Chapter 3.5, Problem 3.7PP

Explanation of Solution

Given program:

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

scale2:

leaq (%rdi, %rdi, 4), %rax

leaq (%rax, %rsi, 2), %rax

leaq (%rax, %rdx, 8), %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; it copies effective address 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.

Example:

The example for a “leaq” instruction is shown below:

leaq 6(%rdx, %rdx,3)

Here, “%rdx” contains value of x. In given code, “%rax” sets register “%rax” to “4x+6”.

Corresponding C code:

// Define method scale2

Void scale2(long x, long y, long z)

{

//Get value

long t = 5*x+2*y+8*z;

//Return

return t;

}

Explanation:

• The register “%rdi”, “%rsi” and “%rdx” contains “x”, “y” and “z” respectively...