代码如下。代码不能在在线编译器上编译,我不知道为什么。它简短而且不言自明,请查看下面的详细信息。
#include <iostream>
#include <cmath>
using namespace std;
int N;
int distance(int a, int b){
if(abs(a-b) > N/2){
return N - abs(a-b);
}
return abs(a-b);
}
bool test(int x, int y){
if(distance(x,y) <=2){
return true;
}
return false;
}
int main()
{
N = 2;
cout << "Hello World" << endl;
cout << test(3,4) << endl;
return 0;
}
以下错误消息:
In file included from /usr/include/c++/4.8.3/bits/stl_algobase.h:65:0,
from /usr/include/c++/4.8.3/bits/char_traits.h:39,
from /usr/include/c++/4.8.3/ios:40,
from /usr/include/c++/4.8.3/ostream:38,
from /usr/include/c++/4.8.3/iostream:39,
from main.cpp:1:
/usr/include/c++/4.8.3/bits/stl_iterator_base_types.h: In instantiation of 'struct std::iterator_
traits<int>':
/usr/include/c++/4.8.3/bits/stl_iterator_base_funcs.h:114:5: required by substitution of 'templ
ate<class _InputIterator> typename std::iterator_traits<_Iterator>::difference_type std::distance
(_InputIterator, _InputIterator) [with _InputIterator = int]'
main.cpp:15:20: required from here
/usr/include/c++/4.8.3/bits/stl_iterator_base_types.h:165:53: error: 'int' is not a class, struct
, or union type
typedef typename _Iterator::iterator_category iterator_category;
^
/usr/include/c++/4.8.3/bits/stl_iterator_base_types.h:166:53: error: 'int' is not a class, struct
, or union type
typedef typename _Iterator::value_type value_type;
^
/usr/include/c++/4.8.3/bits/stl_iterator_base_types.h:167:53: error: 'int' is not a class, struct
, or union type
typedef typename _Iterator::difference_type difference_type;
^
/usr/include/c++/4.8.3/bits/stl_iterator_base_types.h:168:53: error: 'int' is not a class, struct
, or union type
typedef typename _Iterator::pointer pointer;
^
/usr/include/c++/4.8.3/bits/stl_iterator_base_types.h:169:53: error: 'int' is not a class, struct
, or union type
typedef typename _Iterator::reference reference;
答案 0 :(得分:7)
using namespace std;
这是bad idea;它将在std
命名空间中声明的任何内容转储到全局命名空间中,它可能与您在全局命名空间中声明的任何内容冲突。
int distance(int a, int b)
这声明全局命名空间中的一个函数与std
命名空间中同名的函数模板冲突。
if(distance(x,y) <=2)
根据过载分辨率的神秘规则,std::distance
模板比您的函数更匹配。试图实例化它,它失败了,因为它只能为迭代器类型实例化,而不是int
。
最好的选择是删除流氓using-directive,并将std::
添加到您从标准库中使用的任何内容中。如果由于某种原因不想这样做,那么限定函数调用以指定在全局命名空间中声明的函数:
if(::distance(x,y) <=2)