a**********2
发帖数: 340 | 1 求int sqrt(int x)越界如何处理啊？
看到以前的一个帖子，觉得好像不太对， INT_MAX是2^31-1不是2^32,用（1<<16)还是
会越界，请问下该如何处理？
还有判断越界一般用什么方法？多谢了 | a**********2
发帖数: 340 | | i**********e
发帖数: 1145 | 3 Good observation.
You are correct, my previous post has bugs and did not address the overflow
issue properly.
Two methods to address:
i) find the upper bound of SQRT(INT_MAX) once and pass it into that function
. The upper bound can be found using a helper function that tests for
overflow. (ie, as soon as the condition when x > x*x is met, we knew x-1 is
the upper bound).
ii) declare the parameter x as unsigned int. Then initialize the upper bound
,
hi = min(n/2, (1<<16) - 1);
Hope this helps.
【在 a**********2 的大作中提到】
: 求int sqrt(int x)越界如何处理啊？
:
: 看到以前的一个帖子，觉得好像不太对， INT_MAX是2^31-1不是2^32,用（1<<16)还是
: 会越界，请问下该如何处理？
: 还有判断越界一般用什么方法？多谢了
| i**********e
发帖数: 1145 | 4 This should also work too :)
The observation is when it overflows, it must be the case M > M*M, so
readjust the upper bound to M-1 when that happens.
int mysqrt(int n) {
int L = 0;
int H = min(n/2, (1<<16));
while (L < H) {
int M = L+(H-L+1)/2;
int M_sqr = M*M;
if (M <= M_sqr && M_sqr <= n)
L = M;
else /* M > M_sqr || (M_sqr > n) */
H = M-1;
}
return L;
} | k*j
发帖数: 153 | 5 or just test if M*M<0, if so then it overflows.
【在 i**********e 的大作中提到】
: This should also work too :)
: The observation is when it overflows, it must be the case M > M*M, so
: readjust the upper bound to M-1 when that happens.
: int mysqrt(int n) {
: int L = 0;
: int H = min(n/2, (1<<16));
: while (L < H) {
: int M = L+(H-L+1)/2;
: int M_sqr = M*M;
: if (M <= M_sqr && M_sqr <= n)
| l*********8
发帖数: 4642 | 6 这个不能保证吧。
【在 k*j 的大作中提到】
: or just test if M*M<0, if so then it overflows.
|
|
