Why the output is 2 and 0

Question

Int x=0;
Cout<<++x<<" "<<x++<<endl;

Answer 1

I think the output should be undefined behaviour, because you are changing the same value and using it again between the same two sequence points.

If I run it here on Sololearn, I get 1 1 as output. In what situation did you get 2 and 0?

Answer 2

HonFu
You are correct. I didn't put every results of the tests 😁

* In Code Playground:
C++ code
- printf("%d %d\n", ++x, x++); // 2 0
- cout << ++x << " " << x++ << endl; // 1 1

C code
- printf("%d %d\n", ++x, x++); // 2 0

* In C4Droid (C code)
- printf("%d %d\n", ++x, x++); // 1 1

I hope this gives the OP an idea what a mess this sequence point thing was all about.

Answer 3

That is not a definite answer محمود شاكر . Output of such a code where a variable is accessed and modified in a single point of execution sequence is unpredictable. Output in different compiler will most likely be different also.

I tested that line in C4Droid and get 1 1 for output. In Code Playground the result is 2 0, but you will see the sequence point warning.

This is something unexplainable, because the result depends on compiler implementation.

Answer 4

Funny. I got 1 1 right here. Maybe not even the same every time in one environment?