关于一段代码的疑惑

今天需要写一段关于输入字符串的check的代码，突然想到equalIrgoreCase貌似功能上比equals更强大，但是为啥很多时候都只用equals 而不用equalIrgoreCase呢？所以就去看了下源代码，结果发现了String类中的一点额外的东西。

 

源代码如下：

 /**
     * Compares this string to the specified {@code CharSequence}.  The result
     * is {@code true} if and only if this {@code String} represents the same
     * sequence of char values as the specified sequence.
     *
     * @param  cs
     *         The sequence to compare this {@code String} against
     *
     * @return  {@code true} if this {@code String} represents the same
     *          sequence of char values as the specified sequence, {@code
     *          false} otherwise
     *
     * @since  1.5
     */
    public boolean contentEquals(CharSequence cs) {
        if (count != cs.length())
            return false;
        // Argument is a StringBuffer, StringBuilder
        if (cs instanceof AbstractStringBuilder) {
            char v1[] = value;
            char v2[] = ((AbstractStringBuilder)cs).getValue();
            int i = offset;
            int j = 0;
            int n = count;
            while (n-- != 0) {
                if (v1[i++] != v2[j++])
                    return false;
            }
        }
        // Argument is a String
        if (cs.equals(this))
            return true;
        // Argument is a generic CharSequence
        char v1[] = value;
        int i = offset;
        int j = 0;
        int n = count;
        while (n-- != 0) {
            if (v1[i++] != cs.charAt(j++))
                return false;
        }
        return true;
    }

  我很想知道为什么当

 if (cs instanceof AbstractStringBuilder)
{
     ...
}

 成立的时候，为啥结尾不直接返回true呢？如果传入的参数是一个StringBuffer or StringBuilder，则需要遍历两次，这是为什么呢？我感觉添加一个return true 更合理一些，不是吗？如：

 // Argument is a StringBuffer, StringBuilder
        if (cs instanceof AbstractStringBuilder) {
            char v1[] = value;
            char v2[] = ((AbstractStringBuilder)cs).getValue();
            int i = offset;
            int j = 0;
            int n = count;
            while (n-- != 0) {
                if (v1[i++] != v2[j++])
                    return false;
            }
        return true;
        }