How is the evaluation of integer incrementation in chain of std::cout done?

Asked Jul 06 '16 at 12:32

Active Jul 06 '16 at 12:32

Viewed 45 times

Consider

auto a = 0;
std::cout << a;
std::cout << a++;
std::cout << ++a;
std::cout << std::endl;

Output: 002

Which is expected regarding the post and pre incrementation rule.

However the following code

auto a = 0;
std::cout << a << a++ << ++a << std::endl;

Output: 212

Since the following seems to print number in the right order

auto a = 0;
auto b = 1;
auto c = 2;
std::cout << a << b << c << std::endl;

Output: 012

I am trying to figure out why I got 212 in the second case?

asked Jul 06 '16 at 12:32

scoulomb

1

the second example is undefined behaviour. it's illegal to use the same self-modifying variable twice in the same statement – Richard Hodges Jul 06 '16 at 12:35
is it still UB in '17? – xaxxon Jul 06 '16 at 12:35
@xaxxon do you have a particular proposal in mind that would change it? – eerorika Jul 06 '16 at 12:42
@user2079303 there *is* a proposal that defines order of evaluation rules. But I don't know if they affect this case. – Quentin Jul 06 '16 at 12:44

0 Answers0