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main()
{
int i;
printf("%d", &i)+1;
scanf("%d", i)-1;
}
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Answer / abhishek marshetty
Explanation: printf( ) prints address/garbage of i, scanf() dont have & sign, so scans address for i +1, -1 dont have any effect on code.
| Is This Answer Correct ? | 11 Yes | 0 No |
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