String转换Integer原理

好朋友面试被问到这个，回来问我。发现我也不知道。就想着去看看源码，了解一下

String转Integer,我们平时使用时，一般都是 ：

Integer integer = Integer.valueOf("99");

让我们去看下Integer.valueOf的方法吧~

Integer.valueOf方方法如果参数是int时，会先读取缓存的Integer,没有才会去创建一个新的。这个和String转换Integer没有关系，我们主要要去看parseInt(s, 10)方法

// radix  基数  进制 2进制，10进制等
public static int parseInt(String s, int radix)
            throws NumberFormatException
{
    /*
     * WARNING: This method may be invoked early during VM initialization
     * before IntegerCache is initialized. Care must be taken to not use
     * the valueOf method.
     */

    if (s == null) {
        // Android-changed: Improve exception message for parseInt.
        throw new NumberFormatException("s == null");
    }
    // 判断基数要大于等于2且小于等于36. [2,36]
    if (radix < Character.MIN_RADIX) {
        throw new NumberFormatException("radix " + radix +
                                        " less than Character.MIN_RADIX");
    }

    if (radix > Character.MAX_RADIX) {
        throw new NumberFormatException("radix " + radix +
                                        " greater than Character.MAX_RADIX");
    }

    int result = 0;
    boolean negative = false;
    int i = 0, len = s.length();
    int limit = -Integer.MAX_VALUE;
    int multmin;
    int digit;

    if (len > 0) {
        //  判断该字符是正数还是负数 通过 +，- 判断
        char firstChar = s.charAt(0);
        if (firstChar < '0') { // Possible leading "+" or "-"
            if (firstChar == '-') {
                negative = true;
                limit = Integer.MIN_VALUE;
            } else if (firstChar != '+')
                throw NumberFormatException.forInputString(s);

            if (len == 1) // Cannot have lone "+" or "-"
                throw NumberFormatException.forInputString(s);
            i++;
        }
        multmin = limit / radix;
        while (i < len) {
            // Accumulating negatively avoids surprises near MAX_VALUE
            //返回指定基数中字符表示的数值。（此处是十进制数值） 
            //例：int digit = Character.digit('9',10);  dight返回的就是9
            digit = Character.digit(s.charAt(i++),radix);
            if (digit < 0) {
                throw NumberFormatException.forInputString(s);
            }
            if (result < multmin) {
                throw NumberFormatException.forInputString(s);
            }
            // 0, -90
            result *= radix;
            if (result < limit + digit) {
                throw NumberFormatException.forInputString(s);
            }
            // 0-9 = -9, -90-9=-99 
            result -= digit;
        }
    } else {
        throw NumberFormatException.forInputString(s);
    }
    // 根据negative是正数还是负数，正数就前面加-   -(-99) = 99
    return negative ? result : -result;
}

我们再看digit = Character.digit(s.charAt(i++),radix);方法

public static int digit(int codePoint, int radix) {
        //基数必须再最大和最小基数之间
        if (radix < MIN_RADIX || radix > MAX_RADIX) {
            return -1;
        }
        
        if (codePoint < 128) {
            // Optimized for ASCII
            int result = -1;
            //字符在0-9字符之间
            if ('0' <= codePoint && codePoint <= '9') {
                result = codePoint - '0';
            }
            //字符在a-z之间
            else if ('a' <= codePoint && codePoint <= 'z') {
                result = 10 + (codePoint - 'a');
            }
            //字符在A-Z之间
            else if ('A' <= codePoint && codePoint <= 'Z') {
                result = 10 + (codePoint - 'A');
            }
        
            return result < radix ? result : -1;
        }
        return digitImpl(codePoint, radix);
    }

以基数10为例(就是10进制)。看看方法：result = codePoint – ‘0’;  

我们都知道char字符每一个都对应一个ASCII码的数字，那么我们代入上面的方法，例：

当 codePoint = ‘9’时，即 result = ‘9’ – ‘0’;   char 9 对应10进制ASCII码为57， char 0 对应10进制ASCII码为48.

所以：result = ‘9’ – ‘0’;   就等于    result = 57 – 48;   result = 9;

参考：

ASCII码对照表

String 转换成 Integer 的方式及原理