pre & post increment need help

Hi all,

I have below simple code. I tried a lot to analyze but didn't got satisfactory answer.

Code:

---

#include <stdio.h>

int main(void)
{
        int i=10;
        i=((i++) + (++i));
        printf("i= %d\n", i);
        return 0;
}

---

the output is "i = 23"
How come i's value became 23?
appreciate the analysis....


thankx

The simple funda with pre / post inc / dec is that


in pre increment ++a ( the value of a is increment first and then assigned to a)
in post increment a++ ( the value of a is asigned and then increment)

means

Code:

---

#include <stdio.h>

int main() {
  int a = 10;
  int b = a++; // 10 will be assigned to b as first assignment and then increment
  printf("b == %d\n", b);
  int c = ++a; // first increment and then assignment
  printf("c == %d\n", c);

  return 0;
}

---

output 10
12

Because 23 is the number of the magic skidoo.

Code:

---

i=((i++) + (++i));

---

1. first, i is pre-incremented, so i=11
2. then i+i = 22
3. finally, i is post-incremented

Code:

---

        i=((i++) + (++i));

---

The result of the above is undefined behaviour. In other words, there is no guarantee as to what the output will be. Have a look at the following from the comp.lang.c FAQ:

3.12a
3.12b

Especially read the following:

3.8
3.9
3.10a

Quote:

---

Originally Posted by kermit

The result of the above is undefined behaviour. In other words, there is no guarantee as to what the output will be. Have a look at the following from the comp.lang.c FAQ:

---

That was interesting. I would never have guessed that:

Code:

---

a[i] = i++

---

Would be considered undefined. Altho...

Thanks kermi.

I wonder why this question comes up so often. It almost seems to be the first thing everyone decides to try out when first introduced to these operators. Could it be the discovery that something so simple allows one to write something so mind-blowing, since any way you look at it, it doesn't even have any obvious intuitive interpretation?

I guess it's compiled to execute left-to-right in your example.
11 + 12 = 23
Yet if I try printf("i= %d\n", (i++) + (++i));
the result is
11 + 11 = 22
Neato!