向下转换为float：是否保证溢出行为？ [英] Downcasting double to float: is overflow behaviour guaranteed?

问题描述

如果我尝试这个

  float f =（float）numeric_limits< double> :: infinity 
  

或者，尝试将大于float max的任何数据转换为浮点数，

解决方案

它适用于GCC， / div>

float f =（float）numeric_limits< double> :: infinity（）;

如果您的编译平台为浮点计算提供IEEE 754算法，则可以将 f 设置为无穷大它通常做）。

或者，试着将比float max更大的任何值转换为浮点值，无限？

否。在默认的IEEE 754 round-to-nearest模式中，一些 double 的值高于最大有限 float ， FLT_MAX ）转换为 FLT_MAX 。确切的限制是 FLT_MAX （C99十六进制表示中的 0x1.fffffep127 ）和下一个 float 如果单精度格式的指数具有较大范围， 0x2.0p128 ，则可以表示的数字。因此，限额为 0x1.ffffffp127 或约为3.4028235677973366e + 38的十进制。

If I try this

float f = (float)numeric_limits<double>::infinity();

Or indeed, try to cast anything bigger than float max down to a float, am I guaranteed to end up with infinity?

It works on GCC, but is it a standard though?

解决方案

float f = (float)numeric_limits<double>::infinity();

This is guaranteed to set f to infinity if your compilation platform offers IEEE 754 arithmetic for floating-point computations (it usually does).

Or indeed, try to cast anything bigger than float max down to a float, am I guaranteed to end up with infinity?

No. In the default IEEE 754 round-to-nearest mode, a few double values above the maximum finite float (that is, FLT_MAX) convert to FLT_MAX. The exact limit is the number midway between FLT_MAX (0x1.fffffep127 in C99 hexadecimal representation) and the next float number that could be represented if the exponent in the single-precision format had a larger range, 0x2.0p128. The limit is thus 0x1.ffffffp127 or approximately 3.4028235677973366e+38 in decimal.

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