Post increment operator ++

Question

cout<<(x++)++; //fails 
cout<<++(++x); //passes

Why does the post increment fail ? I see it happen but not sure of the technical reason.

What's the type of `x`? That matters a lot. If it's a built-in, `operator++` is not a function call. — MSalters, Sep 13 '11 at 14:12

score 13 · Answer 1 · answered Sep 13 '11 at 06:37

13

x++ returns an rvalue so you can't perform ++ again on it. On the other hand, ++x returns an lvalue so you can perform ++ on it.

answered Sep 13 '11 at 06:37

Mu Qiao

1

also see [this question](http://stackoverflow.com/questions/371503/why-is-i-considered-an-l-value-but-i-is-not) – Benoit Sep 13 '11 at 06:39

Stephan · Answer 2 · 2011-09-13T06:45:50.813

This is how the increment operators work in C/C++.

If you put the ++ after the variable (postfix increment), the whole expression evaluates to the value of the variable before incrementing.

If you put the ++ before the variable (prefix increment), the expression evaluates to the value after the increment operation.

While the prefix operation returns a reference to the passed variable, the postfix version returns a temporary value, which must not be incremented.

score 1 · Answer 3 · answered Sep 13 '11 at 06:40

1

Exactly. yo cannot perform a ++ over an Rvalue. a good explanation about how rvalue works is given here.

answered Sep 13 '11 at 06:40

Stefano

3 Answers3