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#include<stdio.h>
void main()
{
int i=1;
printf("%d%d%d",i++,++i,i);
}

Question Posted / medo

Answers were Sorted based on User's Feedback



#include<stdio.h> void main() { int i=1; printf("%d%d%d",i++,++i,i); }..

Answer / medo

221...
In printf() function compiler calculates the values from
right to left,but prints the values from left to right.

Is This Answer Correct ?    28 Yes 13 No

#include<stdio.h> void main() { int i=1; printf("%d%d%d",i++,++i,i); }..

Answer / abhishek singh

221 is correct answer

Is This Answer Correct ?    12 Yes 5 No

#include<stdio.h> void main() { int i=1; printf("%d%d%d",i++,++i,i); }..

Answer / samrat

The Ans is: 2,3,3

In printf() the evaluation starts from right and the
printing of the values start from left.

Coming from right, the initial value is 1, after that ++i
will increment the value of i to 2. Now i++ will not be
incremented now. It will be incremented after the first "i"
is printed.

So we print 2 first, then the value of i is incremented to 3
(by executing i++). So for the other two i's the value will
be 3. So the ans is 2,3,3

Is This Answer Correct ?    13 Yes 6 No

#include<stdio.h> void main() { int i=1; printf("%d%d%d",i++,++i,i); }..

Answer / venkat

Basically in printf the values are calcualted from right to left...and the output is displayed from left to right.
so the output will be
221
first one will be printed
then one will be incremented by one and made as two..since it is a pre increment and will be printed..
then the value now is 2..so it will be printed then it will be incremented { post increment }..


good question..
a typical example for working of Printf

Is This Answer Correct ?    1 Yes 0 No

#include<stdio.h> void main() { int i=1; printf("%d%d%d",i++,++i,i); }..

Answer / prem_mallappa

The right answer is : Unpredictable/implementation defined behaviour.
Why? : get a C faq's book or visit online at c-faq.org

Reason: variable 'i' is changed more thane once between 'sequence point', a sequence point is a semicolon in 'C'. in such cases the result is unknown or compiler dependent.

Is This Answer Correct ?    1 Yes 0 No

#include<stdio.h> void main() { int i=1; printf("%d%d%d",i++,++i,i); }..

Answer / ricky/sonu dobriyal

I AM RICKY DOBRIYAL
THIS ANSWER IS DEFINETLY CORRECT
221
BECAUSE COMPILER CALCULATE RIGHT TO LEFT

Is This Answer Correct ?    2 Yes 1 No

#include<stdio.h> void main() { int i=1; printf("%d%d%d",i++,++i,i); }..

Answer / medo

what is the the output....?
and why?

Is This Answer Correct ?    5 Yes 5 No

#include<stdio.h> void main() { int i=1; printf("%d%d%d",i++,++i,i); }..

Answer / mementomori76

answer is 221, but you shouldn't use void main
it's better to use int main()

Is This Answer Correct ?    0 Yes 0 No

#include<stdio.h> void main() { int i=1; printf("%d%d%d",i++,++i,i); }..

Answer / rakesh

121
the first value is post increment(++i) and so at the first compilation the value is not incremented.. the next value is pre increment(i++) so it is incremented at the first compile. the third is the same as the input.

Is This Answer Correct ?    0 Yes 0 No

#include<stdio.h> void main() { int i=1; printf("%d%d%d",i++,++i,i); }..

Answer / shahenshah07

its o/p:2 2 1
cause when parameter pass to any f'n its stored in a
stack(lifo).dats y its print i,++i;i++ respectively.

Is This Answer Correct ?    0 Yes 0 No

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