Purpose of using (int) in the code below?

Question

n=b.size()    
n = max(n,(int)a.size());

where a and b are some user-Input strings and n is an integer. would anybody tell me why we use (int)a.size() and what is the purpose of using (int).

if you take the cast out you probably get a compiler warning about signed/unsigned mismatch, casting the `size_t` to `int` would make that disappear. — nick, Mar 16 '20 at 05:26
`std::max` is a template. Take a look at the definition for it and you can probably figure it out. https://en.cppreference.com/w/cpp/algorithm/max — Retired Ninja, Mar 16 '20 at 05:28

score 2 · Answer 1 · answered Mar 16 '20 at 06:02

I am assuming that n is of type int, and your program will be something like this :

#include <iostream>
#include <string>

using namespace std;
int main ()
{
    string a ("Test string");
    string b ("Test two");

    int n = b.size()    ;
    n = max(n,(int)a.size());
    cout << "n : " << n ;

    return 0;
}

Now if you see the documentation for the .size() method of the string class, you will see it returns value of type : size_t

size_t is an unsigned integral type (the same as member type string::size_type) : as per documentation

Now when we look at the documentation for max() you can see it uses templating (you can read more about templating here) what it essentially means is that you can use any type as parameter (int, float, etc..) but both the parameters need to be the same type.

Now since n was declared as an int when calling max(n,x);, x needs to be type of n which basically means int in our case.

Now this is the reason for using (int) before a.size(). What we are doing here is type casting, since a.size() returns in type size_t which is different from int (You can read more about this here), we need to typecast the return value to int which can be done by (int)a.size().

SIDE NOTE

 int n = max(b.size(),a.size());
 cout << "n : " << n << " \n";

would also work, since both are same type so no need to do type casting.

Purpose of using (int) in the code below?

1 Answers1