用兩種方法實(shí)現(xiàn)表達(dá)式自動(dòng)計(jì)算

1、精選優(yōu)質(zhì)文檔-傾情為你奉上一、設(shè)計(jì)思想計(jì)算算數(shù)表達(dá)式并求值，可以采取兩種算法：1.先將算術(shù)表達(dá)式轉(zhuǎn)化為后綴表達(dá)式，然后對(duì)后綴表達(dá)式進(jìn)行計(jì)算。2.直接對(duì)算術(shù)表達(dá)式進(jìn)行計(jì)算。下面依次對(duì)兩種方法進(jìn)行分析：第一種算法有兩步1.先將算數(shù)表達(dá)式轉(zhuǎn)化為后綴表達(dá)式。在計(jì)算過程中，第一，要先建立一個(gè)存放操作符的棧和一個(gè)存放數(shù)字的數(shù)組。首先對(duì)用戶輸入的表達(dá)式進(jìn)行掃面，如果是數(shù)字或者是小數(shù)點(diǎn)，直接存入數(shù)組。如果是操作符，就判斷是否為空或者是否為“（”或者是否它的優(yōu)先級(jí)大于操作符棧頂?shù)膬?yōu)先級(jí)，如果是，就入棧，索引后移；如果它的優(yōu)先級(jí)不大于棧頂操作符，棧頂?shù)牟僮鞣鰲?，進(jìn)入數(shù)組，如此循環(huán)，直到棧頂?shù)膬?yōu)先級(jí)小于掃描到的操

2、作符的優(yōu)先級(jí)的時(shí)候，操作符入棧，索引后移。當(dāng)遇到標(biāo)識(shí)符0時(shí)，掃描結(jié)束。數(shù)組中存放的就是后綴表達(dá)式。2.利用后綴表達(dá)式進(jìn)行計(jì)算。得到后綴表達(dá)式后，要計(jì)算就需要用到數(shù)值棧。首先對(duì)后綴表達(dá)式進(jìn)行掃描，遇到數(shù)字字符，將數(shù)字字符轉(zhuǎn)化為浮點(diǎn)類型，放入數(shù)值棧中，遇到操作符，就從數(shù)值棧中取出兩個(gè)數(shù)，進(jìn)行計(jì)算后將計(jì)算結(jié)果再放入數(shù)值棧中，掃描下一個(gè)，最后的計(jì)算結(jié)果就存到了數(shù)值棧中，直接取出數(shù)值棧棧頂元素，就是計(jì)算的最后結(jié)果。第二種算法首先要建立兩個(gè)棧，一個(gè)存放操作符的棧，一個(gè)是存放數(shù)值的棧。然后開始對(duì)用戶輸入的表達(dá)式進(jìn)行掃描，如果是數(shù)值或是小數(shù)點(diǎn)，就將其轉(zhuǎn)化為浮點(diǎn)型數(shù)據(jù)，然后進(jìn)入數(shù)值棧；如果是操作符，當(dāng)前索引下的

3、操作符的優(yōu)先級(jí)如果不大于棧頂操作符的優(yōu)先級(jí)，就將棧頂操作符取出來，從數(shù)值棧中取出兩個(gè)數(shù)值，進(jìn)行計(jì)算，結(jié)果存入數(shù)值棧。如此循環(huán)，直到棧頂操作符的優(yōu)先級(jí)小于當(dāng)前索引的操作符的優(yōu)先級(jí)是，操作符入棧，然后對(duì)下一個(gè)字進(jìn)行掃描。如果是左括號(hào)，則不進(jìn)行優(yōu)先級(jí)的比較，直接入棧。如果是右括號(hào)，則從數(shù)值棧中取兩個(gè)操作數(shù)，符號(hào)棧中取出一個(gè)符號(hào)，然后進(jìn)行計(jì)算后得數(shù)放入數(shù)棧中，不斷進(jìn)行此類操作，直到從棧中取出的是左括號(hào)為止，左括號(hào)丟棄，掃描下一個(gè)。掃描結(jié)束后，數(shù)值棧中存放的數(shù)值就是計(jì)算產(chǎn)生的結(jié)果，最后把數(shù)值從數(shù)值棧中取出，就是所得的計(jì)算結(jié)果。二、算法流程圖第一種算法:先將表達(dá)式轉(zhuǎn)化為后綴表達(dá)式，然后計(jì)算表達(dá)式的值。 1

4、、中綴表達(dá)式轉(zhuǎn)化為后綴表達(dá)式說明：首先獲得算術(shù)運(yùn)算符，對(duì)其進(jìn)行掃面，如果是數(shù)字或者是小數(shù)點(diǎn)直接進(jìn)入字符數(shù)組。如果是操作符，需要判斷它與操作符棧頂操作符的優(yōu)先級(jí)大小來決定是入棧還是棧頂出棧。如果是括號(hào)，左括號(hào)直接入棧，右括號(hào)從棧中找左括號(hào)，然后拋棄。掃描結(jié)束，字符數(shù)組中存放的就是后綴表達(dá)式，輸出即可。將中綴表達(dá)式轉(zhuǎn)化為后綴表達(dá)式流程圖如下：出棧入數(shù)組獲取表達(dá)式直接入棧不是左括號(hào)右括號(hào)看棧頂左括號(hào)哪種括號(hào)優(yōu)先級(jí)不大于棧頂元素棧頂元素出棧，放入字符數(shù)組與棧頂比較操作符直接放入字符數(shù)組小數(shù)點(diǎn)或者數(shù)字2、后綴表達(dá)式的計(jì)算說明：首先獲取后綴表達(dá)式，對(duì)表達(dá)式進(jìn)行掃描，如果是數(shù)字轉(zhuǎn)換成相應(yīng)的浮點(diǎn)型數(shù)字，放入數(shù)

5、棧，如果是操作符，進(jìn)行相應(yīng)的計(jì)算。后綴表達(dá)式的計(jì)算，實(shí)現(xiàn)的流程圖如下：入棧圖1 中綴表達(dá)式轉(zhuǎn)化為后綴表達(dá)式算法流程圖優(yōu)先級(jí)大于棧頂元素如果是括號(hào)掃描判斷得到后綴表達(dá)式掃描判斷轉(zhuǎn)換成浮點(diǎn)型，入數(shù)棧數(shù)字操作符數(shù)棧中取兩數(shù)字計(jì)算，結(jié)果入棧取出計(jì)算結(jié)果圖2 后綴表達(dá)式計(jì)算算法流程圖第二種算法：直接計(jì)算出結(jié)果直接計(jì)算出結(jié)果的算法流程圖如下：返回?cái)?shù)棧棧頂元素獲取計(jì)算表達(dá)式掃描判斷轉(zhuǎn)換成浮點(diǎn)型，入數(shù)棧數(shù)字與棧頂比較操作符數(shù)棧取出兩個(gè)元素進(jìn)行計(jì)算，結(jié)果入數(shù)棧直接入棧優(yōu)先級(jí)高優(yōu)先級(jí)低括號(hào)類型入棧棧中取出操作符括號(hào)左括號(hào)右括號(hào)是否“（”丟棄是不是數(shù)棧取出兩個(gè)元素進(jìn)行計(jì)算，結(jié)果入數(shù)棧操作符棧是否空是操作符棧頂元素與

6、數(shù)棧取出兩個(gè)元素進(jìn)行計(jì)算，結(jié)果入數(shù)棧不是圖3 直接進(jìn)行表達(dá)式求值算法流程圖說明：首先得到要計(jì)算的表達(dá)式，并對(duì)其進(jìn)行掃描，遇到的字符是數(shù)字，轉(zhuǎn)換成浮點(diǎn)型，入棧。如果是括號(hào)，視其左右，判斷是入?；蛘呤莵G棄。如果是操作符，視其優(yōu)先級(jí)判斷是直接入棧還是進(jìn)行計(jì)算。三、源代碼1、先轉(zhuǎn)換成后綴表達(dá)式，再進(jìn)行表達(dá)式計(jì)算#include<stdio.h>#include<string.h>#include<math.h>#define MAXLENGTH 100 #define EMPTY -1 #define FULL 99typedef struct char op MA

7、XLENGTH;int p ;stack;typedef struct double numMAXLENGTH;int p;number;int Isempty (stack *ps) if (ps->p=EMPTY) return 1; else return 0;int Isfull ( stack *ps)if (ps->p = FULL)return 1;else return 0;int Push ( stack *ps,char ch)if (Isfull(ps)return 0;else ps->p +;ps->opps->p = ch ;retur

8、n 1;char Pop ( stack *ps )if (Isempty(ps)return 0;else return ps->opps->p-;char Top ( stack *ps)if (Isempty(ps)return 0;else return ps->opps->p;int Isemptyn (number *ps) if (ps->p=EMPTY) return 1; else return 0;int Isfulln ( number *ps)if (ps->p = FULL)return 1;else return 0;int Pu

9、shn ( number *ps,double dl)if (Isfulln(ps)return 0;else ps->p +;ps->numps->p = dl ;return 1;double Popn ( number *ps )if (Isemptyn(ps)return 0;else return ps->numps->p-;double Topn ( number *ps)if (Isemptyn(ps)return 0;else return ps->numps->p;int getoplevel(char c)if (c='+&

10、#39;|c='-')return 1;else if (c='*'|c='/'|c='%')return 2;else if (c='(')return 0;else return -1;void init(stack *mystack) mystack->p = EMPTY;void initnum(number *mynum)mynum->p = EMPTY;void calculate(char *mylist,stack *mystack)number mynum;int s=0 ;char

11、mynuMAXLENGTH;int smynu =0;double firstod ,secondod,result;double operand;initnum (&mynum);while(mylists!='0')if (mylists='+'|mylists='-'|mylists='*'|mylists='/'|mylists='%'|mylists=' ')switch(mylists)case '+':firstod = Popn(&my

12、num);secondod = Popn(&mynum);result = secondod + firstod;Pushn(&mynum,result);break;case '-':firstod = Popn(&mynum);secondod = Popn(&mynum);result = secondod - firstod;Pushn(&mynum,result);break;case '*':firstod = Popn(&mynum);secondod = Popn(&mynum);resul

13、t = secondod * firstod;Pushn(&mynum,result);break;case '/':firstod = Popn(&mynum);secondod = Popn(&mynum);if(firstod =0)printf("0 can not be dividorn ");return;result = secondod / firstod;Pushn(&mynum,result);break;case '%':firstod = Popn(&mynum);secondo

14、d = Popn(&mynum);result = (double)(int)secondod % (int)firstod);Pushn(&mynum,result);break;case ' ':break;s +;elsewhile (mylists<='9'&&mylists>='0'|mylists='.')mynusmynu = mylists;s +;smynu +;mynusmynu = '0'operand = atof(mynu);Pushn(&

15、;mynum,operand);smynu =0;result = Popn(&mynum);printf("結(jié)果是:");printf("%f n",result);void getexpression (char *myexp,stack mystack,char *mylist)int i =0;int s=0;int slist=0; while(myexps!='0') if(myexps<='9'&&myexps>='0'|myexps='.'

16、) while(myexps<='9'&&myexps>='0'|myexps='.') mylistslist = myexp s; s+; slist +; mylistslist=' ' slist+; else if (Isempty(&mystack)|myexps='(') Push(&mystack,myexps); s +; else if (myexps =')') while(Top(&mystack)!='('

17、) mylistslist = Pop(&mystack); slist +; mylistslist=' ' slist+; Pop(&mystack); s+; else if (getoplevel(myexps)>getoplevel(Top(&mystack) Push(&mystack,myexps); s+; else while(getoplevel(myexps)<=getoplevel(Top(&mystack) mylistslist = Pop(&mystack); slist +; mylis

18、tslist=' ' slist+; while(!Isempty(&mystack) mylistslist = Pop(&mystack); slist +; mylistslist=' ' slist+; mylistslist = '0' printf("后綴表達(dá)式為: n"); printf("%s",mylist); printf("n"); calculate(mylist,&mystack);void main()char myexp MAXLEN

19、GTH;stack mystack;char mylistMAXLENGTH ;printf("請(qǐng)輸入:n");gets(myexp);init(&mystack);getexpression(myexp,mystack,mylist); getch();2、 用直接計(jì)算表達(dá)式值#include<stdio.h>#include<math.h>#define MAXLENGTH 100 #define EMPTY -1 #define FULL 99 typedef struct char op MAXLENGTH; int p ;stack

20、;typedef struct double numMAXLENGTH;int p;number;int Isempty (stack *ps) if (ps->p=EMPTY) return 1; else return 0;int Isfull ( stack *ps)if (ps->p = FULL)return 1;else return 0;int Push ( stack *ps,char ch)if (Isfull(ps)return 0;else ps->p +;ps->opps->p = ch ;return 1;char Pop ( stack

21、 *ps )if (Isempty(ps)return 0;else return ps->opps->p-;char Top ( stack *ps)if (Isempty(ps)return 0;else return ps->opps->p;int Isemptyn (number *ps) if (ps->p=EMPTY) return 1; else return 0;int Isfulln ( number *ps)if (ps->p = FULL)return 1;else return 0;int Pushn ( number *ps,dou

22、ble dl)if (Isfulln(ps)return 0;else ps->p +;ps->numps->p = dl ;return 1;double Popn ( number *ps )if (Isemptyn(ps)return 0;else return ps->numps->p-;double Topn ( number *ps)if (Isemptyn(ps)return 0;else return ps->numps->p;int getoplevel(char c)if (c='+'|c='-')r

23、eturn 1;else if (c='*'|c='/'|c='%')return 2;else if (c='(')return 0;else return -1;void init(stack *mystack) mystack->p = EMPTY;void initnum(number *mynum)mynum->p = EMPTY;void calculate(char *myexp)stack mystack;number numstack;char mylistMAXLENGTH;int index =0

24、;int indexlist=0;double od1 ,od2;double result,operand ;init(&mystack);initnum(&numstack);while(myexpindex!='0')/if (myexpindex<='9'&&myexpindex>='0'|myexpindex='.')while(myexpindex<='9'&&myexpindex>='0'|myexpindex=&

25、#39;.')mylistindexlist = myexpindex;index +;indexlist +;mylistindexlist = '0'operand = atof(mylist);Pushn(&numstack,operand);indexlist = 0;else if(Isempty(&mystack)|myexpindex='('|getoplevel(myexpindex)>getoplevel(Top(&mystack)Push(&mystack,myexpindex);index +;

26、else if (myexpindex=')')while(Top(&mystack)!='(') od1 = Popn(&numstack);od2 = Popn(&numstack);switch(Top(&mystack)case '+': Pushn(&numstack,od2 + od1);break;case '-':Pushn(&numstack,od2 - od1);break;case '*':Pushn(&numstack,od2 * od

27、1);break;case '/':if(od1 =0)printf("0 can not be divisor n");return ;Pushn(&numstack,od2 / od1);break;case '%':Pushn(&numstack,(double)(int)od2 % (int)od1);break; Pop(&mystack); index +; Pop(&mystack); elsewhile(getoplevel(myexpindex)<=getoplevel(Top(&

28、;mystack)od1 = Popn(&numstack);od2 = Popn(&numstack);switch(Top(&mystack)case '+':Pushn(&numstack,od2 + od1);break;case '-':Pushn(&numstack,od2 - od1);break;case '*':Pushn(&numstack,od2 * od1);break;case '/':if(od1 =0) printf("0 不能做除數(shù)!請(qǐng)重新輸

29、入.n");return ;Pushn(&numstack,od2 / od1);break;case '%':Pushn(&numstack,(double)(int)od2 % (int)od1);break;Pop(&mystack);Push(&mystack,myexpindex);index +;while (!Isempty(&mystack)od1 = Popn(&numstack);od2 = Popn(&numstack);switch(Top(&mystack)case '+

30、':Pushn(&numstack,od2 + od1);break;case '-':Pushn(&numstack,od2 - od1);break;case '*':Pushn(&numstack,od2 * od1);break;case '/':if(od1 =0) printf("0 can not be divisor n");return ;Pushn(&numstack,od2 / od1);break;case '%':Pushn(&numst

31、ack,(double)(int)od2 % (int)od1);break;Pop(&mystack);result = Topn(&numstack);printf("The result is:n");printf("%f n",result);void main()char myexpMAXLENGTH;printf("Please input the expression:n");gets(myexp);calculate(myexp); getch();四、運(yùn)行結(jié)果第一種算法程序運(yùn)行結(jié)果圖：正確輸入表達(dá)式，

32、程序運(yùn)行結(jié)果如圖：圖5 正確表達(dá)式運(yùn)行結(jié)果圖第二種算法程序運(yùn)行結(jié)果圖：正確輸入表達(dá)式，程序運(yùn)行結(jié)果如圖：圖6 正確表達(dá)式運(yùn)行結(jié)果圖五、遇到的問題及解決這部分我主要遇到了如下兩個(gè)問題，其內(nèi)容與解決方法如下所列：一要把后綴表達(dá)式放到了字符數(shù)組中，是如何將后綴表達(dá)式中的數(shù)字字符轉(zhuǎn)換成真正的數(shù)字？如果是單個(gè)的數(shù)字，可以進(jìn)行強(qiáng)制轉(zhuǎn)換，但是數(shù)字不總都是單個(gè)的，我們要計(jì)算的數(shù)字往往是多位的如：12，或者帶有小數(shù)點(diǎn)如：1.2，那么怎么將其轉(zhuǎn)換成真正的浮點(diǎn)型的數(shù)字呢？解決方案：分析問題，我們知道重點(diǎn)是如何將多位的數(shù)字字符轉(zhuǎn)換成浮點(diǎn)型數(shù)字，同時(shí)也知道多位的數(shù)字字符是存儲(chǔ)地址是連續(xù)的，于是根據(jù)這個(gè)特點(diǎn)，我查閱了一下C語(yǔ)言庫(kù)函數(shù)，果然真的有符合條件的庫(kù)函數(shù)atof（）；首先atof（）函數(shù)在“math.h”頭文件下，引用“math.h”就可以使用atof（）函數(shù)直接將地址連續(xù)的多位數(shù)字字符轉(zhuǎn)換成相應(yīng)的浮點(diǎn)型數(shù)字。有了思路，我想到將字符數(shù)組中的數(shù)字按分隔符依次轉(zhuǎn)移到另一個(gè)字符數(shù)組中，并利用atof（）函數(shù)將其轉(zhuǎn)換成相應(yīng)的數(shù)字，但是