Description
Solution: The translated code into C is: f = 2*（&A）
Problem 2 (2.12) Score:
Solution: 1. 0x50000000
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over ow, not the desired answer.
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0xb0000000
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It is the desired answer.
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0xd0000000
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over ow, not the desired answer.
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Problem 3 (2.14) Score: |
0000 0010 0001 0000 1000 0000 0010 0000 |
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Solution: It is in r-type. |
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000000 |
10000 |
10000 |
10000 |
00000 |
100000 |
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opcode-add |
rs |
rt |
rd |
shamt |
funct |
$s0, |
$s0 |
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So the assembly language instruction i, add $s0, |
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Problem 4 (2.16) Score:
Solution: From funct=34 and check the table we can know it is a sub operation. So the instruction in assembly language is:
sub $v1, $v1, $v0 (r-type) And it represent in binary form is: 0000 0000 0110 0010 0001 1000 0010 0010(2)
Problem 5 (2.19) Score:
Solution: 1. The value stored in $t0 is shifted four places to the left and bitwise or with the value stored in $t1.
The answer is: 0xbabefef8.
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The value stored in $t0 is shifted four places to the left and bitwise joined with 1111 1111 1111 1111 1111 1111 1111 1111 The answer is: 0xaaaaaaa0.
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The value stored in $t0 is shifted three places to the right and bitwise joined with 0000 0000 0000 0000 1111 1111 1110 1111 The answer is: 0x00005545.
Problem 6 (2.23) Score:
Solution: The rst instruction set $t2 to 1 because $t0 > 0 is true. Then go into the ‘ELSE’ branch and set $t2 to $t2 plus 2. So after that $t2 equal to 3. 
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