问题描述
以下是测试程序：
void testFunc()
{
double maxValue = DBL_MAX;
double slope = std::numeric_limits<double>::quiet_NaN();
std::cout << "slope is " << slope << std::endl;
std::cout << "maxThreshold is " << maxValue << std::endl;
std::cout << "the_min is " << std::min( slope, maxValue) << std::endl;
std::cout << "the_min is " << std::min( DBL_MAX, std::numeric_limits<double>::quiet_NaN()) << std::endl;
}
int main( int argc, char* argv[] )
{
testFunc();
return 0;
}
在Debug中，我得到：
In Debug, I get:
slope is nan
maxThreshold is 1.79769e+308
the_min is nan
the_min is 1.79769e+308
在发布中，我得到：
slope is nan
maxThreshold is 1.79769e+308
the_min is 1.79769e+308
the_min is nan
为什么我在Release比Debug中得到不同的结果？
Why would I get a different result in Release than Debug?
我已经检查过Stack Overflow href =http://stackoverflow.com/questions/1632145/use-of-min-and-max-functions-in-c>在C ++中使用min和max函数 ，以及它没有提及任何版本/调试差异。
I already checked Stack Overflow post Use of min and max functions in C++, and it does not mention any Release/Debug differences.
我使用Visual Studio 2015。
I am using Visual Studio 2015.
推荐答案
在中比较NAN与任何内容总会产生 false
In IEEE 754 comparing NAN to anything will always yield false, no matter what it is.
slope > 0; // false
slope < 0; // false
slope == 0; // false
更重要的是你
slope < DBL_MAX; // false
DBL_MAX < slope; // false
所以看起来编译器重新排序参数/ use > ; 或< = 而不是< ，这就是为什么你得到不同的结果
So it seems that the compiler reorders the parameters/uses > or <= instead of <, and that's why you get the differing results.
例如，这些函数可以描述为
For example, those functions could be described as such
发布：
double const& min(double const& l, double const r) {
return l <= r ? l : r;
}
调试：
double const& min(double const& l, double const& r) {
return r < l ? r : l;
}
std :: min的需求（LessThanComparable） 旁白，算术具有相同的含义。但是当它们与NaN一起使用时，它们产生不同的结果。
The requirements (LessThanComparable) on std::min aside, those have the same meaning arithmetically. But they yield different results when you use them with NaN.
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