二进制搜索中mid =(beg + end)/2和mid = beg +(end-beg)/2有什么区别? [英] what's the difference between mid=(beg+end)/2 and mid=beg+(end-beg)/2 in binary search?

问题描述

这是C ++入门第五版问题3.26中的问题，我不知道它们之间的区别吗? 可能是第二个可以避免溢出.

It is a problem from C++ primer fifth edition problem 3.26, I don't know the difference between them ? May be the second one can avoid overflow.

推荐答案

第二个可以避免溢出.

May be the second one can avoid overflow.

完全正确.不能保证beg+end是可表示的；但是在第二种情况下，中间值以及预期结果均不大于end，因此没有溢出的危险.

Exactly. There's no guarantee that beg+end is representable; but in the second case the intermediate values, as well as the expected result, are no larger than end, so there is no danger of overflow.

第二种形式也可以用于仿射类型，例如指针和其他随机访问迭代器，可以将它们减去以得出距离，但不能加在一起.

The second form can also be used for affine types like pointers and other random-access iterators, which can be subtracted to give a distance, but not added together.

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