单精度浮点格式范围 [英] Single-precision floating-point format Range

问题描述

符号= 1位，偏置指数= 8位，尾数= 23位

Sign = 1 bit, Biased Exponent = 8 bits, Mantissa = 23 bits

正负可能范围是多少?我的老师告诉我ieee 754的范围如下:

What is the positive and negative possible range? My teacher told me the following range for ieee 754:

-0.5*2^-128 to -(1-2^-24)*2^127 (for negative floating point numbers)

0.5*2^-128 to (1-2^-24)*2^127 (for positive floating point numbers)

但是我发现此范围不正确，因为我无法理解如何将0.5 * 2 ^-128 存储为这种格式.请解释.

But I don't find this range correct because I am not able to understand how to store 0.5 * 2^-128 into this format. Please explain.

推荐答案

首先，浮点数格式对于正数和负数是对称的.所以我们只看积极的情况.

Firstly, the floating-point number format is symmetric for positive and negative numbers. So we will only look at the positive case.

最大正数具有最大尾数1.11111111111111111111111111 ₂ 和最大非无限指数127.因此1.11111111111111111111111111 ₂ ×2 ¹²⁷ =( 2 − 2 ⁻²³ )×2 ¹²⁷ ≈3.402×10 ³⁸ ≈2 ¹²⁸ .

The maximum positive number has the maximum mantissa 1.11111111111111111111111₂ and maximum non-infinite exponent 127. Thus 1.11111111111111111111111₂ × 2¹²⁷ = (2 − 2⁻²³) × 2¹²⁷ ≈ 3.402 × 10³⁸ ≈ 2¹²⁸.

最小正数的尾数为非零，为0.00000000000000000000001 ₂ ，对于次正规/非规格化数，最小指数为-126.因此0.00000000000000000000001 ₂ ×2 ⁻¹²⁶ = 2 ^-23 ×2 ⁻¹²⁶ = 2 ⁻¹⁴⁹ ≈1.401×10 ⁻⁴⁵ .

The minimum positive number has the non-zero mantissa 0.00000000000000000000001₂ and the minimum exponent −126 for subnormal/denormalized numbers. Thus 0.00000000000000000000001₂ × 2⁻¹²⁶ = 2⁻²³ × 2⁻¹²⁶ = 2⁻¹⁴⁹ ≈ 1.401 × 10⁻⁴⁵.

进一步阅读: https://en.wikipedia.org/wiki/Single-precision_floating -point_format

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