Take item position in vector with structure - c++

Question

Let see code:
struct tmp {
int a;
std::vector <second_struct> b;
}
struct second_struct {
int x;
int y;
}
//main.cpp
int main {
std::vector<tmp> test;
(...); //push data to test
}
So when i push data to test, in second function i want to get vector "b" from this vector 'test'. And find vector b by a;
(i.e I have int a and std::vector<tmp> test; , dont have std::vector <second_struct> b;(from vector test) and want to get it. (vector test is a big array, so i need to do it fastest and using little power )
How to do that? (i suppose std::map will be better? But if you tell me yes, tell me too how to do that in std::Vector)

Answer 1

Loop through test vector checking if tmp::a member is equal to your int a. If so, you have your vector<second_struct> b
for (int i=0;i<test.size();i++) {
if (test[i].a == a) {
// do whatever you need to do with test[i].b
break;
}
}
P.S. map would be easier, just
std::map<int, std::vector<second_struct>> map;
//insert data
std::vector<second_struct> b = map[a]; //assuming it's a that you're looking for

Answer 2

The most straightforward approach is to use map (or unordered_map in C++11). Hope this full example helps:
#include <map>
#include <vector>
#include <iostream>
struct str {
str(int _x, int _y) : x(_x), y(_y) {}
int x, y;
};
std::map<int, std::vector<str> > test;
int main() {
std::vector<str> vec;
for (int i = 0; i < 100; ++i) {
vec.clear();
vec.push_back(str(i, 2 * i));
vec.push_back(str(i + 1, i + 2));
test[i] = vec;
}
std::vector<str> result;
// get some element
result = test[10];
std::cout << "Found at position 10:\n";
for (int i = 0; i < result.size(); ++i)
std::cout << result[i].x << ' ' << result[i].y << '\n';
return 0;
}