C |存储类和类型限定符|问题9-yiteyi-C++库

输出

null

                     #include <stdio.h>                   
                     int                               fun()                   
                     {                   
                                         static                               int                               num = 16;                   
                                         return                               num--;                   
                     }                   
                             
                     int                               main()                   
                     {                   
                                         for                               (fun(); fun(); fun())                   
                                         printf                               (                               "%d "                               , fun());                   
                                         return                               0;                   
                     }                   

（A） 无限循环 （B） 13 10 7 4 1 （C） 14 11 8 5 2 （D） 15 12 8 5 2 答复: （C） 说明： 自从号码是静止的乐趣，旧的价值观号码为后续函数调用保留。还有，自从声明返回号码 –是后缀，它返回号码，并更新下一次函数调用的值。

fun() called first time: num = 16 // for loop initialization done;


In test condition, compiler checks for non zero value

fun() called again : num = 15

printf("%d ", fun());:num=14 ->printed

Increment/decrement condition check

fun(); called again : num = 13

----------------

fun() called second time: num: 13 

In test condition,compiler checks for non zero value

fun() called again : num = 12

printf("%d ", fun());:num=11 ->printed

fun(); called again : num = 10

--------

fun() called second time : num = 10 

In test condition,compiler checks for non zero value

fun() called again : num = 9

printf("%d ", fun());:num=8 ->printed

fun(); called again   : num = 7

--------------------------------

fun() called second time: num = 7

In test condition,compiler checks for non zero value

fun() called again : num = 6

printf("%d ", fun());:num=5 ->printed

fun(); called again   : num = 4

-----------

fun() called second time: num: 4 

In test condition,compiler checks for non zero value

fun() called again : num = 3

printf("%d ", fun());:num=2 ->printed

fun(); called again   : num = 1

----------

fun() called second time: num: 1 

In test condition,compiler checks for non zero value

fun() called again : num = 0 => STOP

这个问题的小测验

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