是否将无穷大转换为整数未定义?

问题描述

将无穷大(由float表示)强制转换为整数是未定义的行为吗?

标准说:

但是我不能确定无法表示截断的值"是否包含无穷大.

我试图理解为什么 std :: numeric_limits< int> ::: infinity()和 static_cast< int>(std :: numeric_limits< float> :: infinity())有不同的结果.

  #include< iostream>#include< limits>int main(){std :: cout<<std :: numeric_limits< int> :: infinity()<<std :: endl;std :: cout<<static_cast< int>(std :: numeric_limits< float> :: infinity())<<std :: endl;返回0;} 

输出:

  0-2147483648 

std :: numeric_limits< int> :: infinity()的结果

的C标准语义

由于无穷大的截断仍然是无穷大，并且无穷大不能用 int 表示(我希望这部分没有问题)，所以行为是不确定的.

Is the casting of infinity (represented by float) to an integer an undefined behavior?

The standard says:

but I can't tell whether "truncated value cannot be represented" covers infinity.

I'm trying to understand why std::numeric_limits<int>::infinity() and static_cast<int>(std::numeric_limits<float>::infinity() ) have different results.

#include <iostream>
#include <limits>

int main ()
{
    std::cout << std::numeric_limits<int>::infinity () << std::endl;
    std::cout << static_cast<int> (std::numeric_limits<float>::infinity () ) << std::endl;
    return 0;
}

Output:

0
-2147483648

The result of std::numeric_limits<int>::infinity() is well defined and equal to 0, but I can't find any information about casting infinity.

解决方案

You said

but it all boils down to

The C standard (incorporated into C++ via 26.9) answers that quite plainly:

Since truncation of infinity is still infinity, and infinity cannot be represented in int (I hope there's no question about this part), the behavior is undefined.

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