Kotlin 二进制数与八进制数相互转换的程序

示例 1：二进制转八进制程序

在此程序中，我们将首先将二进制数转换为十进制。然后，将十进制数转换为八进制。

fun main(args: Array<String>) {
    val binary: Long = 101001
    val octal = convertBinarytoOctal(binary)
    println("$binary in binary = $octal in octal")
}

fun convertBinarytoOctal(binaryNumber: Long): Int {
    var binaryNumber = binaryNumber
    var octalNumber = 0
    var decimalNumber = 0
    var i = 0

    while (binaryNumber.toInt() != 0) {
        decimalNumber += (binaryNumber % 10 * Math.pow(2.0, i.toDouble())).toInt()
        ++i
        binaryNumber /= 10
    }

    i = 1

    while (decimalNumber != 0) {
        octalNumber += decimalNumber % 8 * i
        decimalNumber /= 8
        i *= 10
    }

    return octalNumber
}

运行程序后，输出将是

101001 in binary = 51 in octal

此转换如下进行

Binary To Decimal
1 * 25 + 0 * 24 + 1 * 23 + 0 * 22 + 0 * 21 + 1 * 20 = 41

Decimal to Octal
8 | 41
8 | 5 -- 1
8 | 0 -- 5
(51)

示例 2：八进制转二进制程序

在此程序中，首先将八进制数转换为十进制。然后，将十进制数转换为二进制数。

fun main(args: Array<String>) {
    val octal = 67
    val binary = convertOctalToBinary(octal)
    println("$octal in octal = $binary in binary")
}

fun convertOctalToBinary(octalNumber: Int): Long {
    var octalNumber = octalNumber
    var decimalNumber = 0
    var i = 0
    var binaryNumber: Long = 0

    while (octalNumber != 0) {
        decimalNumber += (octalNumber % 10 * Math.pow(8.0, i.toDouble())).toInt()
        ++i
        octalNumber /= 10
    }

    i = 1

    while (decimalNumber != 0) {
        binaryNumber += (decimalNumber % 2 * i).toLong()
        decimalNumber /= 2
        i *= 10
    }

    return binaryNumber
}

运行程序后，输出将是

67 in octal = 110111 in binary

此转换如下进行

Octal To Decimal
6 * 81 + 7 * 80 = 55

Decimal to Binary
2 | 55
2 | 27 -- 1
2 | 13 -- 1
2 | 6  -- 1
2 | 3  -- 0
2 | 1  -- 1
2 | 0  -- 1
(110111)

等效的 Java 代码如下：Java 二进制转八进制及反向转换程序