pre increment or pre decrement

[ronitonline2006]

void main()
{
int i=4,j=7;
printf("%d",i++ +i*j-j*i);
getch();
}


ans--> -2
why?

[vermarajeev]

void main()
{
int i=4,j=7;
printf("%d",i++ +i*j-j*i);
getch();
}


ans--> -2
why?

On my system the answer is 4

[ronitonline2006]

main()
{
int i=4,j=7;
printf("%d",++i+i*j-j*i);
}

output--> -2
why ?

[georges the man]

i compiled the code and the output was 5???
what are you expecting?

[82rob]

Hi

The output is 5.

The process goes like this:
++i+i*j-j*i
++4+4*7-7*4
the process os divided now in three parts
++4 + 4*7 - 7*4
5 + 28 - 28

So answer is 5...how come it is -2 ? Check it out again please...

[geek491]

Hi

The output is 5.

The process goes like this:
++i+i*j-j*i
++4+4*7-7*4
the process os divided now in three parts
++4 + 4*7 - 7*4
5 + 28 - 28

So answer is 5...how come it is -2 ? Check it out again please...

Hi all,

The output is -3 :) well so many ppl are coming up with so many ans so execution of code is must before someone tries to ans it... :)

based on precedence ++ then * and then + -

so now the equation will be
i++ first.
Now in equation replace i++ as 4

4 +i*j-j*i

but since it is a post increment i value will be refleceted only after an operation.

so that operation is i*j = 4 * 7 =28

then the value of i is visible as 5

so the next operation is j * i = 7 * 5 = 35
here i is 5 since a single operation 4*7 occured before it.

now the whole equation becomes.

4 + 28 - 35 = -3

Hope all are happy with ans.... if some one is not satisfied then they can execute the code in a turbo c complier and verify the result.

let me know on how it helped to [e-mail]
thanks..bye

[82rob]

Hii Geek,

If I am not wrong you are speaking about post increment. But it looks like it is pre-increment. I think the process goes like this:-

++i+i*j-j*i
First pre-increment --->>>
++4+i*j-j*i
New value of i will be 5
5 + i*j-j*i
now,
5 + 5*7 - 7*5

So answer is again 5...If I am wrong then I would surely like to hear the correct answer from your side.

Hi all,

The output is -3 :) well so many ppl are coming up with so many ans so execution of code is must before someone tries to ans it... :)

based on precedence ++ then * and then + -

so now the equation will be
i++ first.
Now in equation replace i++ as 4

4 +i*j-j*i

but since it is a post increment i value will be refleceted only after an operation.

so that operation is i*j = 4 * 7 =28

then the value of i is visible as 5

so the next operation is j * i = 7 * 5 = 35
here i is 5 since a single operation 4*7 occured before it.

now the whole equation becomes.

4 + 28 - 35 = -3

Hope all are happy with ans.... if some one is not satisfied then they can execute the code in a turbo c complier and verify the result.

let me know on how it helped to [e-mail]
thanks..bye

[Banfa]

main()
{
int i=4,j=7;
printf("%d",++i+i*j-j*i);
}

output--> -2
why ?

Actually it is undefined behaviour which is why you are all getting different results. You have tried to alter a variable and access it more than once between 2 sequence points.

The result of undefined behaviour is that all bets are off the compiler is free to do what ever it likes.

[zahidkhan]

On my system the answer is 4

modifying same variable in a single expression is generally undefined
It vary from compiler to compiler

[Banfa]

ronitonline2006 please don't double post, you have created much confusion by posting the same problem twice.