简单解析Integer.parseInt()方法

我看到这个知识点是java面试基础中的考点，所以自己为了以后面试打算自己过一遍。我看到别人博客上对源码直接是文字说明，我觉得效果不是很好，我这里直接代数测试这个源码运行流程。

1.我带入一个正正整数 256 注意看注释中的数值

 public static int parseInt(String s, int radix)

    {
        /*
         * WARNING: This method may be invoked early during VM initialization
         * before IntegerCache is initialized. Care must be taken to not use
         * the valueOf method.
         */

        //判断基数是否在 2~36之间
        if (s == null) {
            throw new NumberFormatException("null");
        }

        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; //-2147483647
        int multmin;
        int digit;
        //字符串中的是否有符号
        if (len > 0) {
            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++;
            }
            //计算multmin 值
            multmin = limit / radix;
            // multmin = -214748364

           //开始循环 
            while (i < len) {
                // Accumulating negatively avoids surprises near MAX_VALUE
                //获取字符转换成对应进制的整数
                digit = Character.digit(s.charAt(i++),radix);
                //第一次循环  digit = 2;
                //第二次循环  digit = 5;
                //第三次循环  digit = 6;
                if (digit < 0) {
                    throw NumberFormatException.forInputString(s);
                }
                if (result < multmin) {
                    throw NumberFormatException.forInputString(s);
                }
                result *= radix;
                //第一次循环   result = 0;
                //第二次循环   result = -20;
                //第三次循环   result = -250;
                if (result < limit + digit) {
                //第一次循环  limit + digit = -2147483645;
                //第二次循环   limit + digit = -2147483640;
                //第三次循环   limit + digit = -2147483634;

                    throw NumberFormatException.forInputString(s);
                }
                result -= digit;
                //第一次循环 result = -2;
                //第二次循环 result = -25;
                //第三次循环 result = -256;

            }
        } else {
            throw NumberFormatException.forInputString(s);
        }
        return negative ? result : -result;
        //negative 值为false，所以 -result = -(-256) = 256  返回结果
    }

2.我再带入一个负数 -256

 public static int parseInt(String s, int radix)

    {
        /*
         * WARNING: This method may be invoked early during VM initialization
         * before IntegerCache is initialized. Care must be taken to not use
         * the valueOf method.
         */

        //判断基数是否在 2~36之间
        if (s == null) {
            throw new NumberFormatException("null");
        }

        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; //-2147483647
        int multmin;
        int digit;
        //字符串中的是否有符号
        if (len > 0) {
            char firstChar = s.charAt(0);
            if (firstChar < '0') { // Possible leading "+" or "-"
                if (firstChar == '-') {
                //走这里
                    negative = true;
                    limit = Integer.MIN_VALUE;
               //此时  limit = -2147483648;
                } else if (firstChar != '+')
                    throw NumberFormatException.forInputString(s);

                if (len == 1) // Cannot have lone "+" or "-"
                    throw NumberFormatException.forInputString(s);
                i++;
            }
            //计算multmin 值
            multmin = limit / radix;
            // multmin = -214748364

           //开始循环 
            while (i < len) {
                // Accumulating negatively avoids surprises near MAX_VALUE
                //获取字符转换成对应进制的整数
                digit = Character.digit(s.charAt(i++),radix);
                //第一次循环  digit = 2;
                //第二次循环  digit = 5;
                //第三次循环  digit = 6;
                if (digit < 0) {
                    throw NumberFormatException.forInputString(s);
                }
                if (result < multmin) {
                    throw NumberFormatException.forInputString(s);
                }
                result *= radix;
                //第一次循环   result = 0;
                //第二次循环   result = -20;
                //第三次循环   result = -250;
                if (result < limit + digit) {
                //第一次循环  limit + digit = -2147483646;
                //第二次循环   limit + digit = -2147483641;
                //第三次循环   limit + digit = -2147483635;

                    throw NumberFormatException.forInputString(s);
                }
                result -= digit;
                //第一次循环 result = -2;
                //第二次循环 result = -25;
                //第三次循环 result = -256;

            }
        } else {
            throw NumberFormatException.forInputString(s);
        }
        return negative ? result : -result;
        //negative 值为true，所以 result = -256 = -256  返回结果
    }

从以上代码可以看出 multmin 和result 都为负值 这样设计的原因我猜测是
Accumulating negatively avoids surprises near MAX_VALUE
（累加负值避免超过最大值 最小值：-2147483648 最大值：2147483647）
利用negative 这个标志变量，很巧妙的区分开了正负。

我简单的分析到此为止，有不对的地方，希望留言指正，大家一块学习。