Prob in opreator?

[AnagJohari]

void main()
{
int i=2;
printf("%d %d",++i,++i)
getch();
}

The ans of this program is 4 3
but in my opinion the ans should be 3 3
plz explain why ans 4 3 is comming........ ..

[hype261]

The reason why the answer is 4 3 is because C evaluates arguments from right to left. So...

printf("%d %d, ++i; ++i) <--- second ++i is evaluated first

Also you are using the preincrement operator(++i) which means you increment i before it is evaluated.

if you did

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

your answer would be 3 2

[donbock]

C does not evaluate arguments from right to left.
In fact, the order of evaluation is explicitly labeled implementation-defined. That means C is free to use whatever order it feels like. You might get one result now, but a different one tomorrow. You should consider the output of this program to be unpredictable.

[hype261]

Sorry for the miss information

Sorry for the miss information. Guess I missed that little tid bit of knowledge in portability.

[weaknessforcats]

It's nothing to do with portability. Instead, the rule is that since the order of evaluation is undefined then referring to an variable twice can produce indeterminant results of one or more of the references change the value of the variable.

This is OK:

if (++i)
etc...

but this is not:

if (++i  == --i)
etc...

[hype261]

It's nothing to do with portability. Instead, the rule is that since the order of evaluation is undefined then referring to an variable twice can produce indeterminant results of one or more of the references change the value of the variable.

This is OK:

if (++i)
etc...

but this is not:

if (++i  == --i)
etc...

So are you saying that the compilier itself will not evaluate the function the same way everytime?

[donbock]

So are you saying that the compilier itself will not evaluate the function the same way everytime?

The compiler is not required to evaluate the operators the same way every time, even if you don't change the expression. Apparently inconsequential changes to the expression or the command line switches might cause the optimizer to do the preincrements in a different order.

[hype261]

The compiler is not required to evaluate the operators the same way every time, even if you don't change the expression. Apparently inconsequential changes to the expression or the command line switches might cause the optimizer to do the preincrements in a different order.

Good to know, thank you.

[AnagJohari]

Good to know, thank you.

But in a book the ans of this expression is given 4 3 not give the ans that "it will varry from one compiller to othetr compiller....."

other thing may i know is that if i write
i=2;
i=++i
then the value of i changed or produce an error in a program

[donbock]

But in a book the ans of this expression is given 4 3 not give the ans that "it will varry from one compiller to othetr compiller....."

other thing may i know is that if i write
i=2;
i=++i
then the value of i changed or produce an error in a program

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

What book? Are you certain the example in the book was precisely this?

i = 2;
i = ++i;

Nothing mysterious here. The value of i should end up as 3. Are you saying the compiler gave you an error for these instructions?

[AnagJohari]

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

What book? Are you certain the example in the book was precisely this?

i = 2;
i = ++i;

Nothing mysterious here. The value of i should end up as 3. Are you saying the compiler gave you an error for these instructions?

i m actually finding the concept of this finding the output of an programe
there are four options ,,, one of the option is 4 3
& second one is ans vary from one compiller to other compiller..
in a book ans is given 4 3
but u suggest the ans vary from one compiller to other
so i m in coonfused....

[weaknessforcats]

All you need to remember is that changing a variable more than once in the same statement produces indeterminate results.

Indeterminate means you can't predict the result. Maybe it works, maybe it doesn't.

BAD:

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

GOOD:

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