The assembly language for bar() is: (gdb) disass bar 0x00000000004004f0 : push %

ID: 3854888 • Letter: T

Question

The assembly language for bar() is: (gdb) disass bar 0x00000000004004f0 : push %rbp 0x00000000004004f1 : mov %rsp, %rbp 0x00000000004004f4 s mov %edi, -0xl4 (%rbp) 0x00000000004004f7 : mov -0x14 (%rbp), %eax 0x00000000004004fa : imul -0x14 (%rbp), %eax 0x00000000004004fe : mov %eax, -0x4 (%rbp) 0x0000000000400501 : mov -0x4 (%rbp), %eax 0x0000000000400504 : pop %rbp 0x0000000000400505 : retq End of assembler dump. Give a 1-2 sentence description of the purpose of each instruction. I am more interested in the why than the what.

Explanation / Answer

INSTRUCTION

PURPOSE

Push %rsp

save the base pointer to the stack.

//%rbp is the base pointer, which points to the base of the current stack frame, the stack starts at a high memory address and grows downwards. %rsp is the stack pointer, which points to the top of the current stack frame, points to base pointer.

Mov %rsp,%rbp

put the current stack pointer into the base pointer.

Mov %edi,-0x14(%rdp)

Edi value has copies the value stored in the stack(rdp-14) (stack grows downward)

Mov –ox14(%rdp),%eax

%eax is a general purpose registers , move the content of the next value in stack eax register

Imul -0x14(%rdp),%eax

multiply the contents of rdp by the contents of the memory at location eax. Store the result in eax.(multiply the two values)

Mov %eax , -0x4(%rbp)

Result of multiplication in eax will be stored in stack at the pointer rbp(-4)

Mov -0x4(%rbp),%eax

Reallocating the values to eax

Pop %rdp

Remove the value from the stack

retq

return