C++ Error: Was not declared in the scope - c++

Question

Hey guys I'm working on a project and I was doing pretty well until I hit this wall..
I am getting two errors:
error: 'binarySearch' was not declared in this scope
error: 'addInOrder' was not declared in this scope
Here are my files, I've tried quite a few things with no avail. Help would be much appreciated.
histogram.cpp
#include "histogram.h"
#include "countedLocs.h"
//#include "vectorUtils.h"
#include <string>
#include <vector>
using namespace std;
void histogram (istream& input, ostream& output)
{
// Step 1 - set up the data
vector<CountedLocations> countedLocs;
// Step 2 - read and count the requested locators
string logEntry;
getline (input, logEntry);
while (input)
{
string request = extractTheRequest(logEntry);
if (isAGet(request))
{
string locator = extractLocator(request);
int position = binarySearch (countedLocs,
CountedLocations(locator, 0));
/** Hint - when looking CountedLocations up in any kind
of container, we really don't care if the counts match up
or not, just so long as the URLs are the same. ***/
if (position >= 0)
{
// We found this locator already in the array.
// Increment its count
++countedLocs[position].count;
}
else
{
// This is a new locator. Add it.
CountedLocations newLocation (locator, 1);
addInOrder (countedLocs, newLocation);
}
}
getline (input, logEntry);
}
// Step 3 - write the output report
for (int i = 0; i < countedLocs.size(); ++i)
output << countedLocs[i] << endl;
}
countedLocs.cpp
#include "countedLocs.h"
#include <iostream>
#include <vector>
using namespace std;
int CountedLocations::binarySearch(const vector<CountedLocations> list, CountedLocations searchItem)
{
//Code was here
}
int CountedLocations::addInOrder (std::vector<CountedLocations>& vectr, CountedLocations value)
{
//Code was here
}
countedLocs.h
#ifndef COUNTEDLOCATIONS
#define COUNTEDLOCATIONS
#include <iostream>
#include <string>
#include <vector>
struct CountedLocations
{
std::string url;
int count;
CountedLocations (){
url = "";
count = 0;
}
CountedLocations(std::string a, int b){
url = a;
count = b;
}
int addInOrder (std::vector<CountedLocations>& vectr, CountedLocations value);
int binarySearch (const std::vector<CountedLocations> list, CountedLocations searchItem);
};
inline
std::ostream& operator<< (std::ostream &out, CountedLocations& cL)
{
//out << "URL: " << cL.url << " count: " << cL.count << std::endl;
out << "\"" << cL.url << "\"," << cL.count;
return out;
}
#endif

Answer 1

The methods are member methods of CountedLocations... use something.extractLocator and something.binarySearch or make the histogram() to be also a member method of CountedLocations... (something is of type CountedLocations highly possibly will be countedLocs[position])

Answer 2

You have a free function histogram in which you are trying to use two member functions, addInOrder and binarySearch. In order to use them, you need to have an instance of CountedLocations.
If these are some kind of helper functions, which do not depend on the actual CountedLocations instance, I would turn them into static functions like this (you only need to change the header):
static int addInOrder (std::vector<CountedLocations>& vectr, CountedLocations value);
And then you can call this function by specifying the type of your class:
CountedLocations::addInOrder(...);

Answer 3

You are trying to call member methods of a struct without an object of that type. Strange.

Answer 4

You need to look at what a namespace is.
You declare a class CountedLocations, so far so good. But then you try to use the member functions outside the CountedLocations namespace which will obviously never work.
int position = binarySearch (countedLocs,
CountedLocations(locator, 0));
binarySearch is a member function of the CountedLocations namespace. If you want to call that function you have to create an object that contains a reference to that member function.
CountedLocation myObject;
int position = myObject.binarySearch (countedLocs, CountedLocations(locator, 0));
I dont know if that solves your problem, but you should know this before you even attempt to solve a problem.