中綴表達式轉(zhuǎn)后綴表達式并計算結(jié)果(C語言版).docx

中綴表達式轉(zhuǎn)后綴表達式中綴表達式轉(zhuǎn)后綴表達式的規(guī)則。1.遇到操作數(shù)：直接輸入到后綴表達式棧2.遇到運算符，直接入操作符棧3.遇到左括號：直接將其入棧4.遇到右括號：執(zhí)行出棧操作，并將出棧的元素輸出，直到彈出棧的是左括號，左括號不輸出。5.遇到其他運算符：加減乘除：彈出所有優(yōu)先級大于或者等于該運算符的棧頂元素，然后將該運算符入棧6.最終將操作符棧中的元素依次出棧，輸出到后綴表達式棧。以下是自己寫的代碼。親測沒有問題。（模擬一個計算器，可以帶括號，中間可以空格，只支持整數(shù)輸入，但是輸出結(jié)果精確到小數(shù)后6位）#include stdio.h#define MAX_LEN 100typedef struct calunsigned char isOper;/是否是操作數(shù) 1，操作符 0.操作數(shù)double Num;/值?；蛘呤遣僮鞣腁SCII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;/* 堆棧初始化*/void stackInit(void)int i;for(i=0;i=MAX_LEN)return;stackCaltopCal.Num = num;stackCaltopCal.isOper= isOper;topCal+;/* 操作符出棧*/char stackOperPop(void)if(topOper = 0)return 0;return stackOper-topOper;/* 操作符入棧*/void stackOperPush(char oper)if(topOper =MAX_LEN)return;stackOpertopOper+ = oper;/*比較兩個 sour sour1 的優(yōu)先級*1 sour = sour1 直接入操作符棧*0 sour sour1 直接入計算表達式棧*/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;/* 將輸入的字符串轉(zhuǎn)換為棧*/void StrToCal(char *pStr)int tmpNum = 0;char tmpOper;char islastNum = 0;/判斷上一個字符是什么。如果是符號，現(xiàn)在碰到-，那么數(shù)字就是負數(shù)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) /如果上一個字符是 操作符，那么接下來的數(shù)字是負數(shù)isNumNegative = 1;/ 1代表 是負數(shù)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) /如果最后操作符棧還有數(shù)據(jù)，直接入計算棧stackCalPush(stackOperPop(), IS_OPER);/* 顯示轉(zhuǎn)換后的后綴表達式，用于調(diào)試，看轉(zhuǎn)換是否正確*/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);/* 計算后綴表達式的計算結(jié)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