问题起源

在刷leetcode的时候，看到了discuss里面有一位大佬提出很有意思的解法。

这是一道mini parser的题，题目描述如下：

Given a nested list of integers represented as a string, implement a parser to deserialize it.

Each element is either an integer, or a list -- whose elements may also be integers or other lists.

Note: You may assume that the string is well-formed:

String is non-empty.

String does not contain white spaces.

String contains only digits 0-9, [, - ,, `]

他的解法是：

/**
 * // This is the interface that allows for creating nested lists.
 * // You should not implement it, or speculate about its implementation
 * class NestedInteger {
 *   public:
 *     // Constructor initializes an empty nested list.
 *     NestedInteger();
 *
 *     // Constructor initializes a single integer.
 *     NestedInteger(int value);
 *
 *     // Return true if this NestedInteger holds a single integer, rather than a nested list.
 *     bool isInteger() const;
 *
 *     // Return the single integer that this NestedInteger holds, if it holds a single integer
 *     // The result is undefined if this NestedInteger holds a nested list
 *     int getInteger() const;
 *
 *     // Set this NestedInteger to hold a single integer.
 *     void setInteger(int value);
 *
 *     // Set this NestedInteger to hold a nested list and adds a nested integer to it.
 *     void add(const NestedInteger &ni);
 *
 *     // Return the nested list that this NestedInteger holds, if it holds a nested list
 *     // The result is undefined if this NestedInteger holds a single integer
 *     const vector<NestedInteger> &getList() const;
 * };
 */
class Solution {
public:
    NestedInteger deserialize(string s) {
        istringstream in(s);
        return deserialize1(in);
    }
private:
    NestedInteger deserialize1(istringstream& in){
        int num;
        if(in >> num) return NestedInteger(num);
        in.clear();
        in.get();
        NestedInteger lst;
        while (in.peek()!=']'){
            lst.add(deserialize1(in));
            if(in.peek()==',') in.get();
        }
        in.get();
        return lst;
    }
};

关于istringstream

但是我今天要说的不是leetcode的解法，而是要介绍一种内存I/O——istringstream。

这个类的对象可以直接从string中读取数据，最基本的

istringstream是基于空格来提取字符串的，从而完成从字符串到其它类型的改变

基于空格提取字符串

#include <sstream>
#include <string>
using std::string;
using std::istringstream;
int main()
{
	string a("123a 456 [789]]]");
	istringstream in(a);
	string s1, s2, s3;
  	in >> s1 >> s2 >> s3; //s1:"123a", s2:"456", s3:"[789]]]";
    return 0;
}

从字符串转化成数字（int / double）

#include <sstream>
#include <string>
#include <iostream>
using std::string;
using std::istringstream;
using std::cout;
using std::endl;
int main()
{
	string a("123a ,[456,[789]]]");
	istringstream in(a);
	int i1, i2;
	string s1, s2;
	in >> i1;
	//in.clear();
	in >> s1 >> s2;
	cout << i1 << endl;
	cout << s1 << endl;
  	cout << s2 << endl;
    return 0;
}

这里的i1为123，s1为“a”，s2为“,[456,[789]]]”；
in能顺利将流输入到int类型的i1中，当遇到a时，流输入暂停；开始下一次的输入；

低层的I/O操作

首先引入几个流状态：

iostate value(member constant)	indicates	functions to check state flags
		good()	eof()	fail()	bad()	rdstate()
goodbit	No errors (zero value iostate)	`true`	`false`	`false`	`false`	goodbit
eofbit	End-of-File reached on input operation	`false`	`true`	`false`	`false`	eofbit
failbit	Logical error on i/o operation	`false`	`false`	`true`	`false`	failbit
badbit	Read/writing error on i/o operation	`false`	`false`	`true`	`true`	badbit

引用自：http://www.cplusplus.com/reference/ios/ios/clear/

因此我们来修改一下啊刚才的例子：

#include <sstream>
#include <string>
#include <iostream>
using std::string;
using std::istringstream;
using std::cout;
using std::endl;
int main()
{
	string a("a123a ,[456,[789]]]");
	istringstream in(a);
	int i1, i2;
	string s1;
	in >> i1;
	//in.clear();
	in >> s1;
	cout << i1 << endl;
	cout << s1 << endl;
    return 0;
}

此时，s1，i1都没有被输入正确的数，由于刚开始遇到了'a'，流状态被设置为failbit；此时认为流已经结束；

为了避免这种情况发生，我们使用类内成员clear()去清除该状态，使得流状态处于goodbit；把注释去掉，我们可以看到s1变成了"a123a"，也就是从错误的地方开始；

此时已经解决了部分的问题，但是如果我们仍然想要123这个整数怎么办？

此时引入了另外几个成员函数：get()，peek()；

is.get()：将is的下一个字节作为int返回，并从流中删除它；
is.peek()：将is的下一个字节作为int返回，但不从流中删除它；

观察刚刚的例子，我们可以看到第一个字符为'a'，因此我们可以从流中获取并删除它：

#include <string>
#include <iostream>
using std::string;
using std::istringstream;
using std::cout;
using std::endl;
int main()
{
	string a("a123a ,[456,[789]]]");
	istringstream in(a);
	int i1, i2;
	string s1;
	in >> i1;
	in.clear();
	auto m = in.get();
	cout << m << endl;
	in >> i1;
	in >> s1;
	cout << i1 << endl;
	cout << s1 << endl;
    return 0;
}

此时，输出m为字符'a'的ascii码97，i1为123，s1为“a";