IEEE-754中2到3之间的数字数量 [英] Number of numbers between 2 and 3 in IEEE-754

问题描述

我正在学习IEEE-754的数字表示形式.我知道如何从二进制转换为IEEE，反之亦然.现在，我尝试找出如何找出单精度的数字，例如2到3之间的数字.因此，两者的符号都相同.我认为分数将是一个组合，并且指数取决于适当的数字(因为移位).这样做会是一个聪明的方法吗?我将不胜感激.

解决方案

使用方便在线IEEE-754转换实用程序:

2.0 = 0x40000000
3.0 = 0x40400000

所以:

0x40400000 - 0x40000000 = 0x400000 = 4194304

答案:400万左右.

i am learning IEEE-754 representation of numbers. I know how to convert from binary to IEEE and on vice versa. Now i am trying to figure out how to find out how many numbers in single precision are for instance between 2 and 3. So, the sign will be the same for both. Fraction will be a combination i think and exponent is dependent from a proper number(because of shifts). WHat would be a clever way to do it right ? I would be grateful for any help.

解决方案

Using a handy online IEEE-754 conversion utility:

2.0 = 0x40000000
3.0 = 0x40400000

So:

0x40400000 - 0x40000000 = 0x400000 = 4194304

Answer: 4 million or so.

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