最小和最大运行值 [英] min and max running values
问题描述
嗨
我经常需要阅读数字，只保留最高或最低。
所以我做了类似的事情
int uLimit = 0;
int lLimit = 999999999999; //希望编译器不会抱怨
uLimit = val_read uLimit？ val_read：uLimit;
lLimit = val_read< lLimit？ val_read：lLimit;
如何选择原来的lLimit？
有更多的切割方式吗？
谢谢
Hi
often I need to read numbers and only keep the highest or lowest.
so I do something like
int uLimit = 0;
int lLimit = 999999999999; //hoping the compiler will not complain
uLimit = val_read uLimit ? val_read : uLimit;
lLimit = val_read < lLimit ? val_read : lLimit;
how do I choose the original lLimit?
is there more cleaver way?
thanks
推荐答案
Gary Wessle写道：
Gary Wessle wrote:
嗨
我经常需要阅读数字并且只保留最高或最低。
所以我做了类似
int uLimit = 0;
Hi
often I need to read numbers and only keep the highest or lowest.
so I do something like
int uLimit = 0;
''int''可以是负数，你知道......
''int'' can be negative, you know...
int lLimit = 999999999999 ; //希望编译器不会抱怨
int lLimit = 999999999999; //hoping the compiler will not complain
为什么不只是
int lLimit = INT_MAX; //或者使用std :: numeric_limits< int> :: max
Why not just
int lLimit = INT_MAX; // or use std::numeric_limits<int>::max
?
>
uLimit = val_read uLimit？ val_read：uLimit;
lLimit = val_read< lLimit？ val_read：lLimit;
如何选择原来的lLimit？
有更多的切割方法吗？
>
uLimit = val_read uLimit ? val_read : uLimit;
lLimit = val_read < lLimit ? val_read : lLimit;
how do I choose the original lLimit?
is there more cleaver way?
聪明？我不知道。可靠性更好。
V
-
请在通过电子邮件回复时删除资金'A'
我没有回复最热门的回复，请不要问
Clever? I don''t know. Reliable more like it.
V
-
Please remove capital ''A''s when replying by e-mail
I do not respond to top-posted replies, please don''t ask
On 2007-03-05 10:22:49 -0800，Gary Wessle< ph **** @ yahoo.comsaid：
On 2007-03-05 10:22:49 -0800, Gary Wessle <ph****@yahoo.comsaid:
嗨
我经常需要读取数字，只保留最高或最低。
所以我做了类似
int uLimit = 0;
Hi
often I need to read numbers and only keep the highest or lowest.
so I do something like
int uLimit = 0;
负值怎么样？
What about negative values?
int lLimit = 999999999999; //希望编译器不会抱怨
int lLimit = 999999999999; //hoping the compiler will not complain
为什么希望？为什么不直接获得正确的价值？
Why "hope"? Why not just get the value right to begin with?
>
uLimit = val_read uLimit？ val_read：uLimit;
lLimit = val_read< lLimit？ val_read：lLimit;
如何选择原来的lLimit？
>
uLimit = val_read uLimit ? val_read : uLimit;
lLimit = val_read < lLimit ? val_read : lLimit;
how do I choose the original lLimit?
#include< limits>
#include< algorithm>
.. ..
int uLimit = std :: numeric_limits< int> :: min（）;
int lLimit = std :: numeric_limits< int> :: max（）;
uLimit = std :: max（val_read，uLimit）;
lLimit = std :: min（val_read，uLimit）;
#include <limits>
#include <algorithm>
....
int uLimit = std::numeric_limits<int>::min();
int lLimit = std::numeric_limits<int>::max();
uLimit = std::max(val_read, uLimit);
lLimit = std::min(val_read, uLimit);
有更多的切割方式吗？
谢谢
is there more cleaver way?
thanks
- -
Clark S. Cox III
cl ******* @ gmail。 com
--
Clark S. Cox III
cl*******@gmail.com
如何选择原始lLimit？
how do I choose the original lLimit?
有更多的切割方式吗？
is there more cleaver way?
使用std :: numeric_limits< int> :: max（）和
std :: numeric_limits< int> :: min（） 。
use std::numeric_limits<int>::max() and
std::numeric_limits<int>::min().
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