C++ 如何遍历字符串并知道索引（当前位置）？
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How can I iterate through a string and also know the index (current position)?
提问by pierrotlefou
Often when iterating through a string (or any enumerable object), we are not only interested in the current value, but also the position (index). To accomplish this by using string::iteratorwe have to maintain a separate index:
通常在遍历字符串（或任何可枚举对象）时，我们不仅对当前值感兴趣，还对位置（索引）感兴趣。为了通过使用实现这一点，string::iterator我们必须维护一个单独的索引：
string str ("Test string");
string::iterator it;
int index = 0;
for ( it = str.begin() ; it < str.end(); it++ ,index++)
{
cout << index << *it;
}
The style shown above does not seem superior to the 'c-style':
上面显示的样式似乎并不优于“c-style”：
string str ("Test string");
for ( int i = 0 ; i < str.length(); i++)
{
cout << i << str[i] ;
}
In Ruby, we can get both content and index in a elegant way:
在 Ruby 中，我们可以优雅地获取内容和索引：
"hello".split("").each_with_index {|c, i| puts "#{i} , #{c}" }
So, what is the best practice in C++ to iterate through an enumerable object and also keep track of the current index?
那么，在 C++ 中迭代可枚举对象并跟踪当前索引的最佳实践是什么？
采纳答案by TheUndeadFish
I've never heard of a best practice for this specific question. However, one best practice in general is to use the simplest solution that solves the problem. In this case the array-style access (or c-style if you want to call it that) is the simplest way to iterate while having the index value available. So I would certainly recommend that way.
我从来没有听说过这个特定问题的最佳实践。但是，通常的一种最佳做法是使用解决问题的最简单的解决方案。在这种情况下，数组样式访问（或 c 样式，如果您想调用它）是在索引值可用的情况下进行迭代的最简单方法。所以我肯定会推荐这种方式。
回答by Leandro T. C. Melo
Like this:
像这样：
std::string s("Test string");
std::string::iterator it = s.begin();
//Use the iterator...
++it;
//...
std::cout << "index is: " << std::distance(s.begin(), it) << std::endl;
回答by Andrew
You can use standard STL function distance as mentioned before
您可以使用前面提到的标准 STL 函数距离
index = std::distance(s.begin(), it);
Also, you can access string and some other containers with the c-like interface:
此外，您可以使用类似 c 的接口访问字符串和其他一些容器：
for (i=0;i<string1.length();i++) string1[i];
回答by Alex Reynolds
A good practice would be based on readability, e.g.:
一个好的做法是基于可读性，例如：
string str ("Test string");
for (int index = 0, auto it = str.begin(); it < str.end(); ++it)
cout << index++ << *it;
Or:
或者：
string str ("Test string");
for (int index = 0, auto it = str.begin(); it < str.end(); ++it, ++index)
cout << index << *it;
Or your original:
或者你的原件：
string str ("Test string");
int index = 0;
for (auto it = str.begin() ; it < str.end(); ++it, ++index)
cout << index << *it;
Etc. Whatever is easiest and cleanest to you.
等等。任何对你来说最简单和最干净的东西。
It's not clear there is any one best practice as you'll need a counter variable somewhere. The question seems to be whether where you define it and how it is incremented works well for you.
目前尚不清楚是否有任何最佳实践，因为您需要在某处使用计数器变量。问题似乎是你在哪里定义它以及它是如何增加的。
回答by maxim1000
I would use it-str.begin() In this particular case std::distance and operator- are the same. But if container will change to something without random access, std::distance will increment first argument until it reach second, giving thus linear time and operator- will not compile. Personally I prefer the second behaviour - it's better to be notified when you algorithm from O(n) became O(n^2)...
我会使用 it-str.begin() 在这种特殊情况下 std::distance 和 operator- 是相同的。但是如果容器将更改为没有随机访问的内容，则 std::distance 将增加第一个参数直到它到达第二个，从而给出线性时间并且 operator- 将不会编译。我个人更喜欢第二种行为 - 当你的算法从 O(n) 变成 O(n^2) 时，最好得到通知......
回答by user2015735
Since std::distanceis only constant time for random-access iterators, I would probably prefer explicit iterator arithmetic.
Also, since we're writing C++ code here, I do believe a more C++ idiomatic solution is preferable over a C-style approach.
由于std::distance对于随机访问迭代器来说只是常数时间，我可能更喜欢显式迭代器算法。此外，由于我们在这里编写 C++ 代码，我确实相信 C++ 惯用的解决方案比 C 风格的方法更可取。
string str{"Test string"};
auto begin = str.begin();
for (auto it = str.begin(), end = str.end(); it != end; ++it)
{
cout << it - begin << *it;
}
回答by jscharf
For strings, you can use string.c_str()which will return you a const char*, which can be treated as an array, example:
对于字符串，您可以使用string.c_str()which 将返回一个 const char*，它可以被视为一个数组，例如：
const char* strdata = str.c_str();
for (int i = 0; i < str.length(); ++i)
cout << i << strdata[i];
