Pre and Post Increment in C/C++

Question

Here is my code

#include<stdio.h>
void main()
{
int a=6;
printf ("%d %d %d %d",a,a++,a++,++a);
}

I believe pre-increment added one to the existing the value and use the new values where as post-increment would increase the existing value by 1 but still use the old value.

By using this logic output should be 6 6 7 9 but I am getting 9 8 7 9.

Can someone please explain?

Comments

I am also interested to know the explanation as mine is also coming 9879 for the above program.

Answer 1

Pre-increment operator:
(1) Increments the value of the variable by 1
(2) Returns the variable itself.
Post-increment operator:
(1) Saves the current value in a temp variable.
(2) Increments the value of the variable by 1
(3) Returns the value saved in the temp variable

printf ("%d %d %d %d",a,a++,a++,++a);
In the above printf() statement, execution happens from right to left.
"++a" is executed first, so value of "a" becomes 7 and returns "a"
"a++" is executed next, so value of "a" becomes 8 and returns "7"
"a++" is executed next, so value of "a" becomes 9 and returns "8"
latest value of "a" is 9
so after the above steps printf becomes printf("%d %d %d %d", a, 8, 7, a);
that's why output becomes 9 8 7 9

Comments

Very nice explanation and deserves a recommendation. Was not knowing the concept of return and temp variable.

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I have to admit it beside the of concept of sequence point and its undefined behaviour, your explanation is working exactly the say. I also don't know the concept of temp variable in case of post increment.

I still have doubt on temp variable as we follow the -Wall option says the undefined behaviour and sequence point.
It will be very helpful if have any source, link, textbook, which says the same.

any ways kudos (y).

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Hi Pradeep,
I understand your explanation but I have query, When we look into the above code in perspective of sequence point concept post increment has to happen after sequence point i.e when it encounters ; or && or || or ',' operator. But here there is no such operator in printf statement but still it gets incremented. Could you please tell me the reason?

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Plz refer to the below link, which is having similar explanation about the pre/post-increment operators:
http://alexkalicki.com/blog/programming/pre-vs-post-incrementation/

Answer 2

It is due the concept of sequence point, here is the wiki link on sequence point
Sequence point

There is no defined order of evaluation for the arguments of a function call, and the results might vary for each compiler.

If you compile your code using

 gcc program_name -Wall     // "-Wall" option to generate warning  

you will get warning telling that undefined behaviour on a like this :

 warning: operation on ‘a’ may be undefined [-Wsequence-point]
     printf ("%d %d %d %d\n",a , a++, a++, ++a);
                                       ^