Java - 将字母字符串转换为相应 ascii 的 int？
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Java - Convert a String of letters to an int of corresponding ascii?
提问by AaronAAA
I want to convert a String, lets say "abc", to an int with the corresponding ascii: in this example, 979899.
我想将 String 转换为"abc"具有相应 ascii 的 int：在本例中，979899.
I've run into two problems:
我遇到了两个问题：
1) what I wrote only works for characters whose ascii is two characters long and
1) 我写的只适用于 ascii 为两个字符长的字符，并且
2) since these numbers get very big, I can't use longs and I'm having trouble utilizing BigIntegers.
2) 由于这些数字变得非常大，我不能使用 long，而且我在使用 BigIntegers 时遇到了麻烦。
This is what I have so far:
这是我到目前为止：
BigInteger mInt = BigInteger.valueOf(0L);
for (int i = 0; i<mString.length(); i++) {
mInt = mInt.add(BigInteger.valueOf(
(long)(mString.charAt(i)*Math.pow(100,(mString.length()-1-i)))));
}
Any suggestions would be great, thanks!
任何建议都会很棒，谢谢！
采纳答案by arshajii
What's wrong with doing all the concatenation first with a StringBuilderand then creating a BigIntegerout of the result? This seems to be much simpler than what you're currently doing.
首先使用 a 进行所有连接StringBuilder，然后BigInteger从结果中创建 a有什么问题？这似乎比您目前正在做的要简单得多。
String str = "abc"; // or anything else
StringBuilder sb = new StringBuilder();
for (char c : str.toCharArray())
sb.append((int)c);
BigInteger mInt = new BigInteger(sb.toString());
System.out.println(mInt);
回答by Kent
you don't have to play the number game. (pow 100 etc). just get the number string, and pass to constructor.
你不必玩数字游戏。（战俘 100 等）。只需获取数字字符串，然后传递给构造函数。
final String s = "abc";
String v = "";
final char[] chars = s.toCharArray();
for (int i = 0; i < chars.length; i++) {
v += String.valueOf((int) chars[i]);
}
//v = "979899" now
BigInteger bigInt = new BigInteger(v); //BigInteger
BigDecimal bigDec = new BigDecimal(v); // or BigDecimal
回答by mdgeorge
To handle n-digit numbers, you will have to multiply by a different power of ten each time. You could do this with a loop:
要处理 n 位数字，您必须每次乘以不同的 10 次幂。你可以用一个循环来做到这一点：
BigInteger appendDigits(BigInteger total, int n) {
for (int i = n; i > 0; i /= 10)
total = total.multiply(10);
return total.plus(new BigInteger(n));
}
However, this problem really seems to be about manipulating strings. What I would probably do is simply accumulate the digits int a string, and create a BI from the String at the end:
然而，这个问题似乎真的是关于操作字符串。我可能会做的只是简单地将数字累积为一个字符串，并在最后从字符串创建一个 BI：
StringBuilder result = new StringBuilder();
for (char c : mString.getBytes())
result.append(String.valueOf(c));
return new BigInteger(result.toString());
