将字符（数字）转换为整数的“java”方式是什么
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What's the "java" way of converting chars (digits) to ints
提问by Andrei Ciobanu
Given the following code:
鉴于以下代码：
char x = '5';
int a0 = x - '0'; // 0
int a1 = Integer.parseInt(x + ""); // 1
int a2 = Integer.parseInt(Character.toString(x)); // 2
int a3 = Character.digit(x, 10); // 3
int a4 = Character.getNumericValue(x); // 4
System.out.printf("%d %d %d %d %d", a0, a1, a2, a3, a4);
(version 4 credited to: casablanca)
（第 4 版归功于：卡萨布兰卡）
What do you consider to be the "best-way" to convert a char into an int ? ("best-way" ~= idiomatic way)
您认为将 char 转换为 int的“最佳方式”是什么？（“最佳方式”~=惯用方式）
We are not converting the actual numerical value of the char, but the value of the representation.
我们不是转换字符的实际数值，而是转换表示的值。
Eg.:
例如。：
convert('1') -> 1
convert('2') -> 2
....
采纳答案by casablanca
How about Character.getNumericValue?
回答by Oliver Charlesworth
The first method. It's the most lightweight and direct, and maps to what you might do in other (lower-level) languages. Of course, its error handling leaves something to be desired.
第一种方法。它是最轻量级和最直接的，并且映射到您可能在其他（低级）语言中执行的操作。当然，它的错误处理还有一些不足之处。
回答by jacobm
I'd strongly prefer Character.digit.
我非常喜欢Character.digit.
回答by Peter Lawrey
If speed is critical (rather than validation you can combine the result) e.g.
如果速度很重要（而不是验证，您可以组合结果）例如
char d0 = '0';
char d1 = '4';
char d2 = '2';
int value = d0 * 100 + d1 * 10 + d2 - '0' * 111;
回答by ben
Convert to Ascii then subtract 48.
转换为 Ascii，然后减去 48。
(int) '7' would be 55
((int) '7') - 48 = 7
((int) '9') - 48 = 9
回答by Kevin Ng
The best way to convert a character of a valid digit to an int value is below. If c is larger than 9 then c was not a digit. Nothing that I know of is faster than this. Any digits in ASCII code 0-9(48-57) ^ to '0'(48) will always yield 0-9. From 0 to 65535 only 48 to 57 yield 0 to 9 in their respective order.
将有效数字的字符转换为 int 值的最佳方法如下。如果 c 大于 9，则 c 不是数字。我所知道的没有比这更快的了。ASCII 代码 0-9(48-57) ^ 到 '0'(48) 中的任何数字将始终产生 0-9。从 0 到 65535 只有 48 到 57 按各自的顺序产生 0 到 9。
int charValue = (charC ^ '0');
