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劣质代码评析——《写给大家看的C语言书（第2版）》附录B之21点程序（五） ...

原作者: [db:作者] 来自: [db:来源] 收藏 邀请 
0. #include <stdio.h>
1. #include <time.h>
2. #include <ctype.h>
3. #include <stdlib.h>
4. 
5. #define BELL '\a'
6. #define DEALER 0
7. #define PLAYER 1
8. 
9. #define ACELOW 0
10. #define ACEHIGH 1
11. 
12. int askedForName = 0;
13. 
14. 
15. void dispTitle(void);
16. void initCardsScreen(int cards[52],int playerPoints[2],
17. int dealerPoints[2], int total[2], 
18. int *numCards);
19. int dealCard(int * numCards,int cards[52]);
20. void dispCard(int cardDrawn,int points[2]);
21. void totalIt(int points[2],int tatal[2],int who);
22. void dealerGetsCard(int *numCards,int cards[52],
23. int dealerPoints[2]);
24. void playerGetsCard(int *numCards,int cards[52],
25. int playerPoints[2]);
26. char getAns(char mesg[]);
27. void findWinner(int total[2]);
28. 
29. main()
30. {
31.    int numCards;
32.    int cards[52],playerPoints[2],dealerPoints[2],total[2];
33.    char ans;
34.    
35.    do 
36.    { 
37.       initCardsScreen(cards,playerPoints,dealerPoints,total, &numCards);
38.       dealerGetsCard(&numCards,cards, dealerPoints);
39.       printf("\n");
40.       playerGetsCard(&numCards,cards,playerPoints); 
41.       playerGetsCard(&numCards,cards,playerPoints);
42.       do
43.       {
44.          ans = getAns("Hit or stand (H/S)?");
45.          if ( ans == 'H' )
46.          { 
47.             playerGetsCard(&numCards,cards,playerPoints);
48.          }  
49.       }
50.       while( ans != 'S' );
51.       
52.       totalIt(playerPoints,total,PLAYER);
53.       do
54.       {
55.          dealerGetsCard(&numCards,cards,dealerPoints);
56.       }
57.       while (dealerPoints[ACEHIGH] < 17 );
58.       
59.       totalIt(dealerPoints,total,DEALER);
60.       findWinner(total); 
61.       
62.       ans = getAns("\nPlay again(Y/N)?");  
63.    }
64.    while(ans=='Y');
65.    
66.    return 0;
67. }
68. 
69. void initCardsScreen( int cards[52],int playerPoints[2],
70.                       int dealerPoints[2], int total[2], 
71.                       int *numCards )
72. {
73.    int sub,val = 1 ;
74.    char firstName[15];
75.    *numCards=52;
76.    
77.    for(sub=0;sub<=51;sub++)
78.    {
79.       val = (val == 14) ? 1 : val;
80.       cards[sub] = val;
81.       val++;  
82.    }
83.    
84.    for(sub=0;sub<=1;sub++)
85.    { 
86.       playerPoints[sub]=dealerPoints[sub]=total[sub]=0;
87.    }
88.    dispTitle();
89.    
90.    if (askedForName==0)
91.    { 
92.       printf("What is your first name?");
93.       scanf(" %s",firstName);
94.       askedForName=1;
95.       printf("Ok, %s,get ready for casino action!\n\n",firstName);
96.       getchar();
97.    }
98.    return;        
99. }
100. 
101. void playerGetsCard(int *numCards,int cards[52],int playerPoints[2])
102. {
103.    int newCard;
104.    newCard = dealCard(numCards, cards);
105.    printf("You draw:");
106.    dispCard(newCard,playerPoints);
107. }
108. 
109. 
110. void dealerGetsCard(int *numCards,int cards[52],int dealerPoints[2])
111. {
112.    int newCard;
113.    newCard = dealCard(numCards,cards);
114.    printf("The dealer draws:");
115.    dispCard(newCard,dealerPoints);
116. }
117. 
118. int dealCard(int * numCards,int cards[52])
119. {
120.    int cardDrawn,subDraw;
121.    time_t t;
122.    srand(time(&t));
123.    subDraw = (rand()%(*numCards));
124.    cardDrawn = cards[subDraw];
125.    cards[subDraw] = cards[*numCards -1];
126.    (*numCards)--;
127.    return cardDrawn;
128. }
129. 
130. void dispCard(int cardDrawn, int points[2])
131. {
132.    switch(cardDrawn)
133.    {
134.       case(11): printf("%s\n","Jack");
135.                 points[ACELOW] += 10;
136.                 points[ACEHIGH] += 10;
137.                 break;
138.       case(12): printf("%s\n","Queen");
139.                 points[ACELOW] += 10;
140.                 points[ACEHIGH] += 10;
141.                 break;
142.       case(13): printf("%s\n","King");
143.                 points[ACELOW] += 10;
144.                 points[ACEHIGH] += 10;
145.                 break;
146.       default : points[ACELOW] += cardDrawn;
147.                 if(cardDrawn==1)
148.                 { 
149.                    printf("%s\n","Ace");
150.                    points[ACEHIGH]+= 11;
151.                 }
152.                 else
153.                 {  
154.                   points[ACEHIGH]+=cardDrawn;
155.                   printf("%d\n",cardDrawn); 
156.                 }
157.    }
158.    return ;
159. }
160. 
161. void totalIt(int points[2],int total[2],int who)
162. {
163.    if ( (points[ACELOW] == points[ACEHIGH])
164.       ||(points[ACEHIGH] < 21 ))
165.    { 
166.      total[who] = points[ACELOW];
167.    }
168.    else
169.    { 
170.        total[who] = points[ACEHIGH];
171.    }
172.    
173.    if (who == PLAYER )
174.    {
175.       printf("You have a total of %d\n\n", total[PLAYER]);
176.    }
177.    else
178.    {
179.        printf("The house stands with a total of %d\n\n", 
180.        total[DEALER]);
181.    }
182.    return;
183. }
184. 
185. void findWinner(int total[2])
186. {
187.    if ( total[DEALER] ==  21 )
188.    {
189.        printf("The house wins.\n");
190.        return ;
191.    }
192.    if ( (total[DEALER] > 21) && (total[PLAYER] > 21) )
193.    { 
194.       printf("%s", "Nobody wins.\n");
195.       return ; 
196.    }
197.    if ((total[DEALER] >= total[PLAYER])&& (total[DEALER] < 21))
198.    { 
199.       printf("The house wins.\n");
200.       return ; 
201.    }
202.    printf("%s%c","You win!\n",BELL);
203.    return;
204. }
205. 
206. char getAns(char mesg[])
207. {
208.    char ans;
209.    printf("%s", mesg);
210.    ans = getchar();
211.    getchar();
212.    return toupper(ans);
213. }
214. 
215. void dispTitle(void)
216. {
217.    int i = 0 ;
218.    while(i<25)
219.    { 
220.         printf("\n");
221.         i++; 
222.    }
223.    printf("\n\n*Step right up to the Blackjack tables*\n\n");
224.    return ;
225. }
View Code

　　继续走查dispCard()函数：  

20. void dispCard(int cardDrawn,int points[2]);

130. void dispCard(int cardDrawn, int points[2])
131. {
132.    switch(cardDrawn)
133.    {
134.       case(11): printf("%s\n","Jack");
135.                 points[ACELOW] += 10;
136.                 points[ACEHIGH] += 10;
137.                 break;
138.       case(12): printf("%s\n","Queen");
139.                 points[ACELOW] += 10;
140.                 points[ACEHIGH] += 10;
141.                 break;
142.       case(13): printf("%s\n","King");
143.                 points[ACELOW] += 10;
144.                 points[ACEHIGH] += 10;
145.                 break;
146.       default : points[ACELOW] += cardDrawn;
147.                 if(cardDrawn==1)
148.                 { 
149.                    printf("%s\n","Ace");
150.                    points[ACEHIGH]+= 11;
151.                 }
152.                 else
153.                 {  
154.                   points[ACEHIGH]+=cardDrawn;
155.                   printf("%d\n",cardDrawn); 
156.                 }
157.    }
158.    return ;
159. }

 　　dispCard()函数的功能是显示抽到的牌的点数并计算抽牌者目前的总点数。这个函数最主要的毛病是可读性差，原因主要有两点，第一，在switch语句中蹩脚地嵌套了一句if语句，实际上这个switch语句可以这样写： 

void dispCard(int cardDrawn,int points[]);

void dispCard(int cardDrawn, int points[])
{
   switch(cardDrawn)
   {
      case 11: puts("Jack");
               points[ACELOW]  += 10;
               points[ACEHIGH] += 10;
               break;
      case 12: puts("Queen");
               points[ACELOW]  += 10;
               points[ACEHIGH] += 10;
               break;
      case 13: puts("King");
               points[ACELOW]  += 10;
               points[ACEHIGH] += 10;
               break;
      case 1  :puts("Ace");
               points[ACELOW]  += 1 ;
               points[ACEHIGH] += 11;
               break;                
      default :printf("%d\n",cardDrawn);
               points[ACELOW]  += cardDrawn;
               points[ACEHIGH] += cardDrawn;
               break;
   }
}

显然要好看得多。
　　第二，就是不应把输出牌面点数与计算点数放在这一个函数中同时完成，应该分为两个函数。 

void dispCard(int cardDrawn);

void dispCard(int cardDrawn)
{
   switch(cardDrawn)
   {
      case 11: puts("Jack");
               break;
      case 12: puts("Queen");
               break;
      case 13: puts("King");
               break;
      case 1  :puts("Ace");
               break;                
      default :printf("%d\n",cardDrawn);
               break;
   }
}

void update(int cardDrawn,int points[]);

void update(int cardDrawn, int points[])
{
   switch(cardDrawn)
   {
      case 11: 
      case 12: 
      case 13: points[ACELOW]  += 10;
               points[ACEHIGH] += 10;
               break;
      case 1  :points[ACELOW]  += 1 ;
               points[ACEHIGH] += 11;
               break;                
      default :points[ACELOW]  += cardDrawn;
               points[ACEHIGH] += cardDrawn;
               break;
   }
}

　　这样代码更简单。
　　现在回到main()函数，考察 dealerGetsCard(&numCards,cards, dealerPoints); 之后的代码。 

40.       playerGetsCard(&numCards,cards,playerPoints); 
41.       playerGetsCard(&numCards,cards,playerPoints);

 　　这两行的意思是在dealer（计算机）取得一张牌后，player开始抽牌。由于player至少要抽两张牌，所以连续两处调用playerGetsCard()函数。然而查看一下这个函数的定义就会发现: 

24. void playerGetsCard(int *numCards,int cards[52],
25. int playerPoints[2]);

101. void playerGetsCard(int *numCards,int cards[52],int playerPoints[2])
102. {
103.    int newCard;
104.    newCard = dealCard(numCards, cards);
105.    printf("You draw:");
106.    dispCard(newCard,playerPoints);
107. }

 　　这个函数其实与dealerGetsCard()函数是一样的函数。同样的函数写了两次，无论如何都是很垃圾的写法。

　　写代码有一条基本原则——要“拽”DRY(Don’t repeat yourself)(参见http://www.cnblogs.com/pmer/archive/2011/07/16/2108436.html)。相同的函数定义了两次就是不够“拽”。

　　由于这两个函数基本相同，所以很容易合并为一个。它们的差别只在 

105. printf("You draw:");

 和 

114.    printf("The dealer draws:");

 这两句，可以通过为函数增加一个参数的办法统而为一，也可以把这两句从函数中直接剥离，这时playerGetsCard()和dealerGetsCard()这两个函数本身都没有存在的必要性了。即在main()中直接可调用dispCard()函数： 

int main(void)
{
    /*……*/
    do{
        /*……*/
        printf("The dealer draws:");
        dispCard ( dealCard( numCards , cards ) , dealerPoints ) ;
        
        printf("You draw:");
        dispCard ( dealCard( numCards , cards ) , playerPoints) ;
        /*……*/
    }
    while( getAns("\nPlay again(Y/N)?") == 'Y' );  /*询问是否继续*/
    return 0;
}

 　　main()中的 

42.       do
43.       {
44.          ans = getAns("Hit or stand (H/S)?");
45.          if ( ans == 'H' )
46.          { 
47.             playerGetsCard(&numCards,cards,playerPoints);
48.          }  
49.       }
50.       while( ans != 'S' );

 的作用是让player输入。如果输入为H，则继续抽取一张牌；如果输入为S，则player取牌过程结束；如果输入其他字符，则继续提问。

　　这段代码的逻辑有些复杂化。如果我写，大概会按下面方式写:

      do
      {
         char ans;
         
         ans = getAns("Hit or stand (H/S)?");
         
         if ( ans == 'H' )
             playerGetsCard(&numCards,cards,playerPoints);
         
         if ( ans == ' S ' )
            break;
         
      }
      while( 1 );

 并且把它抽象为一个函数。 

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