为什么 FLT_MIN 等于 0? [英] Why is FLT_MIN equal to zero?

问题描述

limits.h 指定非浮点数学类型的限制，例如INT_MIN 和 INT_MAX.这些值是您可以使用 int 表示的最负值和最正值.

limits.h specifies limits for non-floating point math types, e.g. INT_MIN and INT_MAX. These values are the most negative and most positive values that you can represent using an int.

在 float.h 中，有 FLT_MIN 和 FLT_MAX 的定义.如果您执行以下操作:

In float.h, there are definitions for FLT_MIN and FLT_MAX. If you do the following:

NSLog(@"%f %f", FLT_MIN, FLT_MAX);

你会得到以下输出:

FLT_MIN = 0.000000, FLT_MAX = 340282346638528859811704183484516925440.000000

FLT_MAX 等于一个非常大的数字，如您所料，但为什么 FLT_MIN 等于零而不是一个非常大的负数?

FLT_MAX is equal to a really large number, as you would expect, but why does FLT_MIN equal zero instead of a really large negative number?

推荐答案

它实际上不是零，但如果你使用 printf 或 NSLog 检查它可能看起来像零通过使用 %f.
根据 float.h(至少在 Mac OS X 10.6.2 中)，FLT_MIN 被描述为:

It's not actually zero, but it might look like zero if you inspect it using printf or NSLog by using %f.
According to float.h (at least in Mac OS X 10.6.2), FLT_MIN is described as:

/* Minimum normalized positive floating-point number, b**(emin - 1).  */

注意那句话中的正:FLT_MIN指的是最小(标准化)数大于零.(还有更小的非归一化数字).

Note the positive in that sentence: FLT_MIN refers to the minimum (normalized) number greater than zero. (There are much smaller non-normalized numbers).

如果你想要最小的浮点数(包括负数)，使用-FLT_MAX.

If you want the minimum floating point number (including negative numbers), use -FLT_MAX.

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