大门|大门-CS-2015（第三组）|问题65-yiteyi-C++库

考虑下面的C程序：

null

                     # include <stdio.h>                   
                     int                               main( )                   
                     {                   
                                         int                               i, j, k = 0;                   
                                         j = 2 * 3 / 4 + 2.0 / 5 + 8 / 5;                   
                                         k  -= --j;                   
                                         for                               (i = 0; i < 5; i++)                   
                                         {                   
                                         switch                               (i + k)                   
                                         {                   
                                         case                               1:                   
                                         case                               2:                               printf                               (                               "%d"                               , i + k);                   
                                         case                               3:                               printf                               (                               "%d"                               , i + k);                   
                                         default                               :                               printf                               (                               "%d"                               , i + k);                   
                                         }                   
                                         }                   
                                         return                               0;                   
                     }                   

printf语句被执行的次数为。

（A） 8. （B） 9 （C） 10 （D） 11 答复: （C） 说明： 下面的语句表示j=2

 j = 2 * 3 / 4 + 2.0 / 5 + 8 / 5;

下面的语句使k=-1。

  
 k  -= --j;

在开关中需要注意的一点是，没有中断。让printf语句的计数为“count”

For i = 0, the value of i+k becomes -1, default block 
           is executed, count = 1.
For i = 1, the value of i+k becomes 0, default block
           is executed, count = 2.
For i = 2, the value of i+k becomes 1, all blocks are 
           executed as there is no break, count = 5
For i = 3, the value of i+k becomes 2, three blocks 
           after case 1: are executed, count = 8
For i = 4, the value of i+k becomes 3, two blocks 
           are executed, count = 10

这个问题的小测验

文章版权归作者所有，未经允许请勿转载。

THE END

技术文章