// lower_bound/upper_bound example
#include <iostream> // std::cout
#include <algorithm> // std::lower_bound, std::upper_bound, std::sort
#include <vector> // std::vector
int main () {
int myints[] = {10,20,30,30,20,10,10,20};
std::vector<int> v(myints,myints+8); // 10 20 30 30 20 10 10 20
std::sort (v.begin(), v.end()); // 10 10 10 20 20 20 30 30
std::vector<int>::iterator low,up;
low=std::lower_bound (v.begin(), v.end(), 20); // ^
up= std::upper_bound (v.begin(), v.end(), 20); // ^
std::cout << "lower_bound at position " << (low- v.begin()) << '\n';
std::cout << "upper_bound at position " << (up - v.begin()) << '\n';
return 0;
}
In this code can somebody please explain why we need to do (low - v.begin()) and (up - v.begin()) on the third and fourth lines from the bottom.
If I do
cout << low << endl;
I get the follwoing error which I dont understand
cannot bind ‘std::ostream {aka std::basic_ostream}’ lvalue to ‘std::basic_ostream&&’
ssavi :
*low *is an iterator as you declared. It holds the memory address that generates by your PC and you don't need any memory address to return . By writing \n\nlow- v.begin()\n\n\nYou make a command to your program to return the actual position of your searching query as an answer. \n\nThat's why it returns a value the \n\nAddress Position - Beginning of the Vector Position \n\n\nSuppose your vector starting memory address is FFF1 and your searching value is at FFF8 ... Then it return FFF8 - FFF1 = 7 .. ( Example is just to illustrate )\n\nThat's how I understand .",
2017-01-28T03:04:20