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HW3-Grading Policy - 3.1 0000 0000 0000 0000 0001 0000 0000...

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3.1 0000 0000 0000 0000 0001 0000 0000 0000two 3.2 1111 1111 1111 1111 1111 1000 0000 0001two 3.9 The problem is that A_lower will be sign-extended and then added to $t0. The solution is to adjust A_upper by adding 1 to it if the most significant bit of A_lower is a 1. As an example, consider 6-bit two’s complement and the address 23 = 010111. If we split it up, we notice that A_lower is 111 and will be signextended to 111111 = –1 during the arithmetic calculation. A_upper_adjusted = 011000 = 24 (we added 1 to 010 and the lower bits are all 0s). The calculation is then 24 + –1 = 23. 3.30 a. –1 391 460 350 b. 2 903 506 946 c. –8.18545 * 10 –12 d.sw $s0, 2($t0) 3.42 a. Convert +1 . 1011 * 2 14 + –1 . 11 * 2 –2 1.1011 0000 0000 0000 0000 000 –0.0000 0000 0000 0001 1100 000 ----------------------------------------------------------- 1.1010 1111 1111 1110 0100 000 0100 0110 1101 0111 1111 1111 0010 0000 b. Calculate new exponent: + 111 11 1 + 100 0110 1 +011 1110 1 --------------------------- 1000 0101 0 –011 1111 1 minus bias 1111 1111 --------------------------------------------- 100 0101 1 new exponent Multiply significands: . 1.101 1000 0000 0000 0000 0000 * L L L L g v ( g v ( 1 1.110 0000 0000 0000 0000 0000 ----------------------------------------------------------------------------------- 1 11 11 1 1011 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 11 0110 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 + 1.10 1100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000
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----------------------------------------------------------------------------------- 10.11 1101 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 Normalize and round: exponent 100 0110 0 significand 1.011 1010 0000 0000 0000 0000
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