1、C+ 語言編寫#include<iostream>#include<cmath>#include<string>using namespace std;const int SIZE = 1000;typedef struct node/ 為了處理符號而建立的鏈表 ( 如 : 1+(-2) char data;node *next;node;typedef struct stack_num/ 存儲 數 的棧 double *top;double *base;stack_num;typedef struct stack_char/ 存儲 運算符號 的棧char

2、*top;char *base;stack_char;stack_num S_num;/ 定義stack_char S_char;/ 定義char fu18 = 'n', ')：'+','-', '*', '/', '%', ' A','Q', 'L', 'C', 'S', 'T', 'c', 's', 't', '('int com

3、pare1000;/ 表現出各運算符號的優先級double shu1000;/ 存儲 " 數 " 的數組double dai_result; 運算的結果，是為了處理 M 運算( 簡介函數里有 M 的定義 ) int biao = 0;/ 和 dia_result 一樣，為了處理 M 運算 char lineSIZE;/ 輸入的所要計算的表達式 void init()/ 初始化comparefu0 = -2;/ 用數字的大小表現出符號的優先級comparefu1 = -1;comparefu2 = 2;comparefu3 = 2; comparefu4 = 4;compar

4、efu5 = 4; comparefu6 = 4;comparefu7 = 5;為棧開辟空for(int i = 8; i <= 15; i+) comparefui = 6;comparefu16 = 7;S_num.base = (double*)malloc(sizeof(double)*SIZE);/同上S_char.base = (char*)malloc(sizeof(char)*SIZE);/S_num.top = S_num.base;S_char.top = S_char.base;void push_num(double n)/ 數字進棧* +S_num.top =

5、n;void push_char(char c)/ 運算符號進棧 * +S_char.top = c;double pop_num()/ 數字出棧double m = *S_num.top;S_num.top-;return m;char pop_char()/ 運算符號出棧char cc = *S_char.top;S_char.top-;return cc;char get_top_char()/ 得到運算符號的棧中最頂端的運算符號return *S_char.top;double operate(double y, char c, double x)/ 對兩個數計算 ( 含是雙目運算符

6、如 *, / 等 等)double r;if(c = '-')r = x - y;else if(c = '+')r = x + y;else if(c = '/' && y != 0)r = x / y;else if(c = '*')r = x * y;else if(c = A )r = 1;for(int i = 1; i <= y; i+)r *= x;else if(c = '%')int r0 = (int)x % (int)y;r = double(r0);return r;對

7、一個數運( 含單目運算符 : 如double operate_one(double one, char cc)/log(L), sin(S) 等等)double r;if(cc = 'Q')r = sqrt(one);else if(cc = 'C')r = cos(one);else if(cc = 'S')r = sin(one);else if(cc = 'T')r = tan(one);else if(cc = 'c')r = acos(one); else if(cc = 's')r =

8、asin(one);else if(cc = 't')r = atan(one);return r;求對數的double operate_L(double a, double b, char dian)/double r = log(b) / log(a);return r;double compute()/ 對整個表達式的計算 char c;/ 表示運算符號int p = 0;/ 用于 shu+p, 先初始化 int i, j; init();/ 進行初始化 push_char('n'); linestrlen(line) = 'n' line

9、strlen(line)+1 = '0'續進if(biao)push_num(dai_result);/ 把運算的結果先進棧 , 在這個結果的根底上繼行運算biao = 0;for(i = 0; linei != '0')/把表達式中的數字字符串轉化成可計算的數int flag = 0;int flag1 = 1;/ 標記是否是運算符號/ int flag2 = 1;/ 標記是否出現 '_'double h = 0;int ge;/ 位數int biao_dian = 0;/ 是否是小數的類型while(1)flag1 = 1;for(j = 0

10、; j <= 16; j+)if(linei = fuj)flag1 = 0;if(linei = '_') break;if(linei = '.') i+;ge = 0;biao_dian = 1;if(linei = 'P')shu+p = pi;i+;break;if(linei = 'E')shu+p = e;if(flag1)h = h * 10 + (linei - '0'); flag = 1;i+;if(biao_dian)ge+;elsebreak;if(flag)if(biao_dian

11、)int r = 1;for(int k = 1; k <= ge; k+) r *= 10;h /= r;if(linei = '+')shu+p = h;/ 把轉化而來的數字存于數組 shu+p = -1; elseif(linei = '-')shu+p = -2; elseif(linei = '*')shu+p = -3; elseif(linei = '/')shu+p = -4; else if(linei = '%')shu+p = -5;else if(li nei = ' A

12、9;)shu+p =-6; else if(linei = 'Q')shu+p =-7; else if(linei = 'L')shu+p =-8; else if(linei = 'C')shu+p =-9;else if(linei = 'S')shu+p = -10; else if(linei = 'T')shu+p = -11;else if(linei = 'c')shu+p = -12;else if(linei = 's')shu+p = -13;else if(l

13、inei = 't')shu+p = -14;else if(linei = '(')shu+p = -15;else if(linei = ')')shu+p = -16;else if(linei = 'n')shu+p = -17;i+;i = 1;while(shui != -17 | get_top_char() != 'n') double m = shui;if(m >= 0)push_num(m);elseif(m = -1)c = '+'else if(m = -2) c =

14、 '-'else if(m = -3)1*1.else if(m = -4)c = '/'else if(m = -5) c = '%'else if(m = -6)c = ' A'else if(m = -7)c = 'Q'else if(m = -8)c = 'L'else if(m = -9)c = 'C'else if(m = -10)c = 'S'else if(m = -11)c = 'T'else if(m = -12)c = '

15、c'else if(m = -13)c = 's'else if(m = -14)c = 't'else if(m = -15)c = '('else if(m = -16)c = ')'else if(m = -17)c = 'n'if(comparech <char ch = get_top_char();/ 得到最頂端運算符號comparec)/ 運算符號級別的比擬 push_char(c);else if(ch = '(' && c = ')')

16、pop_char();i+;else if(comparech >= comparec && ch != '(' && ch != 'n')if(ch = 'Q' | ch = 'C' | ch = 'S'| ch = 'T'| ch = 'c' | ch = 's' | ch = 't')double one = pop_num();char dian = pop_char(); push_num(operate

17、_one(one, dian); else if(ch = 'L')double one_L = pop_num();double two_L = pop_num();char dian = pop_char(); push_num(operate_L(two_L, one_L, dian);elsedouble x = pop_num();double y = pop_num();char dian = pop_char();if(dian = '/' && x = 0)/判斷是否除了 " 零"cout<<&qu

18、ot; 由于您除了零，結果將是錯誤的 "<<endl;push_num(operate(x, dian, y);/ 把進行一次計算的結果入 棧elsepush_char(c);i+;得到結果double result = pop_num();/ return result;int check_kuohao()/ 檢查表達式括號是否匹配int i, f = 0;int kuoSIZE, key = 1; memset(kuo, 0, sizeof(kuo); for(i = 0; linei != '0' i+)if(linei = '(')

19、kuo+f = 1; else if(linei = ')') if(kuof = 1)kuof = 0;f-;elsekey = 0;break;if(key && f = 0)return 1;elseretur n 0;int check_char() 檢查運算符號是否合法(女口 : 1 +* 4) int i, ge;ge = 0;for(i = 0; lin ei != '0')-1*1while(li nei='+' | lin ei='-'| lin ei='八II linei='/&

20、#39; | lin ei='%' | lin ei=III linei=Q | linei='L' | lin ei='S'| linei=C | linei=T | lin ei=:='s'| linei='c' | lin ei='t')ge+; i+;if(ge >= 3)return 0;elsereturn 1;void output(double result)/ 打出結果printf(" 所得結果是 : ");cout<<result<&l

21、t;endl;void check()/ 檢查表達式是否合法void introduce();char cc;/ 決定計算器按哪種功能進行計算double result;/ 結果void input();/ 定義if( check_kuohao() && check_char() )/ 看是否合法 , 合法那么計算result = compute();output(result);coutvv"輸入一個字符M'或D或'F',決定是否繼續："vvendl;while(cin>>cc)if(cc = 'M')s

22、ystem("cls");introduce();printf(" 您上次所得結果為 : ");請繼續輸入想計算的表達 cout<<result<<endl;cout<<" 在上次計算結果的根底上"<<endl;dai_result = result;biao = 1;input();/ 輸入表達式break;else if(cc = 'D')system("cls");introduce();cout<<" 計算器已清零 , 請

23、輸入您所要計算的表達式 "<<endl;input();/ 輸入表達式break;else if(cc = 'F')system("cls");cout<<" 計算器關閉 , 謝謝使用 !"<<endl;break;elsecout<<" 所 輸 入 字 符 無 效 , 請 輸 入 一 個 字 符 'M' 或 'D' 或 'F'!"<<endl;continue;else/ 不合法，分兩種不合法if(c

24、heck_kuohao() = 0 && check_char() = 1)cout<<" 您所輸入的表達式括號不匹配 , 請重新輸入 :"<<endl;input();/ 輸入表達式else:"<<endl;cout<<" 您所輸入的表達式不合法 , 請重新輸入 input();/ 輸入表達式void tackle_fuhao()/ 處理負號node *root, *head, *p, *q, *p1;root = head = new node;head->next = NULL;

25、int i;for(i = 0; linei != '0' i+)/ 建立鏈表p = new node;p->data = linei;p->next = head->next;head->next = p;head = p;/ delete p;處理第一個q = (node*)malloc(sizeof(node);head = root;if(root->next->data = '+' | root->next->data = '-')/符p = new node; p->data =

26、'0'p->next = head->next; head->next = p;if(root->next != NULL)for(q = root->next; q; q = q->next)if(q->data = '(' && (q->next->data = '-' | q->next->data = '+')p = new node; p->data = '0' p->next = q->next; q-&

27、gt;next = p;/ delete q;pl = new no de;int qi = -1;for(p1 = root->next; p1; pl = p1- >n ext)lin e+qi = p1->data;lin e+qi = '0'void input() 輸入cin> >li ne;if(biao = 0)tackle_fuhao(); 處理負號check();檢查表達式是否合法void introduce()/對計算器的符號功能的簡要介紹cout<<"計算器簡要介紹"<<endl;cout«"C(cos) S(sin) T(ta n) a(arccos) c(arcs in) "<<e ndl;coutvv"7 89/o