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#include<stdio.h>
main()
{
int i=1;
i=(++i)*(++i)*(++i);
printf("%d",i);
getch();
}
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Answer / joe
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 ? | 13 Yes | 1 No |
Answer / vaseem
++i * ++i * **i
->
2 3 4
now started this way
<-
4 * 4 * 4
=64
| Is This Answer Correct ? | 5 Yes | 5 No |
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