在 C++ 中对 Structure 使用 lower_bound 或 upper_bound

Question

#include <iostream>
#include <algorithm>

using namespace std;

struct arr
{
int a;int b;
}a[1000];

bool comp(arr &lhs, arr &rhs)
{  return lhs.a < rhs.a ; }

int main() 
{
    int n,i  ;

    sort(a,a+n,comp);

    int ind= lower_bound(a,a+n,x,comp)-a;

    return 0;
}

Error message :

/usr/include/c++/4.9/bits/predefined_ops.h: In instantiation of 'bool __gnu_cxx::__ops::_Iter_comp_val<_Compare>::operator()(_Iterator, _Value&) [with _Iterator = arr*; _Value = const int; _Compare = bool ( )(arr&, arr&)]': /usr/include/c++/4.9/bits/stl_algobase.h:965:30:
required from '_ForwardIterator std::__lower_bound(_ForwardIterator, _ForwardIterator, const _Tp&, _Compare) [with _ForwardIterator = arr ; _Tp = int; _Compare = __gnu_cxx::__ops::_Iter_comp_val]' /usr/include/c++/4.9/bits/stl_algo.h:2036:46: required from '_FIter std::lower_bound(_FIter, _FIter, const _Tp&, _Compare) [with _FIter = arr*; _Tp = int; _Compare = bool ( )(arr&, arr&)]' prog.cpp:28:38: required from here /usr/include/c++/4.9/bits/predefined_ops.h:141:37: error: invalid initialization of reference of type 'arr&' from expression of type 'const int' { return bool(_M_comp( __it, __val)); } ^

I wish to use lower_bound over the struct to search for the value x that equals to a[i].a ? I have constructed the comparator function accordingly but getting a long error message of which I am not able to make anything.

What changes are required for the function to run.

Answer 1

lower_bound returns an iterator, not an index of the element found. You'll need to use std::distance to retrieve the index, but usually, the iterator is what you need/want for further processing.

Also note an index is usually returned as an std::size_t, not an int as you seem to assume.

Answer 2

As pointed in previous answers, lower_bound returns an iterator.
The type of x in lower_bound is not clear, x should also be the same type as that of the type in the container and compare function.
In this case, the type of x should be arr .