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#include<stdio.h>
int main()
{
int i=2;
int j=++i + ++i + i++;
printf("%d\n",i);
printf("%d\n",j);
}
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Answer Posted / sanjay
i = 5
j = 11
It is because during the first pre-increment "++i" the compiler gets the value from the memory, increments it and stores it in the memory ie now i = 3. During the second pre-increment "++i" the compiler again gets the value from the memory, increments it, (value in the memory was 3) and so the incremented value is stored again in memory ie i = 4. during the post increment, the value from the memory is received and used in the statement ie) (the whole final statement looks like this ->>( 3 + 4 + 4) ) and then value of i is incremented and stored in memory. thus finally the value of i is 5 and j is 11.
| Is This Answer Correct ? | 2 Yes | 0 No |
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