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One is pass by value, another pass by reference. Pass by value will make a copy of data, possibly much slower and using more memory.

Beware of this so seemingly simple questions. Everybody already knows about "const" and passing by value vs. reference. Answering this I think will be considered newbie answers by a competent interviewer.

The real answer comes along the r-value vs. l-value differences in C++... and temporary objects. The first implementation is a pass-by-value, ok, it involves copies and all that, but it becomes an l-value local to the function, always.

The second one, however WITHOUT "const" is a pass-by-reference and it will be an l-value iff the original is also an l-value, but will be an r-value if a temporary object is involved (perhaps by default conversion into a vector<int>). The C++ standard forbids temporary objects to be l-values, so a compilation error will occur if there needs to be a default conversion.

Adding "const" allows again a conversion on the calling on the function since it will allow for r-values to be passed (remember the "const").

It's a tricky question more easily understood with an example for std::string:

// Pass by value
void doSomething1(std::string str)
{
}

// Pass by reference, non-const
void doSometing2(std::string &str)
{
}

// Pass by reference, const
void doSomething3(const std::string &str)
{
}

// Allowed, a temporary of type std::string is default constructed
// and passed to the function
doSomething1("Whatever");

// Forbidden, function expects a non-const reference and we
// must default construct a temporary, hence an r-value
doSomething2("Whatever");

// Allowed, a temporary of type std::string is default constructed
// and allowed to pass since the function expects a const reference
doSomething3("Whatever");

A constant vector object cannot be passed into the first function whereas that can be passed to second function.
const vector<int> arr;
test1(arr) // compilation error
test2(arr) // valid

#include <iostream>
#include <vector>

using namespace std;

void test1(vector<int> arr);
void test2(const vector<int> &arr);

int main()
{
   vector<int> arr;
   for(int i=0;i<10;i++) {
       arr.push_back(i);
   }
   
   test1(arr);
   test2(arr);
   
   return 0;
}

void test1(vector<int> arr) {
    for(int i=0;i<arr.size();i++) {
        cout<<arr.at(i)<<endl;
    }
}

void test2(const vector<int> &arr) {
    for(int i=0;i<arr.size();i++) {
        cout<<arr.at(i)<<endl;
    }
}

// When testing it, there is no difference in output.