COMPUTER SYSTEMS&MOD MSGT/ET SA AC PKG
COMPUTER SYSTEMS&MOD MSGT/ET SA AC PKG
3rd Edition
ISBN: 9780134671123
Author: Bryant
Publisher: PEARSON
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Chapter 3, Problem 3.58HW

For a function with prototype

long decoda2(long x, long y, long z);

GCC generates the following assembly code:

Chapter 3, Problem 3.58HW, For a function with prototype long decoda2(long x, long y, long z); GCC generates the following

Parameters x, y, and z are passed in registers %rdi, %rsi, and %rdx. The code stores the return value in register %rax.

  Write C code for decode2 that will have an effect equivalent to the assembly code shown.

Expert Solution & Answer
Check Mark

Explanation of Solution

Given assembly code:

x in %rdi, y in %rsi and z in %rdx

decode2:

subq %rdx, %rsi

imulq %rsi, %rdi

movq %rsi, %rax

salq $63, %rax

sarq $63, %rax

xorq %rdi, %rax

ret

Load Effective Address:

  • The load effective address instruction “leaq” is a variant of “movq” instruction.
  • The instruction form reads memory to a register, but memory is not been referenced at all.
  • The first operand of instruction is a memory reference; the effective address is been copied to destination.
  • The pointers could be generated for later references of memory.
  • The common arithmetic operations could be described compactly using this instruction.
  • The operand in destination should be a register.

Data movement instructions:

  • The different instructions are been grouped as “instruction classes”.
  • The instructions in a class performs same operation but with different sizes of operand.
  • The “Mov” class denotes data movement instructions that copy data from a source location to a destination.
  • The class has 4 instructions that includes:
    • movb:
      • It copies data from a source location to a destination.
      • It denotes an instruction that operates on 1 byte data size.
    • movw: 
      • It copies data from a source location to a destination.
      • It denotes an instruction that operates on 2 bytes data size.
    • movl:
      • It copies data from a source location to a destination.
      • It denotes an instruction that operates on 4 bytes data size.
    • movq:
      • It copies data from a source location to a destination.
      • It denotes an instruction that operates on 8 bytes data size.

Comparison Instruction:

  • The “CMP” instruction sets condition code according to differences of their two operands.
  • The working pattern is same as “SUB” instruction but it sets condition code without updating destinations.
  • The zero flag is been set if two operands are equal.
  • The ordering relations between operands could be determined using other flags.
  • The “cmpl” instruction compares values that are double word.

Unary and Binary Operations:

  • The details of unary operations includes:
    • The single operand functions as both source as well as destination.
    • It can either be a memory location or a register.
    • The instruction “incq” causes 8 byte element on stack top to be incremented.
    • The instruction “decq” causes 8 byte element on stack top to be decremented.
  • The details of binary operations includes:
    • The first operand denotes the source.
    • The second operand works as both source as well as destination.
    • The first operand can either be an immediate value, memory location or register.
    • The second operand can either be a register or a memory location.

Corresponding C code:

// Define method decode

long decode(long x, long y, long z)

{

// Declare variable

long tmp = y - z;

//Return

return (tmp * x)^(tmp << 63 >> 63);

}

Explanation:

  • The register “%rdi” has value for “x”, register “%rsi” has value for “y” and register “%rdx” has value for “z”.
  • The details of assembly code is shown below:
    • The instruction “subq %rdx, %rsi” performs operation “y - z” and stores result in register “%rsi”.
      • The statement “long tmp = y - z” corresponds to C code.
    •  The instruction “imulq %rsi, %rdi” multiplies result of operation with “x” and stores result in register “%rdi”.
      • The statement “(tmp * x)” corresponds to C code.
    • The instruction “movq %rsi, %rax” moves value in register “%rsi” to register “%rax”.
    • The instruction “salq $63, %rax” performs left shift on value in register “%rax”.
      • The statement “tmp << 63” corresponds to C code.
    • The instruction “sarq $63, %rax” performs right shift on value in register “%rax”.
      • The statement “tmp << 63 >> 63” corresponds to C code.
    • The instruction “xorq %rdi, %rax” performs “XOR” operation on values in registers “%rax” and “%rdi”.
      • The statement “return (tmp * x)^(tmp << 63 >> 63)” corresponds to C statement.

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Chapter 3 Solutions

COMPUTER SYSTEMS&MOD MSGT/ET SA AC PKG

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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