Mips: store a variable in another variable

Question

I am trying to translate a C program into MIPS assembly code

In my C code I have a line like this:

int base;
int count;

count = base;

In MIPS, how would I store the value of base inside of count? The only instructions I see for loading and storing are lw and sw and their prototypes only go from a register source -> ram destination or a ram source -> register destination.

Any help would be appreciated.

EDIT I was hoping that this could be done in a single instruction, something like

move base acc

but apparently that is impossible (at least I found no example of an instruction similar to that), I opted for this:

lw $t0, base   //load base into $t0
sw $t0, count  //store the value of $t0 in count

If there is a one-line instruction to do this that would be better if anyone knows one.

score 13 · Accepted Answer · 2012-02-02 04:34:09Z

13

MIPS doesn't support direct memory-to-memory moves. (Neither do most common CPUs, actually -- even x86 doesn't.) You'll need to use lw/sw to move data around.

Architecturally, this is because MIPS is designed to only perform a single memory access per cycle -- doing a memory-to-memory move would require two accesses (one read, one write), or stall the pipeline.

edited Feb 2, 2012 at 4:34

answered Feb 2, 2012 at 4:20

user149341

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4 Comments

Hunter McMillen Over a year ago

The values of base and counter aren't in registers, they are stored in the RAM. So when I do this I just get an error message.

user149341 Over a year ago

Oh, if they aren't registers then it's different. Updating.

Hunter McMillen Over a year ago

It'd be nice it they could optimize that instruction, even though it is one more line for lw/sw equivalent, I spent a good 1hr+ searching for the instruction.

user149341 Over a year ago

Like I said, though, there's no way to do it in a single instruction on a classic MIPS-style architecture without pipeline stalls (or a double-ported memory).

markgz · Accepted Answer · 2012-02-02 19:37:06Z

8

Here's how to do it in MIPS

la $t0, base     // load the address of "base"
la $t1, count    // load the address of "count"
lw $t2, 0($t0)   // load the data at location "base"
sw $t2, 0($t1)   // store that data at location "count"

It is not possible to do a memory to memory move in a single instruction.

answered Feb 2, 2012 at 19:37

markgz

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6 Comments

Hunter McMillen Over a year ago

You can actually do it in only 2 instructions: lw $t0, base then sw $t0, count

markgz Over a year ago

lw $t0, base is just MIPS assembler shorthand for lui $at,%hi(base) followed by lw $t0,%lo(base)(at). The load then store sequence is still a total of 4 machine instructions.

Hunter McMillen Over a year ago

but only two physical instructions.

markgz Over a year ago

No. My orignal comment was not well explained. lw $t0, base expands to 2 instructions. sw $t0,count expands to 2 instructions. This gives a total of 4 instructions for the sequence. Physical instructions and machine instructions and instructions are all the same thing on MIPS. Each one occupies a 32 bit word in the executable file and in memory.

entithat Over a year ago

Who can explain what does 0($t1) mean?

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