I was looking at std::numeric_limits<float>::min/max()
but it appears 'min()' returns the smallest absolute value, not the lowest value. Is it safe to use
-std::numeric_limits<float>::max()
, ie is float symmetric in min/max limits?
IEEE 754浮点数使用符号位进行签名(而不是二进制补码),所以如果您确定您的编译器/平台使用该表示(非常常见),那么您可以使用-std::numeric_limits<float>::max()
你怀疑-std::numeric_limits<float>::max()
。
use std::numeric_limits::lowest()
static _Ty __CRTDECL lowest() _THROW0()
{ // return most negative value
return (-(max)());
}
Yes, float
is symmetric in minimum/maximum values.
If you're using the lowest representable value as an initial value in searching a list for its maximum value, consider using infinity instead.
std::numeric_limits<T>::has_infinity()
will return true
for any numeric type that has it and std::numeric_limits<T>::infinity()
will return a value that always evaluates greater than any other non-NaN value for that type. This value can be negated and will evaluate less than anything else.
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.