please ,need a little help on incrementing operators.

[vimukthi]

Hello guys..i am beginner to c programming and i like to know what is happening in this code depth

#include<stdio.h>
main(){
int i=0;
printf("%d %d",++i,i++);
i=0;
printf("\n%d",++i);
i=0;printf("\n%d",i++);
return 0;
}
output is 2 0 1 0

How this happen ...explain to me

[Adak]

Try it this way:

#include<stdio.h>
int main(void){
   int i=0;
   printf("%d %d\n",i++,++i);
   i=0;
   printf("%d\n",++i);
   i=0;
   printf("%d\n",i++);
   return 0;
}

I get
1 1
1
0

Only the first printf() is in doubt. Because you have one variable being incremented twice with no sequence point between them, you can wind up with undefined behavior.

Bottom line, don't do that - it may work, but it may just as likely fail, on another C compiler or system. Also, there's no reason for such shenanigans.

[grumpy]

Only the first printf() is in doubt. Because you have one variable being incremented twice with no sequence point between them, you can wind up with undefined behavior.

The first printf() and anything after it is "in doubt".

Once undefined behaviour is introduced into a program, anything is allowed to happen. Including, in this case, causing a subsequent statement like "i=0;" do do something different than you expect. There is nothing, for example, suggesting that the first expression only affects the value of the variable i.

[Salem]

Same code (with unspecified behaviour, and undefined behaviour), many different compilers, and many different answers.
Post and Pre Increment

[iMalc]

The first printf() and anything after it is "in doubt".

Once undefined behaviour is introduced into a program, anything is allowed to happen. Including, in this case, causing a subsequent statement like "i=0;" do do something different than you expect. There is nothing, for example, suggesting that the first expression only affects the value of the variable i.

Sory, grumpy, this one will have to fall into your 3%.
Whilst something like

printf("%d\n", i++ + ++i);

is undefined behaviour,

printf("%d %d",++i,i++);

is not undefined behaviour. The order of evaluation of function arguments is "unspecified behaviour". There's an important difference.

That means this program is okay, though it may have one of two possible results, depending to the implementation of the compiler and the order the arguments on line 4 are evaluated.
the-difference-between-unspecified-and-undefined-behaviour

[Salem]

> is not undefined behaviour. The order of evaluation of function arguments is "unspecified behaviour". There's an important difference.
It's still undefined behaviour!

printf("%d %d", foo(), bar() );

It is unspecified behaviour as to the order in which foo and bar are called.
This is demonstrated in my test1() code by the differing order of calls to f1 .. f5
The computed answer is consistent (as it should be), but the intermediate steps can vary from one implementation to another.

printf("%d %d",++i,i++);

This is undefined behaviour, because the same object is modified more than once between sequence points.
This is demonstrated in my test2() code producing all sorts of weird and wonderful answers - none of which are meaningful.

[grumpy]

Sory, grumpy, this one will have to fall into your 3%.
Whilst something like

printf("%d\n", i++ + ++i);

is undefined behaviour,

printf("%d %d",++i,i++);

is not undefined behaviour. The order of evaluation of function arguments is "unspecified behaviour". There's an important difference.

This is not within my (coarsely estimated) 3%.

True, the order of evaluation of function arguments is unspecified.

The problem is that i is being modified twice in one statement. In older versions of the standards (the description has now changed) it would be said there is no intervening sequence point between the two increments of i. That is undefined behaviour.

The presence of undefined behaviour (where the standard imposes no bounds whatsoever on what is permitted) trumps the presence of unspecified behaviour (where the standard does bound the permitted results). When both are present - as in this case - the result is undefined, not just unspecified.