Risc-v instruction that i dont understand

Question

I have this risc v code :

lui S0, 0x1234
ori S1, S0, 0x5678
add S2, S1, S1

and the question asks me, "What does the register S2 hold?" The question explains that lui and I quote:

" Load the lower half word of the immediate imm into the upper halfword of register rt. The lower bits of the register are set to 0 "

I don't know how to 'compile this program' and what does 0x1234 mean? Thanks

Answer 1

Take the instructions one at a time. First the load-upper-immediate , take the immediate ( 0x1234 ) and "load" it into the upper half of the S0 register and zero out the lower half:

lui S0, 0x1234 

S0 = 0x12340000

Next the or-immediate , we OR the value in S0 with the value 0x5678 :

ori S1, S0, 0x5678

   0x12340000
OR 0x00005678
   ----------
   0x12345678 = S1

Finally the add , we are adding the value in S1 to itself or, equivalently, multiplying the value in S1 by 2:

add S2, S1, S1

  0x12345678
+ 0x12345678
  ----------
  0x2468ACF0 = S2

So the value in S2 register is 0x2468ACF0 . Note, I am assuming 32-bit words. An immediate is like a constant, so lui is an instruction that places a constant into the upper half of a register. In combination with ori , you can load an entire word-immediate into a register.

Answer 2

"LUI places the U-immediate value in the top 20 bits of the destination register rd, filling in the lowest 12 bits with zeros." -- riscv-spec-v2.2

so, lui s0, 0x1234

s0 should be 0x01234000