std :: ostream {aka std :: basic_ostream< char>}值为'std :: basic_ostream< char>&& [英] std::ostream {aka std::basic_ostream<char>} Ivalue to 'std::basic_ostream<char>&&

问题描述

在此代码中，我尝试将迭代器移动10个元素。

In this code I try to move the iterator by 10 elements.

   #include <iostream>
    #include <string>
    #include <vector>
    int main()
    {
        using namespace std;
       vector<int> v(20);
       auto mid = v.begin() + 10;
        cout<<mid;


    }

运行此代码时，错误在标题中提到。
我是一个初学者。我在几乎每个程序我写这个错误。

On running this code, I get the error mentioned in the title. I'm a beginner. I experience this error in almost every program I write. Where am I going wrong?

推荐答案

迭代器指向一个元素，你想做什么是

An iterator "points" to an element, what you want to be doing is:

cout << *mid;

您必须解引用迭代器打印它指向的内容。尝试直接打印它会给你提到的错误。

You have to "dereference" the iterator to print what it points to. Trying to print it directly gives you the error you mentioned.

编辑：这里有一个演示：

here's a little demo:

#include <iostream>
#include <vector>

int main(int argc, char* argv[])
{
    std::vector<int> numbers;
    numbers.push_back(4);
    numbers.push_back(3);
    numbers.push_back(2);

    auto beg = numbers.begin();
    auto mid = numbers.begin() + 1;
    std::cout << *beg << std::endl;
    std::cout << (beg < mid) << std::endl;      // True because beg (index 0) points to an element earlier than mid (index 1)
    std::cout << (*beg < *mid) << std::endl;    // False because the element pointed-to by beg (4) is bigger than the one pointed-to by mid (3)

    return 0;
}

输出
第一行显示4是第一个元素的值！第二行显示1（所有非零值均为真），最后一行显示0（零是唯一表示false的值）。

Output The first line shows 4 which is the value of the first element! The second line shows 1 (all non-zero values mean true) and the last line shows 0 (zero is the only value that means false).

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