is make_pair on the stack or heap?

Question

Do I need to allocate a pair if I insert it into a map from a different scope?

#include <iostream>
#include <string>
#include <unordered_map>
#include <utility>

using namespace std;
void parseInput(int argc, char *argv[], unordered_map<string, string>* inputs);

int main(int argc, char *argv[])
{
    unordered_map<string, string>* inputs = new unordered_map<string, string>;
    parseInput(argc, argv, inputs);
    for(auto& it : *inputs){
        cout << "Key: " << it.first << " Value: " << it.second << endl;
    }
    return 0;
}

void parseInput(int argc, char *argv[], unordered_map<string, string>* inputs)
{
    int i;
    for(i=1; i<argc; i++){
        char *arg = argv[i];
        string param = string(arg);
        long pos = param.find_first_of("=");
        if(pos != string::npos){
            string key = param.substr(0, pos);
            string value = param.substr(pos+1, param.length()-pos);
            inputs->insert( make_pair(key, value) );//what happens when this goes out of scope
        }
    }
    for(auto& it : *inputs){
        cout << "Key: " << it.first << " Value: " << it.second << endl;
    }
}

Answer 1

make_pair(key, value) returns a temporary object. The lifetime of that object ends at the end of the full-expression in which it is created (at the semicolon, basically).

The function insert creates a new object from that pair, which it puts into the map. The map stores this copy until the map is destroyed or the element is removed from the map.

Answer 2

Its fine:

inputs->insert( make_pair(key, value) );//what happens when this goes out of scope

std::make_pair returns the result by value.

The above has the same affect as:

inputs->insert( std::pair<std::string, std::string>(key, value) );

In both cases the value passed to insert() is copied(or moved) into the map.