当我用g ++编译器编译c ++时，错误是什么意思？ [英] what does the error mean when I am compiling c++ with g++ complier?

问题描述

使用以下代码：

 ＃include< iostream> 
＃include< vector> 
 
 using namespace std; 
 
 int main（）
 {
 vector< int> ivec; 
 for（vector< int> :: size_type ix = 0; ix！= 10; ix ++）
 {
 ivec.push_back（ix）; 
} 
 vector< int> :: iterator mid =（ivec.begin（）+ ivec.end（））/ 2; 
 cout<< * mid<< endl 
 return 0; 
} 
  

使用g ++编译时出现错误：

  iterator_io.cpp：在函数`int main（）'中：
 iterator_io.cpp：13：错误：'operator +'in' & ivec）> std :: vector< _Tp，_Alloc> :: begin [with _Tp = int，_Alloc = std :: allocator< int>]（）+（& ; _Tp，_Alloc> :: end [with _Tp = int，_Alloc = std :: allocator< int>]（）'
 /usr/lib/gcc/x86_64-redhat-linux/3.4.5/ /../../../include/c++/3.4.5/bits/stl_iterator.h:654：注意：候选者是：__gnu_cxx :: __ normal_iterator <_Iterator，_Container> __gnu_cxx :: __ normal_iterator< _Iterator，_Container> :: operator +（const typename std :: iterator_traits< _Iterator> :: difference_type&）const [with _Iterator = int *，_Container = std :: vector< int，std :: allocator< int> ; >] 
 /usr/lib/gcc/x86_64-redhat-linux/3.4.5/../../../../include/c++/3.4.5/bits/stl_bvector.h： 261：note：std :: _ Bit_iterator std :: operator +（ptrdiff_t，const std :: _ Bit_iterator&）
 /usr/lib/gcc/x86_64-redhat-linux/3.4.5/../../。 ./../include/c++/3.4.5/bits/stl_bvector.h:345：note：std :: _ Bit_const_iterator std :: operator +（ptrdiff_t，const std :: _ Bit_const_iterator&）
 / usr / lib / gcc / x86_64-redhat-linux / 3.4.5 /../../../../ include / c ++ / 3.4.5 / bits / stl_iterator.h：765：note：__gnu_cxx :: __ normal_iterator< _Iterator，_Container> ; __gnu_cxx :: operator +（typename __gnu_cxx :: __ normal_iterator< _Iterator，_Container> :: difference_type，const __gnu_cxx :: __ normal_iterator< _Iterator，_Container&）[with _Iterator = int *，_Container = std :: vector& ：allocator< int> >] 
  

我知道ivec.end（）不能用作法线向量元素。但是我不能理解错误信息是什么意思...关于operator +的一些事情。

解决方案

迭代器。

没有为两个迭代器定义运算符+ ，因为该操作不会感。迭代器是对指针的一种泛化 - 它们指向存储在容器中的特定元素。但是，当使用向量时，您可以向迭代器添加整数，例如：

  vec.begin（）+ vec.size（）/ 2 
  pre> 
 
 
这就是为什么你的候选人是：（...）在你的错误信息，一些定义 operator + 。
 
 
 
在你的情况下，最好的，最简洁的方法不会使用迭代器，简单从指定位置获取值：
  int mid = vec [vec.size 2]。 
  
 
using the following code:
#include<iostream>
#include<vector>

using namespace std;

int main()
{
    vector<int> ivec;
    for(vector<int>::size_type ix = 0; ix != 10; ix++)
    {
        ivec.push_back(ix);
    }
    vector<int>::iterator mid = (ivec.begin() + ivec.end()) / 2;
    cout << *mid << endl;
    return 0;
}
I get an error compiling with g++:
iterator_io.cpp: In function `int main()':
iterator_io.cpp:13: error: no match for 'operator+' in '(&ivec)->std::vector<_Tp,               _Alloc>::begin [with _Tp = int, _Alloc = std::allocator<int>]() + (&ivec)->std::vector<_Tp, _Alloc>::end [with _Tp = int, _Alloc = std::allocator<int>]()'
/usr/lib/gcc/x86_64-redhat-linux/3.4.5/../../../../include/c++/3.4.5/bits/stl_iterator.h:654: note: candidates are: __gnu_cxx::__normal_iterator<_Iterator, _Container> __gnu_cxx::__normal_iterator<_Iterator, _Container>::operator+(const typename std::iterator_traits<_Iterator>::difference_type&) const [with _Iterator = int*, _Container = std::vector<int, std::allocator<int> >]
/usr/lib/gcc/x86_64-redhat-linux/3.4.5/../../../../include/c++/3.4.5/bits/stl_bvector.h:261: note:                 std::_Bit_iterator std::operator+(ptrdiff_t, const std::_Bit_iterator&)
/usr/lib/gcc/x86_64-redhat-linux/3.4.5/../../../../include/c++/3.4.5/bits/stl_bvector.h:345: note:                 std::_Bit_const_iterator std::operator+(ptrdiff_t, const std::_Bit_const_iterator&)
/usr/lib/gcc/x86_64-redhat-linux/3.4.5/../../../../include/c++/3.4.5/bits/stl_iterator.h:765: note:                 __gnu_cxx::__normal_iterator<_Iterator, _Container> __gnu_cxx::operator+(typename __gnu_cxx::__normal_iterator<_Iterator, _Container>::difference_type, const __gnu_cxx::__normal_iterator<_Iterator, _Container>&) [with _Iterator = int*, _Container = std::vector<int, std::allocator<int> >]
I know the ivec.end() can not be used as a normal vector element. But I can't understand what the error information means...something about operator+?
解决方案
You cannot add two iterators together.

operator+ is not defined for two iterators, because that operation wouldn't make sense. Iterators are a kind of generalization over pointers - they point to the specific element stored in container. At which element the sum of iterators is pointing?

However, when you use a vector, you can add integers to iterators, like that:
vec.begin() + vec.size() / 2
and that is why you have candidates are: (...) in your error message, followed by some definitions of operator+.

In your case the best, and cleanest way will be not using the iterators, but simple getting the value from specified position:
int mid = vec[vec.size() / 2];


                        
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