vector> sort_a;
vector v2;
vector v3;
for (int i=0; i<4; ++i) {
v2.push_back(i);
for (int j=0; j<4; ++j) {
Don't think of it as a multidimentional vector, think of it as a vector of vectors.
int n = 4;
std::vector<std::vector<int>> vec(n, std::vector<int>(n));
// looping through outer vector vec
for (int i = 0; i < n; i++) {
// looping through inner vector vec[i]
for (int j = 0; j < n; j++) {
(vec[i])[j] = i*n + j;
}
}
I included parentheses in (vec[i])[j]
just for understanding.
Edit:
If you want to fill your vector via push_back
, you can create a temporary vector in the inner loop, fill it, and then push_back it to your vector:
for (int i = 0; i < n; i++) {
std::vector<int> temp_vec;
for (int j = 0; j < n; j++) {
temp_vec.push_back(j);
}
vec.push_back(temp_vec);
}
However, push_back
calls result in slower code, since not only you need to reallocate your vector all the time, but also you have to create a temporary and copy it.
a vector<vector<int>>
is not the best implementation for a multidimensional storage. The following implantation works for me.
template<typename T>
class array_2d {
std::size_t data;
std::size_t col_max;
std::size_t row_max;
std::vector<T> a;
public:
array_2d(std::size_t col, std::size_t row)
: data(col*row), col_max(col), row_max(row), a(data)
{}
T& operator()(std::size_t col, std::size_t row) {
assert(col_max > col && row_max > row)
return a[col_max*col + row];
}
};
use case:
array_2d<int> a(2,2);
a(0,0) = 1;
cout << a(0,0) << endl;
This solution is similar to the one described here.