为了说明这是如何工作的，我将使用函数foo作为示例：

def foo(x,y):
  return x + y

main:
   z = foo(7,10) * 3

下面是z的值是如何计算的。（实际上可能会发生的情况是，编译器只是预先计算了17*3，并将其作为常量加载，但这对我试图演示的内容没有帮助。）

main:
  mov  $7, %edi       # put integer args in 64-bit registers efficiently
  mov  $10, %esi      # with zero-extension of 32-bit
  call foo      

#the ret instruction at the end of "foo" takes us back here
#the result of "foo" is now in %rax, and the rest of the program can use it.

  mov %rax, %rdx    # Save a copy of %rax so we can multiply by 3.
  shl $1, %rax      # multiply by 2
  add %rdx, %rax    # add the original, this means we effectively multiplied by 3.

# lea (%rax, %rax, 2), %rax   # efficient way to multiply by 3

  ret              # exit program by returning from main()

####################################################

foo:
  add %rsi, %rdi    # add the two numbers
  mov %rsi, %rax    # return value in %rax
  ret

# compilers would have used  lea (%rdi, %rsi), %rax  to copy-and-add