Postfix and prefix increment [duplicate] - c++

Question

This question already has answers here:
Why are these constructs using pre and post-increment undefined behavior?
(14 answers)
Undefined behavior and sequence points
(5 answers)
Closed 9 years ago.
Why the output of next code is 2 1 2?
#include "iostream"
int main(int argc, const char *argv[])
{
int i = 0;
std::cout << i << std::endl << i++ << std::endl << ++i << std::endl;
return 0;
}
Because first i is equal 2 but not zero, it means that the whole like of cout is evaluated first
and then printed (not part by part). If so, then first value should be 1, but not 2, because i++ should increment i after printing. Could you clarify?
EDIT:
The output of next code is 2 2 0.
#include "iostream"
int main(int argc, const char *argv[])
{
int i = 0;
std::cout << i << std::endl << ++i << std::endl << i++ << std::endl;
return 0;
}
why?

Answer 1

There is no sense reasoning in the output of your code because as it stands your program exhibits Undefined Behavior.
Per paragraph 1.9/15 of the C++11 Standard:
The value computations of the operands of an operator are sequenced before the value computation of the result of the operator. If a side effect on a scalar object is unsequenced relative to either another side effect on the same scalar object or a value computation using the value of the same scalar object, the behavior is undefined.
Because there is no sequence point separating both mutations of i, Undefined Behavior ensues. Your compiler might not output anything, and the program might output differently on different compilers. But arguing about the output is unnecessary in this context.
If you separate the statements, the result will then come out as expected:
std::cout << i << std::endl; // 0
std::cout << i++ << std::endl; // 0
std::cout << ++i << std::endl; // 2

Answer 2

the evalution goes from right to left.
i = 0
++i -> i = 1
i++ -> i = 1, post incrementation, a copy occurs. then i = 2
i -> i = 2
As all this occurs before being send to cout, the value of i is 2, and the middle one have been copied and its value is 1.

Answer 3

Please tell me if I don't understand your question clearly:
cout << i++;
is the equivalent of
cout << i;
i+=1;
while cout << ++i
is the equivalent of
i += 1;
cout << i;
in otherwords any time you use i++, post-increment it returns the current value then changes while ++i means increment first then return the new value. It has nothing to do with cout