data n:word 0x5350494d Pr word Ox20693265 qz word Ox61207573 r:word 0x0072206c s

Question

data n:word 0x5350494d Pr word Ox20693265 qz word Ox61207573 r:word 0x0072206c s:word 0x2073696d tword 0x756c6174 u:word 0x37412066 v; word 0x64636261 w;_word 0xc561726e x:word 0x9350494d yzword Ox4d495033 IZ? MOrd 0x20617373 MIPS Registers R8 (t0) R9 (tl): R10 (t2): R11 (t3) R12 (t4): R13 (t5) R14 (t6) R15 (t7): R16 (s0) R24 (t8) 0x00000000 R25 (t9): RO (r0): 0x00000000 0x00000000 0xff00ff00 R17 (s1): Oxff03f77 R18 (s2) R1 (at): 0x10010000 0x0000ffff R2 (v0): 0x00000000 0x00000000 0x00000023 0x00000000 R3 (v1): 0x0000000c 0x10010020 0x10010000 0x00000000 R4 (a0): 0x00000023 0x10010030 0x00400020 0x10008000 R5 (a1): 0x10010010 0x10010020 0x000000ff 0x7fffeff0 R6 (a2) 0x0000000a0x803000100x800c10010x00000000 R7 (a3): 0x00000050 0xffff0000 7a0x00000000 R19 (s3) R20 (s4) R21 (s5) R22 (s6) R23 (s7): R26 (k0): R27 (k1): R28 (gp) R29 (SP) R30 (Ep) R31 (ra) str-_ascüz "hello worldln" 0x00000010 0x00400060 Data [0x10000000 J0x1000fffc] 0x00000000 [0x10010000] 0x5350494d 0x206932650x612075730x0072206c [0x100100100x2073696d0x756c6174 0x374120660x64636261 [0x10010020]0xc561726e0x9350494d0x4d4950330x20617373 0x100100301 10x68656c6c 10x6120776f Ox726c640a0x00000000 [0x100100401.0x10020000] 0x00000000 Remember that user data declarations start at location 0x 1001 0000, and the data is inserted into memory exactly in the order that it is declared. Fon ry readouts use Big urposes of this test, assume that the memo convention Endian

Explanation / Answer

a)Values of the Given instruction

sra $t0,$t6,20    ----> What is the value in t0 register.

Ans) Hex = 0xfffff803

             Explanation:-

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ror $t1,$t0,6    ----> What is the value in t1 register.

Ans) Hex = 0x00000000

             Explanation:-

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b) Address and value of a label

        and $t0,$s1,$s5

        sw $t0,x

Explanation:-

*) X value is specified above the first table's data declaration section.

*) Address of the data section starts at 0x10010000, So all data declarations will be reside in that address.

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c) Find the value of $t0 and check branch taken or not

          lw $t0,12($a1)

   Ans) $t0 value=0x64636261

Explanation:-

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bltz $t0,loop

Ans)No

Explanation:-

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d) Byte value and output

     lb $a0,19($s3)

What byte is stored in $a0-->3 byte is stored in a0

Explanation:-

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li $v0,11----------->System call for byte display.

Ans)49

Explanation:-

       

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