Call to a constructor while executing (obj < 3) . Kindly explain?

[User 3971400]

class CA{
private:
int m_Var;
public:
CA(int x=0): m_Var(x){cout<<"1";}
bool operator < (const CA &obj2) { cout<<"2"; return (m_Var < obj2.m_Var) ? true : false;}
};
int main(void) {
CA obj1=2;
if(obj1 < 3){cout << "In If block" << endl;}
}

the output of this code is 112In If block .. i can understand the reason behind the first 1 and "2In If block" .. i dont understand when the second 1 is being called? as per my understanding when (obj1<3) executes it may call obj1.operator<(3) But constructor is being called but whats makes it to call? kindly explain why the second 1 is being displayed?

[Naveen]

Member 3974347 wrote:

as per my understanding when (obj1<3) executes it may call obj1.operator<(3)

Not actually. It is actually called like obj1.operator<( CA(3));// A local temporary object is created here. Here the integer 3 is automatically converted to a CA object by involking the constructor. Such conversions are called implicit conversions. You can prevent such conversions using the explicit keyword[^]

nave [OpenedFileFinder] [My Blog]

[N a v a n e e t h]

Member 3974347 wrote:

But constructor is being called but whats makes it to call?

Because constructor accepts a int.

Member 3974347 wrote:

as per my understanding when (obj1<3) executes it may call obj1.operator<(3)

Your operator overload accepts const CA & but you specified the expression obj1<3, since type CA has a constructor which takes int, a temporary will be created by passing 3 to the constructor of CA and this is the reason for the second time constructor call. Use explicit keyword to avoid the confusion. :)

Navaneeth How to use google | Ask smart questions

[N a v a n e e t h]

You are so fast! :-D

Navaneeth How to use google | Ask smart questions

[Naveen]

Well, I borrowed some monkeys[^] from Rajesh. :-D

nave [OpenedFileFinder] [My Blog]

[User 3971400]

Thanks. :)

[User 3971400]

Thanks :)