int16_t浮动转换怪事 [英] int16_t to float conversion weirdness

问题描述

我在跟这里发生了什么损失。我需要一个浮点转换为int16_t和背部。下面是语法：

I am at a loss with what is happening here. I need to convert a float to an int16_t and back. Here is the syntax:

int16_t val = (int16_t)round((float)0xFFFE/100 * angle);
//and back
float angle = ((float)100/0xFFFE * val;

当我使用的-0.093081初始角度值，将其转换回来。但是当我使用它182.241211转换回-17.764824？

When I use an initial angle value of -0.093081, it converts back. But when I use 182.241211 it converts back to -17.764824?

任何想法是怎么回事？

推荐答案

0xFFFE 是几乎的最大的16位数字;和仅一个的无符号的16位数字，在那。如果除以100，然后由182乘，它肯定会溢出。

0xFFFE is almost the maximum 16-bit number; and only for an unsigned 16-bit number, at that. If you divide it by 100 and then multiply by 182, it's definitely going to overflow.

让我们完全做到在基地10清晰度（0xFFFE为65534）：

Let's do it fully in base 10 for clarity (0xFFFE is 65534):

65534 / 100 *  -0.093081 =    -60.99970254
65534 / 100 * 182.241211 = 119429.95521674

您签署的16位整数的全系列几乎肯定[-32768，32767]。那最后的结果将不适合。

The full range of your signed 16-bit integer is almost certainly [-32768, 32767]. That last result won't fit.

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