MIPS lw语义：“ lw $ t2，$ t0”和“ lw $ t2，（$ t0）”之间的区别？

Question

I'm picking up some MIPS in this quick tutorial , where the author distinguishes between these two lw instructions:

lw $t2, $t0 # copy word (4 bytes) at source RAM location to destination register.

lw $t2, ($t0) # load word at RAM address contained in $t0 into $t2

I feel like the author's two comments mean the same...
When I think of these registers as loosely pointers in C++ (of course they're not since registers contain both memory addresses and actual data), both statements seem to do the same thing: copying $t0's "pointee" into $t2, so that $t2's actual value is $t0's "pointee", basically:

Word * $t0, $t2; //some hypothetical pointers
Word someData=1111000011110000.... //some hypothetical type (32 bits in total)
someData = *$t0; //de-reference $t0 and copy its value into someData
$t2 = someData; //impossible in real C++ but you know what I mean

Is there any difference between those two instructions at all? What about lw $t2, 0($t0) and lw $t2, (0)$t0 ? I'm so confused...

Answer 1

lw $t2, $t0 isn't a MIPS instruction --- I suspect you may have misread the page.

In general, in assembler-land, (thing) or [thing] is the common convention for value at address thing . So lw $t2, ($t0) means load the word at the address in $t0 . Yes, this is exactly how pointers work. It's the equivalent of, in C:

t2 = *(uint32_t*)t0;

It is also the same instruction as lw $t2, 0($t0) . The assembler just allows you to emit the number if it's 0; the number is an offset applied to $t0 , so the address accessed is $t0+1234 . lw $t2, 1234($t0) is:

t2 = *(uint32_t*)(t0 + 1234);

MIPS is very regular, unlike x86, and the only instructions that touch memory are the load and store instructions. You won't see this construction anywhere else.