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. }

View Code

 　　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 );