Understanding boot loader code to set the A20 gate

Question

Okay guys , I just tried to understand the code that helps in accessing the memory above 1Mb . I am writing the required snippet below . I will write what I understood , I just need the confirmation that what I understood is right or not ( I am self-studying OS ) . Thanks in advance

seta20.1:
  inb $0x64,%al # Wait for not busy
  testb $0x2,%al
  jnz seta20.1

  movb $0xd1,%al # 0xd1 -> port 0x64
  outb %al,$0x64

seta20.2:
  inb $0x64,%al # Wait for not busy
  testb $0x2,%al
  jnz seta20.2

  movb $0xdf,%al # 0xdf -> port 0x60
  outb %al,$0x60

In seta20.1 first we are getting status from keyboard-controller ( mkbc ) to check whether it os okay to write something to it . Once it is free we write 0xd1 to it which goes in it's buffer which will ultimately set the output port to write .

In seta20.1 we are checking first whether the mkbc has read it and set the output port or not . Once done we send the data 0xdf which will set the A20 gate which will ultimately allow us to access the memory above 1Mb .

Is my understanding right ? If not please correct me at the right places .