中缀表达式转后缀表达式并计算结果(C语言版).docx

中缀表达式转后缀表达式中缀表达式转后缀表达式的规则。1.遇到操作数：直接输入到后缀表达式栈2.遇到运算符，直接入操作符栈3.遇到左括号：直接将其入栈4.遇到右括号：执行出栈操作，并将出栈的元素输出，直到弹出栈的是左括号，左括号不输出。5.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，然后将该运算符入栈6.最终将操作符栈中的元素依次出栈，输出到后缀表达式栈。以下是自己写的代码。亲测没有问题。（模拟一个计算器，可以带括号，中间可以空格，只支持整数输入，但是输出结果精确到小数后6位）#include stdio.h#define MAX_LEN 100typedef struct calunsigned char isOper;/是否是操作数 1，操作符 0.操作数double Num;/值。或者是操作符的ASCII值STRUCT_CAL;#define IS_NUM 0x00#define IS_OPER0x01STRUCT_CAL stackCalMAX_LEN;STRUCT_CAL stackCalBackMAX_LEN;unsigned char topCal;char stackOperMAX_LEN;unsigned char topOper;/* 堆栈初始化*/void stackInit(void)int i;for(i=0;i=MAX_LEN)return;stackCaltopCal.Num = num;stackCaltopCal.isOper= isOper;topCal+;/* 操作符出栈*/char stackOperPop(void)if(topOper = 0)return 0;return stackOper-topOper;/* 操作符入栈*/void stackOperPush(char oper)if(topOper =MAX_LEN)return;stackOpertopOper+ = oper;/*比较两个 sour sour1 的优先级*1 sour = sour1 直接入操作符栈*0 sour sour1 直接入计算表达式栈*/unsigned char comparPrior(char sour, char sour1)if(sour =0 |sour1 = 0) return 1;switch(sour)case +:case -:if(sour1 = * |sour1 = /|sour1 = + |sour1 = - )return 0;elsereturn 1;break;case *:case /:if(sour1 = * |sour1 = /|sour1 = + |sour1 = -|sour1 = ()return 1;elsereturn 0;break;default:return 1;break;/* 将输入的字符串转换为栈*/void StrToCal(char *pStr)int tmpNum = 0;char tmpOper;char islastNum = 0;/判断上一个字符是什么。如果是符号，现在碰到-，那么数字就是负数char isNumNegative = 0;while(*pStr != 0)switch(*pStr)case +:while(comparPrior(+,stackOpertopOper-1) = 0)stackCalPush(stackOperPop(), IS_OPER);if(topOper1)break;stackOperPush(+);pStr+;islastNum = 0;break;case -:if(islastNum = 0) /如果上一个字符是 操作符，那么接下来的数字是负数isNumNegative = 1;/ 1代表 是负数pStr+;break;while(comparPrior(-,stackOpertopOper-1) = 0)stackCalPush(stackOperPop(), IS_OPER);if(topOper1)break;stackOperPush(-);pStr+;islastNum = 0;break;case *:while(comparPrior(*,stackOpertopOper-1) = 0)stackCalPush(stackOperPop(), IS_OPER);if(topOper1)break;stackOperPush(*);pStr+;islastNum = 0;break;case /:while(comparPrior(/,stackOpertopOper-1) = 0)stackCalPush(stackOperPop(), IS_OPER);if(topOper=0 )& (*pStr =9)tmpNum = tmpNum *10 +*(pStr+) - 0;if(isNumNegative)isNumNegative = 0;tmpNum = -tmpNum;stackCalPush(tmpNum, IS_NUM);tmpNum = 0;islastNum = 1;break;case :pStr+;break;default:pStr+;break;while(topOper) /如果最后操作符栈还有数据，直接入计算栈stackCalPush(stackOperPop(), IS_OPER);/* 显示转换后的后缀表达式，用于调试，看转换是否正确*/void dispCalc(void)int i = 0;printf(n);for(i = 0;itopCal;i+)if(stackCali.isOper = IS_NUM)printf(%d,(int)stackCali.Num);else if(stackCali.isOper = IS_OPER)printf(%c,(char)stackCali.Num);/* 计算后缀表达式的计算结果*/void suffixExpression(void)int i = 0;int tmpTop;double num1,num2;for(i = 0;iNum;num2 = stackCalPop()-Num;switch(char)stackCalBacki.Num)case +:stackCalPush(num2+num1, IS_NUM);break;case -:stackCalPush(num2-num1, IS_NUM);break;case *:stackCalPush(num2*num1, IS_NUM);break;case /:stackCalPush(int)num2/num1, IS_NUM);break;default :break;elsestackCalPush(stackCalBacki.Num, IS_NUM);i+;if(i=tmpTop) break;printf(nThe result is : %fn, stackCal0.Num);stackInit();int main