Increment does not work as expected

Question

I took this exercise and put it to compile, but it’s going wrong I think because the values I see would be 5 and 9, but it’s going 6 and 8, I believe the IDE is not reading correctly.

#include <stdio.h>

 int main (){
 int i = 5, j = 3, *p, *q;
 p = &i;
 q = &j;
 printf("%d %d \n", *p, (*p)++ + *q);
    return 0;
    }

Answer 1

First, you must read What is a programming language, IDE and compiler?.

If you think you should give another value you should justify it. Programming is justification, otherwise it turns to bullshit.

You took a undefined behavior. So don’t do it, there’s no need. You have a compiler who won’t even let you compile it if you don’t force it. Don’t use operators you own side effects, as is the case with the increment (++) along with other expressions, anything can happen without the code being wrong, and therefore you have no control over it. If you really want to make the increment then do it in a separate line and control the exact moment that should happen.

#include <stdio.h>

int main (){
    int i = 5, j = 3, *p = &i, *q = &j, x = (*p)++;
    printf("%d %d \n", *p, x + *q);
}

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

So you have control and give the result you got, that is, the compiler chose to make the increment first and then do the rest, which is what I would do since the increment operator takes precedence over the others, including the comma that has less precedence, then it runs before everything and then uses the result already obtained. Although the expected result could be achieved on some compiler or platform, it would make less sense.

In fact 5 and 9 I can’t even imagine why it would, you should really justify it.

And you can’t change anything in the code, think it looked like it, and then give it the expected result, another code is another code. A code that has no side effect will give different result, are completely different operations.

Answer 2

#include <stdio.h>

 int main (){
 int i = 5, j = 3, *p, *q;
 p = &i;
 q = &j;

 printf("%d %d \n", *p,(*p + 1 + *q ));
    return 0;
    }

I did it this way, and it came out what you wanted ... it’s kind of a weird way, but it was