pre & post increment need help

This is a discussion on pre & post increment need help within the C Programming forums, part of the General Programming Boards category; Hi all, I have below simple code. I tried a lot to analyze but didn't got satisfactory answer. Code: #include ...

  1. #1
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    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

  2. #2
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    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

  3. #3
    spurious conceit MK27's Avatar
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    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
    C programming resources:
    GNU C Function and Macro Index -- glibc reference manual
    The C Book -- nice online learner guide
    Current ISO draft standard
    CCAN -- new CPAN like open source library repository
    3 (different) GNU debugger tutorials: #1 -- #2 -- #3
    cpwiki -- our wiki on sourceforge

  4. #4
    ... kermit's Avatar
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    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
    Last edited by kermit; 10-09-2009 at 08:17 AM.

  5. #5
    spurious conceit MK27's Avatar
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    Quote Originally Posted by kermit View Post
    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.
    C programming resources:
    GNU C Function and Macro Index -- glibc reference manual
    The C Book -- nice online learner guide
    Current ISO draft standard
    CCAN -- new CPAN like open source library repository
    3 (different) GNU debugger tutorials: #1 -- #2 -- #3
    cpwiki -- our wiki on sourceforge

  6. #6
    The larch
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    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?
    Last edited by anon; 10-09-2009 at 08:58 AM.
    I might be wrong.

    Thank you, anon. You sure know how to recognize different types of trees from quite a long way away.
    Quoted more than 1000 times (I hope).

  7. #7
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    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!

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