please give me answer with details #include<stdio.h> main() { int

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Question

please give me answer with details
#include<stdio.h>
main()
{
int i=1;
i=(++i)*(++i)*(++i);
printf("%d",i);
getch();
}

Question Submitted By :: Sv.mallesh

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Answer Posted By

Re: please give me answer with details #include<stdio.h> main() { int i=1; i=(++i)*(++i)*(++i); printf("%d",i); getch(); }

Answer
# 1

Answer is :64

Is This Answer Correct ?

3 Yes

3 No

Gita

Re: please give me answer with details #include<stdio.h> main() { int i=1; i=(++i)*(++i)*(++i); printf("%d",i); getch(); }

Answer
# 2

++i * ++i * **i
->
2       3     4
now started this way
<-
4   *   4  *   4
=64

Is This Answer Correct ?

1 Yes

3 No

Vaseem

Re: please give me answer with details #include<stdio.h> main() { int i=1; i=(++i)*(++i)*(++i); printf("%d",i); getch(); }

Answer
# 3

The precedence of the operations, should be (reading from
left to right in the equation)

++i   <first ++i i=2>
++i   <second ++i i=3>
*     <first product yields 3*3=9>
++i   <third ++i i=4>
*     <giving the second product 3*4=36>

Thus, the first product (*) is computed before the third ++i
is computed.  Once the first product is completed, i is
incremented to i=4 and the second product can occur now.


Now, if you add some parentheses to the expression giving

++i * (++i * ++i)

then you will get 64, as the other replies suggest.  Tracing
through the order of operations in this one

++i <first ++i i=2> 
++i <second ++I i=3>
++i <third ++I i=4>
*   <the product in the parentheses now yields 4*4=16>
*   <the first * yields 4*16=64>

Here, the first product (*) cannot occur until it knows the
result of the product in the parenthesis.  Thus, all three
increments must occur before the multiplications take place.

Is This Answer Correct ?

4 Yes

0 No

Joe

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