數(shù)據(jù)結(jié)構(gòu)課程設(shè)計(jì)算術(shù)表達(dá)式求值計(jì)算器Word

1、高級(jí)語言程序設(shè)計(jì)算術(shù)表達(dá)式求值課程設(shè)計(jì)報(bào)告 第一章 系統(tǒng)目標(biāo) 算術(shù)表達(dá)式求值 系統(tǒng)可以實(shí)現(xiàn)實(shí)現(xiàn)對(duì)算術(shù)四則混合運(yùn)算表達(dá)式求值，并打印求值過程中運(yùn)算符棧、操作數(shù)棧的變化過程。 第二章 系統(tǒng)分析 開始運(yùn)行時(shí)界面如下：你可以輸入一個(gè)表達(dá)式，按e對(duì)其進(jìn)行求值。 第三章 系統(tǒng)設(shè)計(jì) 開始輸出菜單界面，輸入表達(dá)式選c清除選e計(jì)算，求值選q確定否是結(jié)束，退出 第四章 系統(tǒng)實(shí)現(xiàn) #include <stdio.h>#include <conio.h>#include <stdlib.h>#include <string.h>#define n 100double n

2、umstackn=0;/操作數(shù)棧int numtop;char opstackn;/運(yùn)算符棧int optop;void print_num(double str1,int n) int i; printf("n操作數(shù)棧：n"); for(i=0;i<n;i+) printf("%g ",str1i);void print_op(char str2,int m) int j; printf("n運(yùn)算符棧：n"); for(j=0;j<m;j+) printf("%c ",str2j);int op(ch

3、ar ch)/判斷運(yùn)算符優(yōu)先級(jí) if(ch='+'|ch='-') return 2; if(ch='*'|ch='/') return 3; if(ch='(') return -1; return 0;double result(double num1,char op,double num2)/計(jì)算 if(op='+') return num1+num2; if(op='-') return num1-num2; if(op='*') return num1*nu

4、m2; if(op='/') return num1/num2; return 0;int compute(char str) double num=0; int i=0,j=1,k=1; numtop=optop=0; while(stri!='0'|optop>0) if(stri>='0'&&stri<='9') num=num*10+stri-'0' else if( k=1&&stri='-'&&(i=0|op(stri-1

5、) ) k=-1; else if(i>0&&!op(stri-1)&&stri!='('&&stri-1!=')') numstacknumtop+=num*k; if(optop!=0&&numtop!=0) print_num(numstack,numtop); num=0; j=1; k=1; if(optop=0|stri='(') opstackoptop+=stri;print_op(opstack,optop); else if(stri=')'

6、) while(optop>0&&opstack-optop!='(') numstacknumtop-2=result(numstacknumtop-2,opstackoptop,numstacknumtop-1); if(optop!=0&&numtop!=0) print_num(numstack,numtop); print_op(opstack,optop); numtop-; if(opstackoptop!='(') return 0; else if(stri='0'&&num

7、top=0) return 0; while(optop>0&&op(stri)<=op(opstackoptop-1) numstacknumtop-2=result(numstacknumtop-2,opstack-optop,numstacknumtop-1); if(optop!=0&&numtop!=0) print_num(numstack,numtop-1); print_op(opstack,optop); numtop-; if(stri!='0') opstackoptop+=stri; if(optop!=0&a

8、mp;&numtop!=0) print_op(opstack,optop); if(stri!='0') i+; if(numtop!=1|optop!=0) return 0; return 1;void menu() system("cls"); printf("_n"); printf(" clear(c) | equal(e) | quit(q) n"); printf("-n");int main(void) int i=0,j=0,k; char strn="0&qu

9、ot; char numn="0" char saven="0" char ch; double temp; unsigned long temp2; menu(); printf("input an expression,press key 'e' to computen"); ch=getch(); while( 1 ) if(ch=')'|op(ch)|ch>='0'&&ch<='9') stri+=ch; stri='0'

10、; menu(); printf("input an expression,press key 'e' to computen"); printf("%s",str); if( ch='-'&&(i=1|op(stri-2)|ch>='0'&&ch<='9' ) numj+=ch; numj='0' else j=0; if(ch='c'|ch='c') if(strlen(str) str-i=&#

11、39;0' menu(); printf("input an expression,press key 'e' to computen"); printf("%s",str); if(ch='e'|ch='e') if(compute(str) printf("n=%gn",numstack0); j=0; temp=numstack0; if(temp<0) temp=-temp; numj+='-' numj='0' temp2=(uns

12、igned long)temp; k=1; while(temp2/k>=10) k*=10; while(k) numj+=temp2/k+'0' numj='0' temp2=temp2%k; k/=10; temp=temp-(int)temp; if(temp!=0) numj+='.' numj='0' temp+=0.0000005; for(k=6;k>0;k-) if(temp=0) break; temp*=10; numj+=(int)temp+'0' numj='0'

13、; temp=temp-(int)temp; i=0; j=0; str0='0' if(ch='q'|ch='q') printf("nare you sure to quit?（y/n）n"); ch=getch(); if(ch='y'|ch='y') break; else menu(); printf("input an expression,press key 'e' to computen"); printf("%s",str); ch=getch(); return 0; 第五章 系統(tǒng)測試 1. 先輸入： 3+2*5 后按e求值2. 再輸入：12/4-5 后按e求