考虑以下 C函数，输出是什么?
#include
int f(int n)
{
static int r = 0;
if (n <= 0) return 1;
if (n > 3)
{
r = n;
return f(n-2)+2;
}
return f(n-1)+r;
}
int main()
{
printf("%d", f(5));
}
(一) 5
(乙) 7
(三) 9
(四) 18答案： (D)
解释：
f(5) = f(3)+2
The line "r = n" changes value of r to 5. Since r
is static, its value is shared be all subsequence
calls. Also, all subsequent calls don't change r
because the statement "r = n" is in a if condition
with n > 3.
f(3) = f(2)+5
f(2) = f(1)+5
f(1) = f(0)+5
f(0) = 1
So f(5) = 1+5+5+5+2 = 18 这个问题的测验