Question

I use the stream operator << and the bit shifting operator << in one line.
I am a bit confused, why does code A) not produce the same output than code B)?

A)

int i = 4;  
std::cout << i << " " << (i << 1) << std::endl;   //4 8

B)

myint m = 4;
std::cout << m << " " << (m << 1) << std::endl;   //8 8

class myint:

class myint {
    int i;
public:
    myint(int ii) {
        i = ii;
    }
    inline myint operator <<(int n){
        i = i << n;
        return *this;
    }
    inline operator int(){
        return i;
    }
};

thanks in advance
Oops

Answer 1

Your second example is undefined behavior.

You have defined the << operator on your myint class as if it were actually <<=. When you execute i << 1, the value in i is not modified, but when you execute m << 1, the value in m is modified.

In C++, it is undefined behavior to both read and write (or write more than once) to a variable without an intervening sequence point, which function calls and operators are not, with respect to their arguments. It is nondeterministic whether the code

std::cout << m << " " << (m << 1) << std::endl;  

will output the first m before or after m is updated by m << 1. In fact, your code may do something totally bizarre, or crash. Undefined behavior can lead to literally anything, so avoid it.

One of the proper ways to define the << operator for myint is:

myint operator<< (int n) const
{
   return myint(this->i << n);
}

(the this-> is not strictly necessary, just my style when I overload operators)