Question

我的任务是将浮点数从小数转换为32位二进制表示（IEEE 754。）

以下是我尝试解决任务的问题;它导致了几个在线转换工具让我产生的相同结果，但是有人告诉我，我的解决方案并不合适，因为它编码了一个不正确的指数。＆＃34;

我是否可以深入了解我的哪些步骤正是导致此问题（如果有）以及为什么我仍然得到正确的（？）结果？

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要转换的浮点数（十进制表示法）：6.4

二进制小数点前的位置：

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

=> 110

二进制的小数位数：

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

标准化：

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

特点：

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

标志：0（正面）

结果：

01000000110011001100110011001101

可能导致同样问题的代码段：

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

将小数位转换为二进制时，第8行表示

0.2 * 2 = 0.4 -> 0

然后继续

0.8 * 2 = 1.6 -> 1

而不是

0.4 * 2 = 0.8 -> 0

行 12,15,18,21和24

上存在同样的问题

修复这个应该给你正确的尾数（标准化后）给你正确的结果

01000000110011001100110011001101

您还可以查看this IEEE 754 converter

将float从十进制表示转换为32位二进制（IEEE 754浮点）

1 个答案: