1. Given: int a = 2; unsigned int b = -1; When a is compared with b, since unsig

Question

1. Given: int a = 2; unsigned int b = -1;
When a is compared with b, since unsigned is considered the higher type, C promotes a to an unsigned and does the comparison (a > b) as unsigned. Which best describes the result.

A. The result is false because 2 < unsigned interpretation of -1.

B. The result is true because 2 > -1.

C. The result is true because 2 > 1.

D. Signed positive values are always < an unsigned interpretation of a negative value.

E. A and D above are correct.

2. Given: int a = -7;

unsigned int b = a; What can we say about the bit sequences for a and b?

A. The highest order bit in a will be 1.

B. The highest order bit in b will be 1.

C. Both A and B are correct.

D. Most of the bits in bothaandbwill be 1.

E. ll of the above are correct.

Please briefly explain why, that would be helpful.

Explanation / Answer

1. Ans: E. A and D above are correct.

unsigned interpretation of -1 => 1111 1111 1111 1111 (considering 8 bit length)

clearly 2 < 1

2.

Ans C. Both A and B are correct.

highest order bit in a negative number in always 1 and this only signifies that number is negative.

But in unsigned number if highest order bit is 1 then it is the part of value of number

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