I am practicing things described Here to learn the concept of lvalue and rvalue.
However, when I contrive my own example as follows, I find it compiles and runs without any error (using VS2017). I have learnt that in case 1, it is the same as calling
produceA(10).operator=(A())
and thus is legal. But, I still don't understand why case 2 and 3 are allowed. In fact, case 2 even contradicts the example given in the article. Am I actually getting the addresses of rvalues in these 2 cases? Do they lead to undefined behavior?
#include <string>
class A
{
int data = 1;
public:
A() = default;
A(int in)
: data(in)
{}
A& operator=(const A& rhs)
{
data = rhs.data;
return *this;
}
};
A produceA(int i)
{
A a(i);
return a;
}
int main()
{
// case 1
produceA(10) = A();
// case 2
A* pa = &produceA(10); // rvalue?
// case 3
std::string* pstr = &std::string("Temp"); // rvalue?
return 0;
}
