Using auto in cpp

With C++11 came a new keyword called auto. This has made writing and reading programs so much simpler. For example previously we used to do something like this:

int x = 50;
string s = "hello";

Now we can do something like this:

auto x = 50;
auto s = "hello";

Well you'll say that's basic stuff. What's the big deal? Well it's kind of a big deal. Let's say you are working with non primitive data types like std::vector. To get the iterator to the vector you needed to do something like:

std::vector<int>::iterator it = vector<int> my_vec {1,2,3};

Well that does not seem that bad. Let's see the next example. Here we are using std::map of vector<vector<int>> and std::vector<int>::iterator it.

// my_map is defined as : std::map<vector<vector<int>>, std::vector<int>::iterator> my_map
std::map<vector<vector<int>>, std::vector<int>::iterator>::iterator my_iter = my_map.begin();

Using auto it simplifies to:

auto my_iter = my_map.begin();

Using lambdas

Another great use of auto is with lambdas.

std::function<int(int, int)> sum = [](int a, int b){return a+b;}; // without auto 
auto sum = [](int a, int b){return a+b;}; // using auto

It is also great to use for automatically deducing argument types

auto sum = [](auto const a, auto const b){return a+b;}; // using auto

// another example
template<class A, class B>
auto product(A a, B b){
    return A.value * B. value;
}

Best use of auto

In my opinion the best use of auto is in initializing variables. It makes sure that every variable gets initialized.

auto x; // will throw an error

auto x = 1; // no error and a good practice too!

string s = "Hello"
int x = s.size(); // loss of data since unsigned int being converted to int
auto x = s.size(); // no loss of data

Few caveats

Be very careful while working with references. auto in general is not able to deduce referenced return type.

static int val = 5;
int & get_val(){
    return val;
}

int main(){
    auto y = get_val();
    val = 6;
    cout<<y<<endl; // gives 5
                   // new value of val not propogated to y
}

In the above program the data type of y is int and not &int as expected. The compiler is not able to deduce the return type as &int. This issue is easily solved by using &auto instead of auto.

static int val = 5;
int & get_val(){
    return val;
}

int main(){
    auto &y = get_val();
    val = 6;
    cout<<y<<endl; // gives 6
}

For a more expansive overview visit auto. Another great post by Herb Setter

Signing out

-G