tmp2;num.push(tmp1);tmp1=0;tmp2=0;}else if(exp[position]=='a'){num.push(a);position++;}else if(exp[position]=='b'){num.push(b);position++;}else if(exp[position]=='c'){num.push(c);position++;}else{if (ope.empty()){ope.push(exp[position]);position++;}else{if (judge(ope.top(),exp[position])){while (!ope.empty()&&judge(ope.top(),exp[position])){//cout<<"end"<<endl;if (exp[position]==')'){while(ope.top()!='('){//tmp[i++]=ope.top();operator1=ope.top();lastnum=num.top();num.pop();firstnum=num.top();num.pop();tmp3=count(firstnum,lastnum,operator1);ope.pop();num.push(tmp3);}ope.pop();//cout<<"end"<<endl;break;}else{//tmp[i++]=ope.top();operator1=ope.top();lastnum=num.top();num.pop();firstnum=num.top();num.pop();tmp3=count(firstnum,lastnum,operator1);ope.pop();num.push(tmp3);}}  if (exp[position]!=')')  {  ope.push(exp[position]);  }  position++;}else{ ope.push(exp[position]);position++;}}}}}return sum;}

以下是main函数：

<span style="font-size:14px;">// Expression.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include <iostream>#include <cstdlib>#include <cassert>#include <math.h>#include<stack>using  namespace std; float a,b,c;<span style="white-space:pre"></span>//全局变量a,b,c bool IsValid(char exp[],int& position){ bool flag=false;for(position;position<100;position++){if(exp[position]=='\0'){if(!(exp[0]=='(')&&!isdigit(exp[0])&&!(exp[0]=='a')&&!(exp[0]=='b')&&!(exp[0]=='c')){//判断表达式是否以+，-，*，/等运算符开头，如是，则表达式不合格cout<<"the expression can not end of operator"<<endl;return false;}if(!(exp[position-1]==')')&&!isdigit(exp[position-1])&&!(exp[position-1]=='a')&&!(exp[position-1]=='c')&&!(exp[position-1]=='b')){cout<<"the expression can not end of operator"<<endl;//判断表达式是否以+，-，*，/等运算符结尾，如是，则表达式不合格return false;}break;}else{if (!isdigit(exp[position])){if (exp[position]=='('){operatorstack.push(exp[position]);}else{if(exp[position]==')'){if (operatorstack.empty()){cout<<"the expression is unvalid!"<<endl;//若表达式有多余的"(",")"，则不合格return false;}else{operatorstack.pop();}}else{if (exp[position]=='a'||exp[position]=='b'||exp[position]=='c'){continue;}if(exp[position-1]=='+'||exp[position-1]=='-'||exp[position-1]=='*'||exp[position-1]=='^'||exp[position-1]=='/'||exp[position-1]=='%'||exp[position-1]=='&'||exp[position-1]=='|'||exp[position-1]=='>'||exp[position-1]=='<'){cout<<"the expression can not contain two continuous operators"<<endl;//若表达式中有两个以上的+，-，*，/等运算符出现，则不合格return false;}continue;}}}}}cout<<"the expression is valid!"<<endl;return true;}int judge(char op1,char op2){if (op2=='('){return 0;}if (op2==')'){return 1;}if (op2=='+'||op2=='-'){switch (op1){case '+':case '-':case '(':return 0;break;case '*':case '/':case '^':return 1;break;}}if (op2=='*'||op2=='/'){switch (op1){case '+':case '-':case '(':case '*':case '/':return 0;break;case '^':return 1;break;}}if (op2=='^'){switch (op1){case '+':case '-':case '(':case '*':case '/':case '^':return 0;break;}}}void In2Pos(char exp[],int& position ,char* tmp){ int i=0,j;for (j=0;j<100;j++){tmp[j]='e';}stack<char> ope;for (position;position<100;position++){if (exp[position]=='\0'){while (!ope.empty()){char exchange;exchange=ope.top();ope.pop();tmp[i++]=exchange;}return ;}else{if (isdigit(exp[position])||exp[position]=='.'||exp[position]=='a'||exp[position]=='b'||exp[position]=='c'){tmp[i++]=exp[position];}else{if (ope.empty()){ope.push(exp[position]);}else{if (judge(ope.top(),exp[position])){while (!ope.empty()&&judge(ope.top(),exp[position])){//cout<<"end"<<endl;if (exp[position]==')'){while(ope.top()!='('){tmp[i++]=ope.top();ope.pop();}ope.pop();//cout<<"end"<<endl;break;}else{tmp[i++]=ope.top();ope.pop();}}  if (exp[position]!=')')  {  ope.push(exp[position]);  }}else{ ope.push(exp[position]);}}}}}return ;}float count(float first,float last,char ope){float cot;switch (ope){case '+':cot=first+last;break;case '-':cot=first-last;break;case '*':cot=first*last;break;case '/':cot=first/last;break;case '^':cot=pow(first,last);break;}return cot;}float Evaluate(char exp[],int& position){int j;float sum=0,tmp1=0,tmp2=0,tmp3=0;float firstnum,lastnum;char operator1;stack<float> num;stack<char> ope;while(true){if (exp[position]=='\0'){while (!ope.empty()){operator1=ope.top();ope.pop();lastnum=num.top();num.pop();firstnum=num.top();num.pop();tmp3=count(firstnum,lastnum,operator1);num.push(tmp3);}sum=num.top();return sum;}else{if (isdigit(exp[position])||exp[position]=='.'){while (isdigit(exp[position])||exp[position]=='.'){if (exp[position]=='.'){position++;int point=0;float pnum=0;while(isdigit(exp[position])){point++;pnum=(float)(exp[position]-48);for (j=0;j<point;j++){pnum=pnum/10;}tmp2+=pnum;position++;}}else{tmp1=tmp1*10+(int)(exp[position]-48);position++;}}tmp1=tmp1+tmp2;num.push(tmp1);tmp1=0;tmp2=0;}else if(exp[position]=='a'){num.push(a);position++;}else if(exp[position]=='b'){num.push(b);position++;}else if(exp[position]=='c'){num.push(c);position++;}else{if (ope.empty()){ope.push(exp[position]);position++;}else{if (judge(ope.top(),exp[position])){while (!ope.empty()&&judge(ope.top(),exp[position])){//cout<<"end"<<endl;if (exp[position]==')'){while(ope.top()!='('){//tmp[i++]=ope.top();operator1=ope.top();lastnum=num.top();num.pop();firstnum=num.top();num.pop();tmp3=count(firstnum,lastnum,operator1);ope.pop();num.push(tmp3);}ope.pop();//cout<<"end"<<endl;break;}else{//tmp[i++]=ope.top();operator1=ope.top();lastnum=num.top();num.pop();firstnum=num.top();num.pop();tmp3=count(firstnum,lastnum,operator1);ope.pop();num.push(tmp3);}}  if (exp[position]!=')')  {  ope.push(exp[position]);  }  position++;}else{ ope.push(exp[position]);position++;}}}}}return sum;} int main(){cout<<"My procedure can calculate integer type and float-point type equation."<<endl;cout<<"welcome to operate my procedure."<<endl;cout<<"please input the value of a,b,c:"<<endl;cin>>a>>b>>c;cout<<"your input is \na="<<a<<"\nb="<<b<<"\nc="<<c<<endl;cin.ignore();int position=0;char expression[100];cout<<"please input your expression"<<endl;cin.getline(expression,100); IsValid(expression,position);cout<<endl; position=0; char t[100];In2Pos(expression,position,t);cout<<"the postfix expression of this expression is:"<<endl;int i=0;while(t[i]!='e'){cout<<(char)t[i];i++;}cout<<endl;cout<<endl;position=0;cout<<"the answer of the expression is :"<<endl;cout<<Evaluate(expression,position)<<endl;cout<<endl;cout<<"thanks for operating my procedure."<<endl;return 0;}</span>