Why "&" is needed during calling this list? - c++

Question

I was studying LISTs and playing with function, this program gives 10 number and each time if the entered number is the bigger than the maximum value in our list, that number will be added to our list and finally after 10 tries all members will show up. The program works fine, but what I don't understand is why do I need use "&" in the line 6: " void insertMax(list &lst, int n) {"??
#include <iostream>
#include <list>
using namespace std;
void insertMax(list<int> &lst, int n) {
lst.sort();
int max = lst.back();
if (n > max) {
lst.push_back(n);
}
}
void disply(list<int> lst) {
list<int>::iterator iter = lst.begin();
while (iter!=lst.end()){
cout << *iter << endl;
iter++;
}
}
int main()
{
list<int> numbers;
numbers.push_back(0);
int input=0;
for (int j = 1; j < 11; j++){
cout << "Enter a number: " << endl;
cin >> input;
insertMax(numbers, input);
}
cout << "Now that's all: " << endl;
disply(numbers);
return 0;
}
Thanks in advance.

Answer 1

So you pass a reference to the list rather than a copy of it.
Google "pass by reference" and "pass by value".
Pass by reference means you don't have to make a copy of the whole data structure you are passing (which could be slow - especially if you have a big list)
Having said that, you're question is not quite clear: "Why is & needed during calling this list?" - Line 6 is not a call, it's the declaration of the function signature. So it's saying "When you call me I expect you to pass a reference to a list of ints"

Answer 2

By putting the ampersand (&) in, you specify that the list be taken as a reference, instead of being copied into the function scope. By taking it as a reference you can manipulate the external object.
http://www.cprogramming.com/tutorial/references.html

Answer 3

If I have understood correctly the line 6 is the starting line of the function definition
void insertMax(list<int> &lst, int n) {
lst.sort();
int max = lst.back();
if (n > max) {
lst.push_back(n);
}
}
Symbol & in the first parameter declaration means that the parameter will be a refence to the original argument. So any changes of the list in the function impact on the original argument.
If to remove this symbol & as for example
void insertMax(list<int> lst, int n) {
//...
the it will mean that the function will deal with a copy of the original argument. In this case any changes of the parameter that is pf the copy of the argument will not impact the original argument.
So the new item will be added to a copy of the list but the list itself will not be changed. It is its copy that will be changed.

Answer 4

If you don't add the '&' (pass by reference), any changes you make to the List inside the InsertMax function will not impact the list in your main method.
This is the reason why you'll sometimes see C++ method signatures declared as
void DoSomething(const std::string &value)
{
/*Method Body*/
}
This is done so all that data in the value string is not copied to a new place in memory. If the DoSomething method needs to modify the value string, it will need to make a copy of it first inside of the function. The const modifier makes sure that the reference is read-only by the method.
For example:
std::string DoSomething(const std::string &value)
{
std:string result = value + "some other data";
return result;
}