Post fix increment

[ssoffel]

Can somebody help explain the internals of what's happening here.

int num = 5;
int num1 = -num++;
cout << num << " " << num1;

The above I understand, -num makes a copy which gets assigned to num1, then num gets incremented. The output is -5 6.

The below code doesn't compile and I don't understand why

int num3 = 5;
int num4 = (-num)++;

compiler: error lvalue required as increment operand. However, the above code does work with an overloaded operator ++ for my user defined types?

[keskiverto]

int num = 5;
int num1 = -num++;
cout << num << " " << num1;

Is approximately:

int num = 5;
const int tmp = num;
int num1 = -tmp;
++num;
cout << num << " " << num1;

int num = 7;
int num1 = (-num)++;
cout << num << " " << num1;

Is approximately:

int num = 7;
const int tmp = -num;
int num1 = tmp;
++tmp; // error
cout << num << " " << num1;

The (-num) is a temporary value. Effectively -7. A constant. There is no increment for integer constants.

Your custom types:

Foo num = 7;
Foo num1 = (-num)++;

What type does Foo::operator- () return?

[ssoffel]

Foo::operator-()

returns a negative copy of type foo

[ssoffel]

keskiverto thanks for the reply,

So the brackets around (-num) pretty much say increment this value, which is a copy. Unlike in the first example where the copy gets returned then the original value gets incremented?

[keskiverto]

Yes.

See: http://en.cppreference.com/w/cpp/language/operator_incdec
and this for canonical implementations: http://en.cppreference.com/w/cpp/language/operators