堆棧計(jì)算器實(shí)驗(yàn)報(bào)告

1、一、實(shí)驗(yàn)?zāi)康谋敬螌?shí)驗(yàn)為了學(xué)習(xí)用基本的數(shù)據(jù)結(jié)構(gòu)解決實(shí)際應(yīng)用中的問題，將學(xué)習(xí)的理論知識(shí)應(yīng)用于實(shí)踐，增強(qiáng)解決實(shí)際問題的能力。學(xué)會(huì)利用棧結(jié)構(gòu)，按照算術(shù)運(yùn)算優(yōu)先級(jí)順序，實(shí)現(xiàn)基本算術(shù)表達(dá)式的運(yùn)算過程。二、實(shí)驗(yàn)要求功能要求：要求使用堆棧實(shí)現(xiàn)算術(shù)表達(dá)式的求值功能。數(shù)據(jù)結(jié)構(gòu)要求：使用兩個(gè)工作棧，一個(gè)為optr,用于寄存運(yùn)算符；另一個(gè)為opnd,用于寄存操作數(shù)和運(yùn)算結(jié)構(gòu)。數(shù)據(jù)要求：輸入：操作數(shù)棧的數(shù)據(jù)為整型，運(yùn)算符棧的數(shù)據(jù)位字符型 輸出：整型數(shù)據(jù)三、設(shè)計(jì)實(shí)現(xiàn)數(shù)據(jù)結(jié)構(gòu)設(shè)計(jì)：設(shè)置兩個(gè)棧，一個(gè)為操作數(shù)棧opnd,一個(gè)為運(yùn)算符棧optr。輸入的表達(dá)式，遇到如果是操作數(shù)就壓入到操作數(shù)堆棧中，如果是運(yùn)算符就將優(yōu)先級(jí)與當(dāng)前堆棧

2、運(yùn)算符的優(yōu)先級(jí)比較，誰的優(yōu)先級(jí)數(shù)大就壓入堆棧，否則將棧頂運(yùn)算符彈出來進(jìn)行運(yùn)算。程序流程設(shè)計(jì)：定義鏈棧，判斷運(yùn)算符優(yōu)先級(jí)，實(shí)現(xiàn)具體計(jì)算，錯(cuò)誤處理、算法流程：主功能程序代碼實(shí)現(xiàn)：#include stdafx.h#include stdlib.h#include math.h#include string.h#include conio.h#include iostream.h#include float.hstruct optr /存放操作符的棧char * base;char* top;int size;struct opnd /存放操作數(shù)的棧 int* base;int* top;int s

3、ize;void initstack(struct optr* s)/申請(qǐng)操作符棧的空間及長度s-base = (char*)malloc(100*sizeof(char);s-top = s-base;s-size =100;void initstack(struct opnd* s)/申請(qǐng)操作數(shù)棧的空間及長度s-base = (int*)malloc(100*sizeof(int);s-top = s-base;s-size =100;struct optr* push(struct optr* stack , char opp)/插入操作符opp*stack-top = opp;stac

4、k-top+;return stack;struct opnd* push(struct opnd* stack , int num , int i)*stack-top = num;stack-top+;return stack;char gettop(struct optr* stack)/取棧頂元素return *(stack-top-1);int gettop(struct opnd * stack) /取棧頂元素return *(stack-top-1);char precede(char oldchar , char newchar)int oldint;int newint;sw

5、itch (oldchar)case +:oldint = 4;break;case -:oldint = 4;break;case *:oldint = 6;break;case /:oldint = 6;break;case (:oldint = 2;break;case ):oldint = 7;break;case =:oldint = 0;break;switch (newchar)case +:newint = 3;break;case -:newint = 3;break;case *:newint = 5;break;case /:newint = 5;break;case (

6、:newint = 7;break;case ):newint = 2;break;case =:newint = 0;break;if(oldint newint)return ; if(oldint newint)return ;if(oldint = newint)return =;int add(int a , int b)return (a+b);int minute(int a ,int b)return (a-b);int multiply (int a ,int b)return (a*b);int division (int a,int b)return (a/b);int

7、operate(int a , char theta ,int b)switch (theta)case +:return add(a,b);case -:return minute(a ,b);case *:return multiply(a,b);case /:return division(a,b);int main(int argc, char* argv)char a80; /建立一個(gè)字符串用來存放表達(dá)式char temp;int stringcount = 0;struct optr* stackoptr=(struct optr*)malloc(sizeof(optr);stru

8、ct opnd* stacknopnd=(struct opnd*)malloc(sizeof(opnd);printf(請(qǐng)輸入算數(shù)表達(dá)式); scanf(%s,a); initstack(stackoptr);stackopp = push(stackopnd,= );initstack(stackopnd);temp = astringcount;stringcount+;while(temp != = | gettop(stackopptr) != = )/如果不是等于號(hào)執(zhí)行下面的操作if(!in(temp)/將輸入字符轉(zhuǎn)化成數(shù)字，并入棧，然后加一int value = (int)tem

9、p-48;stacknum = push(stackoptr,value,0); coutgettop(stackopnd) ;temp = astringcount;stringcount= stringcount+1;elseswitch(precede(gettop(stackopnd),temp)case ;int a = (int)*(-stackoptr-top);coutgettop(stackopnd)top);char theta = *(-stackopnd-top);stackopnd = push(stackopnd,operate(b,theta,a),0);coutgettop(stackopnd)top-;temp = astringcount;stringcount+;break;printf(感謝您的使用:n);printf(n運(yùn)算結(jié)果=：%dn,gettop(stackopnd); return 0;四、實(shí)驗(yàn)結(jié)果輸入輸出結(jié)果 1. 輸入：5+2（7-4）=輸出：數(shù)據(jù)棧棧頂元素：3，7，2，7，5，3，15最后結(jié)果為1