以下代码有什么问题:
#include <ctime>
#include <vector>
#include <utility>
#include <algorithm>
#include <iostream>
int main()
{
std::vector< std::pair< char, unsigned > > vec;
for( unsigned i = 0; i < 100; ++i )
{
char ch = 0;
unsigned number = 0;
do {
ch = i;
number = i;
} while( std::find( vec.begin(), vec.end(), std::make_pair< char, unsigned >( ch, number ) ) != vec.end() );
std::cout << ch << number << '\n';
vec.push_back( std::make_pair< char, unsigned >( ch, number ) );
}
}
它可以很好地编译:
g++ test.cxx
但失败了:
$ g++ -std=c++11 test.cxx /tmp
test.cxx: In function 'int main()':
test.cxx:21:98: error: no matching function for call to 'make_pair(char&, unsigned int&)'
test.cxx:21:98: note: candidate is:
In file included from /usr/include/c++/4.7/bits/stl_algobase.h:65:0,
from /usr/include/c++/4.7/vector:61,
from test.cxx:3:
/usr/include/c++/4.7/bits/stl_pair.h:268:5: note: template<class _T1, class _T2> constexpr std::pair<typename std::__decay_and_strip<_Tp>::__type, typename std::__decay_and_strip<_T2>::__type> std::make_pair(_T1&&, _T2&&)
/usr/include/c++/4.7/bits/stl_pair.h:268:5: note: template argument deduction/substitution failed:
test.cxx:21:98: note: cannot convert 'ch' (type 'char') to type 'char&&'
test.cxx:25:69: error: no matching function for call to 'make_pair(char&, unsigned int&)'
test.cxx:25:69: note: candidate is:
In file included from /usr/include/c++/4.7/bits/stl_algobase.h:65:0,
from /usr/include/c++/4.7/vector:61,
from test.cxx:3:
/usr/include/c++/4.7/bits/stl_pair.h:268:5: note: template<class _T1, class _T2> constexpr std::pair<typename std::__decay_and_strip<_Tp>::__type, typename std::__decay_and_strip<_T2>::__type> std::make_pair(_T1&&, _T2&&)
/usr/include/c++/4.7/bits/stl_pair.h:268:5: note: template argument deduction/substitution failed:
test.cxx:25:69: note: cannot convert 'ch' (type 'char') to type 'char&&'
答案 0 :(得分:14)
<强> SOLUTION:强>
而不是以这种方式明确指定make_pair<>()
的模板参数:
std::make_pair< char, unsigned >( ch, number )
让他们推断出来:
std::make_pair( ch, number )
<强>说明强>
本指南背后的基本原理可以通过定义std::make_pair<>()
的方式找到,以及模板参数推导适用于通用引用的方式。从C ++ 11标准的第20.3.3 / 8-9段开始:
template <class T1, class T2>
pair<V1, V2> make_pair(T1&& x, T2&& y);
返回:
pair<V1, V2>(std::forward<T1>(x), std::forward<T2>(y));
其中V1和V2的确定如下:对于每个Ui
,decay<Ti>::type
为Ti
。如果Vi
等于X&
,则每个Ui
为reference_wrapper<X>
,否则Vi
为Ui
。 [示例:代替:
return pair<int, double>(5, 3.1415926); // explicit types
C ++程序可能包含:
return make_pair(5, 3.1415926); // types are deduced
- 结束示例]
此处,T1
和T2
意味着要推断。通过自己明确指定模板参数,您可以使用make_pair<>()
的类型推导机制来生成正确的返回类型,从而强制实例化一个接受char
的右值引用的函数和rvalue对unsigned
的引用:
... make_pair(char&&, unsigned&&)
但是,您没有在输入中提供rvalues,因为ch
和number
是左值。这就是编译器所抱怨的。
<强> ALTERNATIVE:强>
另请注意,您可以隐式构建 std::pair
个对象,这样就可以免于调用make_pair<>()
:
vec.push_back( { ch, number } );