kotlin - 为什么将单个char和 “single char String”转换为long(.toLong())时不相等

Question

我想对Long变量的数字求和并将其添加到它自己的变量中，我附带了下一个工作代码:

private fun Long.sumDigits(): Long {
    var n = this
    this.toString().forEach { n += it.toString().toLong() }
    return n
}

用法:assert(48.toLong() == 42.toLong().sumDigits())
我必须使用it.toString()才能使其正常工作，因此我进行了下一个测试，但没有得到它的结果:

@Test
fun toLongEquality() {
    println("'4' as Long = " + '4'.toLong())
    println("\"4\" as Long = " + "4".toLong())
    println("\"42\" as Long = " + "42".toLong())

    assert('4'.toString().toLong() == 4.toLong())
}

输出:

'4' as Long = 52
"4" as Long = 4
"42" as Long = 42

使用char.toString().toLong()是个好习惯还是有更好的方法将char转换为Long？
"4"是否由char表示？为什么它不等于char表示形式？

Answer 1

从文档中:

class Char : Comparable (source) Represents a 16-bit Unicode character. On the JVM, non-nullable values of this type are represented as values of the primitive type char.

fun toLong(): Long

Returns the value of this character as a Long.

当您使用'4' as Long时，您实际上会获得字符'4'的Unicode(ASCII)代码