1 / 7

MIPS Floating Point Instructions Exploration

Understand floating point instructions in MIPS architecture from Chapter 3. Learn about floating point numbers, conversions, coprocessor usage, and guard and round bits. Includes interactive examples.

zuwena
Télécharger la présentation

MIPS Floating Point Instructions Exploration

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CS/COE0447Computer Organization & Assembly Language Chapter 3 Part 3

  2. Floating Point Instructions in MIPS .data nums: .float 0.75,15.25,7.625 .text la $t0,nums lwc1 $f0,0($t0) lwc1 $f1,4($t0) add.s $f2,$f0,$f1 #0.75 + 15.25 = 16.0 = 10000 binary = 1.0 * 2^4 #f2: 0 10000011 000000... = 0x41800000 swc1 $f2,12($t0) #1001000c now contains that number # Click on coproc1 in Mars to see the $f registers

  3. Another Example .data nums: .float 0.75,15.25,7.625 .text loop: la $t0,nums lwc1 $f0,0($t0) lwc1 $f1,4($t0) c.eq.s $f0,$f1 # cond = 0 bc1t label # no branch c.lt.s $f0,$f1 # cond = 1 bc1t label # does branch add.s $f3,$f0,$f1 label: add.s $f2,$f0,$f1 c.eq.s $f2,$f0 bc1f loop # branch (infinite loop) #bottom of the coproc1 display shows condition bits

  4. nums: .double 0.75,15.25,7.625,0.75 #0.75 = .11-bin. exponent is -1 (1022 biased). significand is 1000... #0 01111111110 1000... = 0x3fe8000000000000 la $t0,nums lwc1 $f0,0($t0) lwc1 $f1,4($t0) lwc1 $f2,8($t0) lwc1 $f3,12($t0) add.d $f4,$f0,$f2 #{$f5,$f4} = {$f1,$f0} + {$f2,$f1}; 0.75 + 15.25 = 16 = 1.0-bin * 2^4 #0 10000000011 0000... = 0x4030000000000000 # value+0 value+4 value+8 value+c # 0x00000000 0x3fe80000 0x00000000 0x402e8000 # float double # $f0 0x00000000 0x3fe8000000000000 # $f1 0x3fe80000 # $f2 0x00000000 0x402e800000000000 # $f3 0x402e8000 # $f4 0x00000000 0x4030000000000000 # $f5 0x40300000

  5. Guard and Round bits • To round accurately, hardware needs extra bits • IEEE 274 keeps extra bits on the right during intermediate additions • guard and round bits

  6. Example (in decimal)With Guard and Round bits • 2.56 * 10^0 + 2.34 * 10^2 • Assume 3 significant digits • 0.0256 * 10^2 + 2.34 * 10^2 • 2.3656 [guard=5; round=6] • Round step 1: 2.366 • Round step 2: 2.37

  7. Example (in decimal)Without Guard and Round bits • 2.56 * 10^0 + 2.34 * 10^2 • 0.0256 * 10^2 + 2.34 * 10^2 • But with 3 sig digits and no extra bits: • 0.02 + 2.34 = 2.36 • So, we are off by 1 in the last digit

More Related