本文介绍了如何使用std :: vector :: emplace_back for vector< vector< int> >?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
vector<vector<int> > res;
res.emplace_back({1,2}); // change to res.push_back({1,2}); would work
这给我错误
main.cpp:61:25: error: no matching function for call to ‘std::vector<std::vector<int> >::emplace_back(<brace-enclosed initializer list>)’
main.cpp:61:25: note: candidate is:
In file included from /usr/include/c++/4.7/vector:70:0,
from /usr/include/c++/4.7/bits/random.h:34,
from /usr/include/c++/4.7/random:50,
from /usr/include/c++/4.7/bits/stl_algo.h:67,
from /usr/include/c++/4.7/algorithm:63,
from miscalgoc.hpp:1,
from main.cpp:1:
/usr/include/c++/4.7/bits/vector.tcc:92:7: note: void std::vector<_Tp, _Alloc>::emplace_back(_Args&& ...) [with _Args = {}; _Tp = std::vector<int>; _Alloc = std::allocator<std::vector<int> >]
如何使这项工作?另外,为什么在这里需要分配器?
How to make this work? Also, why an allocator is needed here?
推荐答案
问题是函数模板参数不会导出 std :: initializer_list
从一个braced-init-list(如 {1,2}
)。
The problem is that function template arguments doesn't deduce std::initializer_list
from a braced-init-list (like { 1, 2 }
).
/ p>
Example:
#include <initializer_list>
#include <type_traits>
template<typename T>
void func(T arg) {
}
int main() {
auto init_list = {1, 2}; // This works because of a special rule
static_assert(std::is_same<decltype(init_list), std::initializer_list<int>>::value, "not same");
func(std::initializer_list<int>{1, 2}); // Ok. Has explicit type.
func({1, 2}); // This doesn't because there's no rule for function
// template argument to deduce std::initializer_list
// in this form.
}
是一个函数模板,它的参数被推导出来。因此,传递 {1,2}
将无法工作,因为它无法推导出来。为其添加显式类型
std::vector::emplace_back()
is a function template with its arguments being deduced. So passing it {1, 2}
will not work because it couldn't deduce it. Putting an explicit type to it
res.emplace_back(std::initializer_list<int>{1,2});
会使其正常工作。
a href =http://coliru.stacked-crooked.com/a/ef151ed35da1e4d0>实例
Live example
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