If the following program fragment ( assume negative numbers are stored in 2's complement form )
unsigned i=1;
int j = -4;
printf( " %u ", i + j);
prints x, then printf( " %d ", 8* sizeof( int ));
outputs an integer that is same as (log in the answers are to the base two)
| A. | an unpredictable value |
| B. | 8 * log( x + 3 ) |
| C. | log( x+ 3 ) |
| D. | none of above |
Answer : C Explanation : Let size of ( int ) = 1. So, -4 will be stored as 1 1 1 1 1 1 0 0. Since we are adding unsigned and signed integers, the signed gets converted to unsigned. So. i + j will become 1 1 1 1 1 1 0 1. We are trying to print this as an unsigned integer. So, what is printed will be 28 - 1 - 2. So, log (x + 3) = 8 (i.e.. 8 * sizeof ( int)}. |
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Option: A Explanation : Explanation will come here. Explanation will come here. Explanation will come here. Explanation will come here. Explanation will come here. |
