不正確指數（？）從十進制表示轉換浮子32位二進制（IEEE 754浮點）時

Question

我的任務是浮法從其十進制轉換成其32位的二進制表示（IEEE 754）

以下是我的解決任務的嘗試;它導致了幾個在線轉換工具讓我產生同樣的結果，但是我被告知我的解決方案並不好，因爲「它編碼的指數不正確。」

我可以深入瞭解哪些步驟完全是導致此問題（如果有的話）以及爲什麼我仍然得到正確的結果？

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浮法（十進制表示），以轉換：6.4

預小數點位置以二進制：

6/2 = 3 -> 0 
3/2 = 1.5 -> 1 
1/2 = 0.5 -> 1 

=> 110 

二進制小數位數：

0.4 * 2 = 0.8 -> 0 
0.8 * 2 = 1.6 -> 1 
0.6 * 2 = 1.2 -> 1 
0.2 * 2 = 0.4 -> 0 
0.4 * 2 = 0.8 -> 0 
0.8 * 2 = 1.6 -> 1 
0.6 * 2 = 1.2 -> 1 
0.2 * 2 = 0.4 -> 0 
0.4 * 2 = 0.8 -> 0 
0.8 * 2 = 1.6 -> 1 
0.6 * 2 = 1.2 -> 1 
0.2 * 2 = 0.4 -> 0 
0.4 * 2 = 0.8 -> 0 
0.8 * 2 = 1.6 -> 1 
0.6 * 2 = 1.2 -> 1 
0.2 * 2 = 0.4 -> 0 
0.4 * 2 = 0.8 -> 0 
0.8 * 2 = 1.6 -> 1 
0.6 * 2 = 1.2 -> 1 
0.2 * 2 = 0.4 -> 0 
0.4 * 2 = 0.8 -> 0 
0.8 * 2 = 1.6 -> 1 
0.6 * 2 = 1.2 -> 1 
0.2 * 2 = 0.4 -> 0 
0.4 * 2 = 0.8 -> 0 
0.8 * 2 = 1.6 -> 1 
0.6 * 2 = 1.2 -> 1 
0.2 * 2 = 0.4 -> 0 
0.4 * 2 = 0.8 -> 0 
0.8 * 2 = 1.6 -> 1 

=> 011001100110011001100110011001 

規範化：

110,011001100110011001100110011001 * 2^0 
1,10011001100110011001100110011001 * 2^2 

特性：

2+127 = 129 (Exponent + Bias) 

129/2 = 64.5 -> 1 
64/2 = 32 -> 0 
32/2 = 16 -> 0 
16/2 = 8 -> 0 
8/2 = 4 -> 0 
4/2 = 2 -> 0 
2/2 = 1 -> 0 
1/2 = 0.5 -> 1 

=> 10000001 

星座：0（正）

結果：

01000000110011001100110011001101 

代碼片段假想導致同樣的問題：

public static void main(String[] args) { 

    float input = IO.readFloat("Please enter a float!\n"); 
    String sign = input < 0 ? "1" : "0"; 
    if (input < 0) { 
     input *= -1; 
    } else if (input == 0) { 
     IO.println("00000000000000000000000000000000"); 
     return; 
    } 

    float preDecimalPointPositions = input - (input % 1); 
    String preDecimalPointPositionsBinary = ""; 
    while (preDecimalPointPositions >= 1) { 
     if (preDecimalPointPositions % 2 == 0) { 
      preDecimalPointPositionsBinary = "0" + preDecimalPointPositionsBinary; 
     } else { 
      preDecimalPointPositionsBinary = "1" + preDecimalPointPositionsBinary; 
     } 
     preDecimalPointPositions /= 2; 
     preDecimalPointPositions = preDecimalPointPositions - (preDecimalPointPositions % 1); // Abschneiden der decimalPlacesstellen 
    } 

    float decimalPlaces = input % 1; 
    String decimalPlacesBinary = ""; // 
    for (int i = 0; i < 23; i++) { // 
     decimalPlaces *= 2; 
     if (decimalPlaces >= 1) { 
      decimalPlacesBinary += "1"; 
      decimalPlaces -= 1; 
     } else { 
      decimalPlacesBinary += "0"; 
     } 
    } 

    int exponent; 
    if (input < 1) { 
     IO.println("Negative exponents ain't allowed here! =/"); 
     return; 
    } else { 
     exponent = preDecimalPointPositionsBinary.length() - 1; 
    } 
    String normalized = preDecimalPointPositionsBinary.substring(1) + decimalPlacesBinary; 
    float characteristic = exponent + 127; 
    String characteristicBinary = ""; 
    while (characteristic >= 1) { 
     if (characteristic % 2 == 0) { 
      characteristicBinary = "0" + characteristicBinary; 
     } else { 
      characteristicBinary = "1" + characteristicBinary; 
     } 
     characteristic /= 2; 
     characteristic = characteristic - (characteristic % 1); 
    } 

    String result = sign + characteristicBinary + normalized.substring(0, 23); 
    IO.println(result); 
} 

Answer 1

當轉換小數爲二進制的8號線說

0.2 * 2 = 0.4 -> 0 

但你去與

0.8 * 2 = 1.6 -> 1 

，而不是

0.4 * 2 = 0.8 -> 0 

上線存在同樣的問題12,15,18,21和24

那麼要解決這個應該給你正確的尾數（歸一化後）給你的

01000000110011001100110011001101 

正確的結果還可以用this IEEE 754 converter