A simple MIPS programming question [array, lw]

Question

So, in this code snippet, I am trying to find out the values of $t2, and $t3 in HEX. I get the answer to be $t2 = 0x30 and $t3 to be 0x3C. However, the answer in the back is $t2 = 0x130, $t3 = 0x13C. Can someone explain??

    .data
x:  .byte 1, 2, 3, 4, 5
y:  .word 19, 20, 25, 30, 35

    .text
    addi $t0, $0, 8
    lw $t1, x($t0)
    sll $t2, $t1, 4
    ori $t3, $t2, 12

we would like to see how you derived at your answers.

codaddict
– codaddict

2010-12-22 07:14:04 +00:00
Commented Dec 22, 2010 at 7:14 — codaddict
– codaddict, Commented Dec 22, 2010 at 7:14

asdfjklqwer · Accepted Answer · 2010-12-22 07:37:11Z

2

How did you derive your answer? I'm pretty rusty with MIPS and this is based on what I remember from uni.

x points to the byte with data 1. On a little-endian machine, x+5 marks the end of the 4 byte word (with data 19) and x+8 marks the beginning. So 19 (10011) is loaded into $t1, shifted left by 4 to 100110000 (304 or 0x130). Finally, 12 is added to get 0x13C.

Let me know if you don't understand something. And it helps to draw a diagram :)

edited Dec 22, 2010 at 7:37

answered Dec 22, 2010 at 7:31

asdfjklqwer

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4 Comments

smooth_smoothie Over a year ago

Doesn't the line lw $t1, x($t0) make $t1 equal to 3 (i.e 8/4 = 2 and x[2] equals 3). Therefore $t1 is shifted by 4 to give you 110000 or 0x30?

asdfjklqwer Over a year ago

There's no need to divide by 4 - don't forget that MIPS uses byte addressing (i.e. each memory address points to a byte in memory). I made a little diagram here which should help: i41.photobucket.com/albums/e299/Seltzer100/Misc/…

Thomas Pornin Over a year ago

actually this also works on a big-endian system. The assembler aligns "y" on a 32-bit boundary, so the word at address x+8 (from x+8 to x+11) has the numerical value 19. You just need the assembler to use the same convention than the CPU when it executes the lw opcode.

smooth_smoothie Over a year ago

So you are saying that memory is continuous, and it's allowed to go over into y, like that? I thought, x and y are totally separate from each other, and in order to access 19 from y, you have to have something like lw $t1, y($t0). Thanks for the pic.

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