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CP I/O Q52

52If the following program fragment(assume negative numbers are stored in 2's complement form)
unsigned i=1;
int j=-4;
printf("%d",8*size of(int));
output is (log in answers are to the base two)
  • Option : C
  • Explanation :
    Let size of (int)=1. So, -4 will be stored as 11111100. Since we are adding unsigned and signed integers the signed gets converted to unsigned.
    So, i + j will become 11111101. 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,*size of (int)).
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