Exam 1 Answers: Section 59

 

  1. smallest number: 1.000 0000 0000 0000 0000 0001 * 2-127


  2. Underflow occurs when a number is too small in magnitude to fit into a floating point number. There are some numbers close to zero that cannot be represented in floating point because the exponent is too small. These are the numbers x where |x| <= (min significand) * 2min exponent. An attempt to represent a number in that range results in an underflow.


  3. smallest number: 000 0000 0000 00002 or 010
largest number:  111 1111 1111 11112 or 3276710

    1.      - 472.84210 = - 111011000.1101012
               = - 1.110110001101012 * 28
     IEEE format: 1 1000 0111 110 1100 0110 1010 0000 0000
     
    2.      0.0458510 = 0.000010112
               = 1.0112 * 2-5
     IEEE format: 0 0111 1010 011 0000 0000 0000 0000 0000
     

    1.      0 1000 1000 010 0110 1001 1100 0000 0000
	 = 1.01001101001112 * 29
               = 1010011010.01112
               = (2 + 8 + 16 + 128 + 512 + .25 + .125 + .0625)10
               = 666.437510
     
    2.      1 0111 1011 011 0000 0000 0000 0000 0000
	 = - 1.0112 *  2-4
               = - 0.00010112
               = - (.0625 + .015625 + .0078125)10
               = - .085937510
     

  6.     check bits:
      C1   0
      C2   1
      C4   0
      C8   0

    stored word: 1010 0100 0110

    1.      fetched word: 1101 0011 1011
     fetched check bits:
        C1 = 1
        C2 = 1
        C4 = 1
        C8 = 0
     recalculated check bits:
        C1 = 1
        C2 = 0
        C4 = 1
        C8 = 1
     syndrome:
        1010
     error in d6, position 1010

    2.      fetched word: 0110 1111 1110
     fetched check bits:
        C1 = 0
        C2 = 1
        C4 = 1
        C8 = 1
     recalculated check bits:
        C1 = 0
        C2 = 1
        C4 = 1
        C8 = 0
     syndrome:
        1000
     error in c8, position 1000
  8. Two errors


  9. $t1 = 4, $t2 = 1, $t7 = 100


  10.        sub.s    $f4, $f4, $f4            # $f4 = 0 to use in check for loop end
       sub.s    $f5, $f5, $f5            # $f5 = 0 to initialize sum
  loop:
       la       $a0, prompt              # prepare to print prompt
       li       $v0, 4
       syscall                           # print prompt
       li       $v0, 6                   # prepare to read float
       syscall                           # read float
       c.eq.s   $f0, $f4                 # if user entered 0
       bc1t     loopend                  #    exit loop
       add.s    $f5, $f5, $f0            # add num to sum
       b        loop                     # back to top of loop
  loopend:
       la       $a0, maxmsg              # prepare to print sum msg
       li       $v0, 4
       syscall                           # print sum  msg
       mov.s    $f12, $f2                # prepare to print sum
       li       $v0, 2
       syscall                           # print sum
       li       $v0, 10
       syscall                           # exit from program

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