Pre vs. Post-Increment

[haris_iftikhar]

I have a question related to Pre vs. Post-Increment Operations in Programming.
Having read the C++ for Programmers Article related to Operators,
I learned that Pre and Post-Increment operators behave differently in nature.

The code snippet below does not follow the concepts discussed in the Pre vs. Post-Increment section in the Operators Article (C++ for Programmers):
Operators Article (C++ for Programmers)

Code Snippets:

#include <iostream> int main() { // This code returns 4 as answer, where it should be 3 (according to the article) int x= 1; x = x++ + ++x ; std::cout<< x << std::endl; return 0; }

#include <iostream> int main() { //This code returns 5 as answer, where it should be 3 (according to the article) int x= 1; x = ++x + x++; std::cout<< x << std::endl; return 0; }

Please explain why do I get these outputs, and if there’s any misunderstanding from me with learning the concepts, then also let me know.
Thanks.

[tgrtim]

If you double check the examples they’re using different variables in the examples. As written this is something you’d never do so it’s not too much to worry about.

Edit: Of course I find it ten seconds later, see the examples of undefined behaviour matching yours at- https://en.cppreference.com/w/c/language/eval_order

[haris_iftikhar]

Well if you consider the results to be “undefined” then, why does it give me fixed values of 4 and 5 respectively?

[tgrtim]

You’ll want to have a wee look into what ‘undefined behaviour’ actually means, that page even hyperlinks the definition. If it’s not clear then there’s plenty of guidance online to be found with a quick search e.g. c++ - Undefined, unspecified and implementation-defined behavior - Stack Overflow

[haris_iftikhar]

Thanks, I got it now! Really appreciate it.