diff --git a/docs/chapter_data_structure/number_encoding.md b/docs/chapter_data_structure/number_encoding.md index 95cc5ca36..4b8b073f8 100644 --- a/docs/chapter_data_structure/number_encoding.md +++ b/docs/chapter_data_structure/number_encoding.md @@ -105,7 +105,7 @@ $$ 二进制数 `float` 对应值的计算方法为： $$ -\text {val} = (-1)^{b_{31}} \times 2^{\left(b_{30} b_{29} \ldots b_{23}\right)_2-127} \times\left(1 . b_{22} b_{21} \ldots b_0\right)_2 +\text {val} = (-1)^{b_{31}} \times 2^{\left(b_{30} b_{29} \ldots b_{23}\right)_2-127} \times\left(1 + b_{22} b_{21} \ldots b_0\right)_2 $$ 转化到十进制下的计算公式为： @@ -119,7 +119,7 @@ $$ $$ \begin{aligned} \mathrm{S} \in & \{ 0, 1\}, \quad \mathrm{E} \in \{ 1, 2, \dots, 254 \} \newline -(1 + \mathrm{N}) = & (1 + \sum_{i=1}^{23} b_{23-i} 2^{-i}) \subset [1, 2 - 2^{-23}] +(1 + \mathrm{N}) = & (1 + \sum_{i=0}^{22} b_{i} 2^{i-23}) \subset [1, 2 - 2^{-23}] \end{aligned} $$ @@ -141,8 +141,8 @@ $$ | 指数位 E | 分数位 $\mathrm{N} = 0$ | 分数位 $\mathrm{N} \ne 0$ | 计算公式 | | ------------------ | ----------------------- | ------------------------- | ---------------------------------------------------------------------- | -| $0$ | $\pm 0$ | 次正规数 | $(-1)^{\mathrm{S}} \times 2^{-126} \times (0.\mathrm{N})$ | -| $1, 2, \dots, 254$ | 正规数 | 正规数 | $(-1)^{\mathrm{S}} \times 2^{(\mathrm{E} -127)} \times (1.\mathrm{N})$ | +| $0$ | $\pm 0$ | 次正规数 | $(-1)^{\mathrm{S}} \times 2^{-126} \times (0+\mathrm{N})$ | +| $1, 2, \dots, 254$ | 正规数 | 正规数 | $(-1)^{\mathrm{S}} \times 2^{(\mathrm{E} -127)} \times (1+\mathrm{N})$ | | $255$ | $\pm \infty$ | $\mathrm{NaN}$ | | 值得说明的是，次正规数显著提升了浮点数的精度。最小正正规数为 $2^{-126}$ ，最小正次正规数为 $2^{-126} \times 2^{-23}$ 。 diff --git a/en/docs/chapter_data_structure/number_encoding.md b/en/docs/chapter_data_structure/number_encoding.md index 41703f756..a9c6f6aa8 100644 --- a/en/docs/chapter_data_structure/number_encoding.md +++ b/en/docs/chapter_data_structure/number_encoding.md @@ -105,7 +105,7 @@ According to the IEEE 754 standard, a 32-bit `float` consists of the following t The value of a binary `float` number is calculated as: $$ -\text{val} = (-1)^{b_{31}} \times 2^{\left(b_{30} b_{29} \ldots b_{23}\right)_2 - 127} \times \left(1 . b_{22} b_{21} \ldots b_0\right)_2 +\text {val} = (-1)^{b_{31}} \times 2^{\left(b_{30} b_{29} \ldots b_{23}\right)_2-127} \times\left(1 + b_{22} b_{21} \ldots b_0\right)_2 $$ Converted to a decimal formula, this becomes: @@ -119,7 +119,7 @@ The range of each component is: $$ \begin{aligned} \mathrm{S} \in & \{ 0, 1\}, \quad \mathrm{E} \in \{ 1, 2, \dots, 254 \} \newline -(1 + \mathrm{N}) = & (1 + \sum_{i=1}^{23} b_{23-i} \times 2^{-i}) \subset [1, 2 - 2^{-23}] +(1 + \mathrm{N}) = & (1 + \sum_{i=0}^{22} b_{i} 2^{i-23}) \subset [1, 2 - 2^{-23}] \end{aligned} $$ @@ -141,8 +141,8 @@ As shown in the table below, exponent bits $\mathrm{E} = 0$ and $\mathrm{E} = 25 | Exponent Bit E | Fraction Bit $\mathrm{N} = 0$ | Fraction Bit $\mathrm{N} \ne 0$ | Calculation Formula | | ------------------ | ----------------------------- | ------------------------------- | ---------------------------------------------------------------------- | -| $0$ | $\pm 0$ | Subnormal Numbers | $(-1)^{\mathrm{S}} \times 2^{-126} \times (0.\mathrm{N})$ | -| $1, 2, \dots, 254$ | Normal Numbers | Normal Numbers | $(-1)^{\mathrm{S}} \times 2^{(\mathrm{E} -127)} \times (1.\mathrm{N})$ | +| $0$ | $\pm 0$ | Subnormal Numbers | $(-1)^{\mathrm{S}} \times 2^{-126} \times (0+\mathrm{N})$ | +| $1, 2, \dots, 254$ | Normal Numbers | Normal Numbers | $(-1)^{\mathrm{S}} \times 2^{(\mathrm{E} -127)} \times (1+\mathrm{N})$ | | $255$ | $\pm \infty$ | $\mathrm{NaN}$ | | It's worth noting that subnormal numbers significantly improve the precision of floating-point numbers. The smallest positive normal number is $2^{-126}$, and the smallest positive subnormal number is $2^{-126} \times 2^{-23}$. diff --git a/zh-hant/docs/chapter_data_structure/number_encoding.md b/zh-hant/docs/chapter_data_structure/number_encoding.md index ae64c83a1..c162fa81a 100644 --- a/zh-hant/docs/chapter_data_structure/number_encoding.md +++ b/zh-hant/docs/chapter_data_structure/number_encoding.md @@ -105,7 +105,7 @@ $$ 二進位制數 `float` 對應值的計算方法為： $$ -\text {val} = (-1)^{b_{31}} \times 2^{\left(b_{30} b_{29} \ldots b_{23}\right)_2-127} \times\left(1 . b_{22} b_{21} \ldots b_0\right)_2 +\text {val} = (-1)^{b_{31}} \times 2^{\left(b_{30} b_{29} \ldots b_{23}\right)_2-127} \times\left(1 + b_{22} b_{21} \ldots b_0\right)_2 $$ 轉化到十進位制下的計算公式為： @@ -119,7 +119,7 @@ $$ $$ \begin{aligned} \mathrm{S} \in & \{ 0, 1\}, \quad \mathrm{E} \in \{ 1, 2, \dots, 254 \} \newline -(1 + \mathrm{N}) = & (1 + \sum_{i=1}^{23} b_{23-i} 2^{-i}) \subset [1, 2 - 2^{-23}] +(1 + \mathrm{N}) = & (1 + \sum_{i=0}^{22} b_{i} 2^{i-23}) \subset [1, 2 - 2^{-23}] \end{aligned} $$ @@ -141,8 +141,8 @@ $$ | 指數位 E | 分數位 $\mathrm{N} = 0$ | 分數位 $\mathrm{N} \ne 0$ | 計算公式 | | ------------------ | ----------------------- | ------------------------- | ---------------------------------------------------------------------- | -| $0$ | $\pm 0$ | 次正規數 | $(-1)^{\mathrm{S}} \times 2^{-126} \times (0.\mathrm{N})$ | -| $1, 2, \dots, 254$ | 正規數 | 正規數 | $(-1)^{\mathrm{S}} \times 2^{(\mathrm{E} -127)} \times (1.\mathrm{N})$ | +| $0$ | $\pm 0$ | 次正規數 | $(-1)^{\mathrm{S}} \times 2^{-126} \times (0+\mathrm{N})$ | +| $1, 2, \dots, 254$ | 正規數 | 正規數 | $(-1)^{\mathrm{S}} \times 2^{(\mathrm{E} -127)} \times (1+\mathrm{N})$ | | $255$ | $\pm \infty$ | $\mathrm{NaN}$ | | 值得說明的是，次正規數顯著提升了浮點數的精度。最小正正規數為 $2^{-126}$ ，最小正次正規數為 $2^{-126} \times 2^{-23}$ 。