Why does that code does not compile due to an error:
#include <iostream>
using namespace std;
int main()
{
int i = 0;
cout << ++(i++) << " " << i << endl;
return 0;
}
While that code does compile:
#include <iostream>
using namespace std;
int main()
{
int i = 0;
cout << (++i)++ << " " << i << endl;
return 0;
}
I do not understand that. From my point of view it would be pretty reasonable for the first chunk to compile. The expression ++(i++) would just mean take i, increment it and output, then increment it again.
I am not asking about an undefined behavior in int overflow. I do not know about r and l value at all at the time of writing the question and I do not care why is ++i considered an l-value, but i++ is not.
This is because the post increment and pre increment operators return values with different types. Result of post increment is a so-called 'rvalue', meaning it can not be modified. But pre-increment needs a modifiable value to increment it!
On the other hand, result of pre-increment is an lvalue, meaning that it can be safely modified by the post increment.
And the reason for above rules is the fact that post-increment needs to return a value of the object as it was before increment was applied. By the way, this is why in general case post-incrememts are considered to be more expensive than pre-increments when used on non-builtin objects.