如何在C ++中将向量迭代器转换为int [英] How to convert vector iterator to int in C++

问题描述

我正在寻找C ++向量中的元素，当我找到它时，我想以数字形式(整数，浮点数)获取元素的索引.

I am looking for an element in a C++ vector, and when I find it, I want to get found element's index in a numerical form(integer, float).

我的天真尝试是这样

int x;
int index;
vector<int> myvector;
vector<int>::iterator it;
it = find(myvector.begin(), myvector.end(), x);  
index = (int) * it;

此代码给出了错误.您能否告诉我如何将迭代器转换为int(如果可能)，或者可以告诉我如何以其他方式获取元素的索引?谢谢.

This code is giving error. Can you tell me how I can convert iterator to int(if possible), or can you tell me how I can get found element's index in other way? Thanks.

推荐答案

您需要使用标准函数std::distance

index = std::distance( myvector.begin(), it );

if ( index < myvector.size() )
{
    // do something with the vector element with that index
}

即使随机访问迭代器，也请尝试始终使用std::distance.在新的和旧的C ++标准中都可以使用此功能.

Try always to use std::distance even with random access iterators. This function is available in the new and old C++ Standards.

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