main()
{
signed int bit=512, i=5;
for(;i;i--)
{
printf("%d\n", bit >> (i - (i -1)));
}
}
a. 512, 256, 0, 0, 0
b. 256, 256, 0, 0, 0
c. 512, 512, 512, 512, 512
d. 256, 256, 256, 256, 256
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Answer / shanker
Query : When bits cant be changed, then option c is also
correct?
Is This Answer Correct ? | 3 Yes | 0 No |
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