Default values for a user-defined types

[tomy12]

Hi all,

Please consider this snipped-code:

int main()
{
   cout << int() << endl;
   return 0;
}

This will usually result in showing 0 as the output.

Consider that we have a user-defined type like this:

class myvals
{
public:
   myval() { };
   int x = 5;
   double y = 12.5;
};

Now consider a function that receives an int and a double as its arguments:

void foo(int i, double d) {
 cout << "i = " << i <<" and d = " << d << endl;
}

Now how to use the function foo this way:

foo(myvals());

?

[laserlight]

Perhaps you just need to do some function overloading:

void foo(int i, double d) {
    cout << "i = " << i <<" and d = " << d << endl;
}

void foo(const myvals& vals) {
    foo(vals.x, vals.y);
}

Now this will work as expected:

foo(myvals());

[AvDav]

Another way is to instantiate the class's object and pass it to the function parameter.

#include <iostream>
#include <string>

class myvals
{
public:
   myvals() : x(5), y(12.4) {}
   int x;
   double y;
};
void foo(int i, double d) {
 std::cout << "i = " << i <<" and d = " << d << std::endl;
}

int main()
{
  const myvals& ref = myvals();
  foo(ref.x, ref.y);
}