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（六）

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. }

 　　main()函数中player完成抽牌之后，立刻计算了player的点数：  

52.       totalIt(playerPoints,total,PLAYER);

 　　这个计算结果基于ACE的点数被作为1或11两种可能性，取最好一种作为最后的结果。 

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. }

 　　这个函数让我们得以领略什么叫思路含糊和废话连篇。首先 

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.    }

 　　它的意思是当较高点数超过21点时把较低作为最终的点数，否则把较高点数作为最后的点数。显而易见这可以更简洁地表述为 

      if ( points[ACEHIGH] > 21 )
      { 
         total[who] = points[ACELOW];
      }
      else
      { 
         total[who] = points[ACEHIGH];
      }

　　原来的代码把“(points[ACELOW] == points[ACEHIGH])||”写出来是思路不清导致的拖泥带水。

　　更简洁的写法是： 

total[who] = ( points[ACEHIGH] > 21 )? points[ACELOW]: points[ACEHIGH];

  　　“?:”这个三目运算在这里应用得恰到好处。

　　有些人对三目运算有一种无名的恐惧，鼓吹所谓“尽量不要用三目运算符”。这是毫无道理的，这种无理源自无知。他们自己不会用刀，于是就骗人骗己地宣称使用木棍强于用刀。 

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.    }

 　　这一段同样拖泥带水，其实它的效果和下面写法没有本质区别： 

   printf( "%s a total of %d\n\n", 
           who == PLAYER ? "You have" : "The house stands with",
           total[who] );

 　　所以totalIt()函数应改写为 

      void totalIt(int [],int [],int );
      
      void totalIt(int points[],int total[],int who)
      {
           
         total[who] = ( points[ACEHIGH] > 21 )? points[ACELOW]: points[ACEHIGH];
         
         printf( "%s a total of %d\n\n", 
                 who == PLAYER ? "You have" : "The house stands with",
                 total[who] );
      
      }

 　　计算了完player的点数之后，按照规则在main()中由dealer继续抽牌（前面已抽过第一张牌）。dealer抽牌的策略是不到17点则继续，据代码作者说现实中的庄家的策略也是如此。

53.       do
54.       {
55.          dealerGetsCard(&numCards,cards,dealerPoints);
56.       }
57.       while ( dealerPoints[ACEHIGH] < 17 );

 　　dealerGetsCard ()函数的功能与playerGetsCard()函数重叠，前面已经提到过，甚至可以说这两个函数都是多余的。

　　此外这里还有一个更严重的问题，那就是“dealerPoints[ACEHIGH] < 17”这个表达式的逻辑问题。这个表达式要求dealer的点数达到17点或以上时停止抽牌，但问题在于点数有两种计算方法，一种是把Ace作为11点（Soft hand），另一种是把Ace作为1点。“dealerPoints[ACEHIGH] ”的意义是Soft hand点数，但是原作者在对程序的说明中压根就没有明确dealer的Soft hand点数达到或超过17点时停牌，只是泛泛地说了一句“the dealer stands on 17”。按软件工程的说法，这叫需求不清，是比代码错误更加严重的错误。

　　紧接着，矛盾出现了。 

59.       totalIt(dealerPoints,total,DEALER);

　　totalIt()函数计算dealer的点数却是按照最好成绩计算的，这就发生了矛盾。比如dealer为15点时又取了一张牌Ace，按照Soft hand规则，dealer的点数是25点，但最后的成绩却是按照硬牌规则为16点，而如果dealer的点数为16点，那么前面他根本就不应该停牌。这是“双重标准”的C语言版。

　　这里较为合理的写法应该是

       do
       {
          dealerGetsCard(&numCards,cards,dealerPoints);
             totalIt(dealerPoints,total,DEALER);
       }
       while ( total [DEALER] < 17 );