Currently when I have to use vector.push_back()
multiple times.
The code I'm currently using is
std::vector<int> TestVector;
TestVector.push_back(2);
TestVector.push_back(5);
TestVector.push_back(8);
TestVector.push_back(11);
TestVector.push_back(14);
Is there a way to only use vector.push_back()
once and just pass multiple values into the vector?
These days (c++17
) it's easy:
auto const pusher([](auto& v) noexcept
{
return [&](auto&& ...e)
{
(
(
v.push_back(std::forward<decltype(e)>(e))
),
...
);
};
}
);
pusher(TestVector)(2, 5, 8, 11, 14);
You can do it with initializer list:
std::vector<unsigned int> array;
// First argument is an iterator to the element BEFORE which you will insert:
// In this case, you will insert before the end() iterator, which means appending value
// at the end of the vector.
array.insert(array.end(), { 1, 2, 3, 4, 5, 6 });
You can also use vector::insert.
std::vector<int> v;
int a[5] = {2, 5, 8, 11, 14};
v.insert(v.end(), a, a+5);
Edit:
Of course, in real-world programming you should use:
v.insert(v.end(), a, a+(sizeof(a)/sizeof(a[0]))); // C++03
v.insert(v.end(), std::begin(a), std::end(a)); // C++11
Yes you can, in your case:
vector<int>TestVector;`
for(int i=0;i<5;i++)
{
TestVector.push_back(2+3*i);
}
using vector::insert (const_iterator position, initializer_list il); http://www.cplusplus.com/reference/vector/vector/insert/
#include <iostream>
#include <vector>
int main() {
std::vector<int> vec;
vec.insert(vec.end(),{1,2,3,4});
return 0;
}
These are the three most straight forward methods:
1) Initialize from an initializer list:
std::vector<int> TestVector = {2,5,8,11,14};
2) Assign from an initializer list:
std::vector<int> TestVector;
TestVector.assign( {2,5,8,11,14} ); // overwrites TestVector
3) Insert an initializer list at a given point:
std::vector<int> TestVector;
...
TestVector.insert(end(TestVector), {2,5,8,11,14} ); // preserves previous elements
You can also use Boost.Assignment:
const list<int> primes = list_of(2)(3)(5)(7)(11);
vector<int> v;
v += 1,2,3,4,5,6,7,8,9;
Source: Stackoverflow.com