C++ 将迭代器转换为 int
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Convert iterator to int
提问by thetux4
int i;
vector<string> names;
string s = "penny";
names.push_back(s);
i = find(names.begin(), names.end(), s);
cout << i;
I'm trying to find index of an element in vector. It's ok with iterators, but I want it as int. How can I do it?
我正在尝试查找向量中元素的索引。没关系iterators，但我想要它作为int. 我该怎么做？
回答by CB Bailey
You can use std::distancefor this.
您可以std::distance为此使用。
i = std::distance( names.begin(), std::find( names.begin(), names.end(), s ) );
You will probably want to check that your index isn't out of bounds, though.
不过，您可能想要检查您的索引是否超出范围。
if( i == names.size() )
// index out of bounds!
It's probably clearer to do this with the iterator before using std::distance, though.
不过，在使用 std::distance 之前使用迭代器执行此操作可能更清楚。
std::vector<std::string>::iterator it = std::find( names.begin(), names.end(), s );
if( it == names.end() )
// not found - abort!
// otherwise...
i = std::distance( names.begin(), it );
回答by Georg Fritzsche
std::vector<string>::iterator it = std::find(names.begin(), names.end(), s);
if (it != names.end()) {
std::cout << std::distance(names.begin(), it);
} else {
// ... not found
}
回答by iolo
try
尝试
i = (find( names.begin(), names.end(), s ) - names.begin());
Edit: Although you should consider using vector::size_type instead of an int.
编辑：尽管您应该考虑使用 vector::size_type 而不是 int。
回答by Johnsyweb
Assumptions that I am making about your code:
我对您的代码所做的假设：
using std::vector;
using std::cout;
using std::string;
If my assumptions are correct, then you can findthe distancebetween the beginning of the vectorand the iterator(essentially the index into the vectorat which you can find such element):
如果我的假设是正确的，那么你可以find在distance的开始之间vector和iterator（基本的索引，vector在这里您可以找到这样的元素）：
using std::distance;
Like so...
像这样...
vector<string>::iterator i = find(names.begin(), names.end(), s);
if (i != names.end())
{
cout << "Index " << std::distance(names.begin(), i);
}
else
{
cout << s << "not found";
}
回答by shaina
You can just dereference your iterator
您可以取消引用您的迭代器
int i = *name;
