输出?
#include
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答案： (C)
说明：由于NUM是乐趣(静态)，NUM的旧值保存为后续函数的调用。另外，由于语句return num –是后缀，因此它返回num的旧值，并更新下一个函数调用的值。
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 \n", 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 \n", 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 \n", 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 \n", 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 \n", 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 这个问题的测验
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