Scala char 到 int 的转换
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Scala char to int conversion
提问by flavian
def multiplyStringNumericChars(list: String): Int = {
var product = 1;
println(s"The actual thing + $list")
list.foreach(x => { println(x.toInt);
product = product * x.toInt;
});
product;
};
This is a function that takes a String like 12345and should return the result of 1 * 2 * 3 * 4 * 5. However, I'm getting back doesn't make any sense. What is the implicit conversion from Charto Intactually returning?
这是一个接受类似 String 的函数，12345并且应该返回1 * 2 * 3 * 4 * 5. 然而，我回来没有任何意义。从Char到Int实际返回的隐式转换是什么？
It appears to be adding 48to all the values. If instead I do product = product * (x.toInt - 48)the results are correct.
它似乎增加48了所有的价值。相反，如果我这样做product = product * (x.toInt - 48)，结果是正确的。
回答by om-nom-nom
It does make sense: that is the way how characters encoded in ASCII table: 0 char maps to decimal 48, 1 maps to 49 and so on. So basically when you convert char to int, all you need to do is to just subtract '0':
这是有道理的：这就是在 ASCII 表中编码字符的方式：0 字符映射到十进制 48，1 映射到 49 等等。所以基本上当你将 char 转换为 int 时，你需要做的就是减去“0”：
scala> '1'.toInt
// res1: Int = 49
scala> '0'.toInt
// res2: Int = 48
scala> '1'.toInt - 48
// res3: Int = 1
scala> '1' - '0'
// res4: Int = 1
Or just use x.asDigit, as @Reimer said
或者只是使用x.asDigit，正如@Reimer 所说
scala> '1'.asDigit
// res5: Int = 1
