Computer Systems: A Programmer's Perspective Plus Mastering Engineering With Pearson Etext -- Access Card Package (3rd Edition)
Computer Systems: A Programmer's Perspective Plus Mastering Engineering With Pearson Etext -- Access Card Package (3rd Edition)
3rd Edition
ISBN: 9780134123837
Author: Randal E. Bryant, David R. O'Hallaron
Publisher: PEARSON
Expert Solution & Answer
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Chapter 3, Problem 3.73HW

Explanation of Solution

Given assembly code:

x in %xmm0

find_range:

vxorps %xmm1, %xmm1, %xmm1

vucomiss %xmm1, %xmm0

ja .L5

vucomiss %xmm1, %xmm0

jp .L8

movl $1, %eax

je .L3

.L8:

vucomiss %xmm1, %xmm0

setbe %al

movzbl %al, %eax

addl $2, %eax

ret

.L5:

movl $0, %eax

.L3:

Rep;ret

Explanation:

  • The assembly function “find_range” illustrates conditional branching in floating point.
  • It compares a value “x” with 0.
  • If value is less than 0, then result is negative.
  • If value is greater than 0, then result is positive.
  • If value equals 0, then result is zero.
  • The instruction “vxorps %xmm1, %xmm1, %xmm1” sets value of “%xmm1” to 0.
  • The instruction “vucomiss %xmm1, %xmm0” compares values.
  • The instruction “jp .L8” jumps to label “.L8”.
  • The instruction “ja .L5” jumps to label “.L5” if first value is greater than second.
  • The instruction “movl $0, %eax” moves immediate value to register “%eax”.
  • The instruction “movl $1, %eax” moves immediate value to register “%eax”.

    Function find_range:

    #Define function

range_t find_range(float x) {

  __asm__(

      "vxorps %xmm1, %xmm1, %xmm1\n\t"

      "vucomiss %xmm1, %xmm0\n\t"

      "jp .P\n\t"

      "ja .A\n\t"

      "jb .B\n\t"

      "je .E\n\t"

      "...

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Chapter 3 Solutions

Computer Systems: A Programmer's Perspective Plus Mastering Engineering With Pearson Etext -- Access Card Package (3rd Edition)

Ch. 3.5 - Prob. 3.11PPCh. 3.5 - Prob. 3.12PPCh. 3.6 - Prob. 3.13PPCh. 3.6 - Prob. 3.14PPCh. 3.6 - Prob. 3.15PPCh. 3.6 - Prob. 3.16PPCh. 3.6 - Practice Problem 3.17 (solution page 331) An...Ch. 3.6 - Practice Problem 3.18 (solution page 332) Starting...Ch. 3.6 - Prob. 3.19PPCh. 3.6 - Prob. 3.20PPCh. 3.6 - Prob. 3.21PPCh. 3.6 - Prob. 3.22PPCh. 3.6 - Prob. 3.23PPCh. 3.6 - Practice Problem 3.24 (solution page 335) For C...Ch. 3.6 - Prob. 3.25PPCh. 3.6 - Prob. 3.26PPCh. 3.6 - Practice Problem 3.27 (solution page 336) Write...Ch. 3.6 - Prob. 3.28PPCh. 3.6 - Prob. 3.29PPCh. 3.6 - Practice Problem 3.30 (solution page 338) In the C...Ch. 3.6 - Prob. 3.31PPCh. 3.7 - Prob. 3.32PPCh. 3.7 - Prob. 3.33PPCh. 3.7 - Prob. 3.34PPCh. 3.7 - Prob. 3.35PPCh. 3.8 - Prob. 3.36PPCh. 3.8 - Prob. 3.37PPCh. 3.8 - Prob. 3.38PPCh. 3.8 - Prob. 3.39PPCh. 3.8 - Prob. 3.40PPCh. 3.9 - Prob. 3.41PPCh. 3.9 - Prob. 3.42PPCh. 3.9 - Practice Problem 3.43 (solution page 344) Suppose...Ch. 3.9 - Prob. 3.44PPCh. 3.9 - Prob. 3.45PPCh. 3.10 - Prob. 3.46PPCh. 3.10 - Prob. 3.47PPCh. 3.10 - Prob. 3.48PPCh. 3.10 - Prob. 3.49PPCh. 3.11 - Practice Problem 3.50 (solution page 347) For the...Ch. 3.11 - Prob. 3.51PPCh. 3.11 - Prob. 3.52PPCh. 3.11 - Practice Problem 3.52 (solution page 348) For the...Ch. 3.11 - Practice Problem 3.54 (solution page 349) Function...Ch. 3.11 - Prob. 3.55PPCh. 3.11 - Prob. 3.56PPCh. 3.11 - Practice Problem 3.57 (solution page 350) Function...Ch. 3 - For a function with prototype long decoda2(long x,...Ch. 3 - The following code computes the 128-bit product of...Ch. 3 - Prob. 3.60HWCh. 3 - In Section 3.6.6, we examined the following code...Ch. 3 - The code that follows shows an example of...Ch. 3 - This problem will give you a chance to reverb...Ch. 3 - Consider the following source code, where R, S,...Ch. 3 - The following code transposes the elements of an M...Ch. 3 - Prob. 3.66HWCh. 3 - For this exercise, we will examine the code...Ch. 3 - Prob. 3.68HWCh. 3 - Prob. 3.69HWCh. 3 - Consider the following union declaration: This...Ch. 3 - Prob. 3.71HWCh. 3 - Prob. 3.72HWCh. 3 - Prob. 3.73HWCh. 3 - Prob. 3.74HWCh. 3 - Prob. 3.75HW
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