Let $s6 be the base address of an array A and $t0 contains some value. When you add:

 

add $t0, $s6, $t0

 

Does register $s6 refer to a value or an address? More importantly, what does $t0 contain? Does $t0 contain a value or an address? How does the computer know whether the register refers to values or addresses?

If it is a value, then how is it possible to do this afterward:

 

lw $s0, 0($t0)

 

How can a register have a set of offset addresses? Does the address of A carry over to $t0? If so where are they stored, because obviously $t0 is used to store the data. I thought offsets only applied to arrays. So does $s6 and $t0 both refer to an address and a value?

 

Is $s6 by default equivalent to 0($s6)?

Edited February 25, 201510 yr by DylsexicChciken

Let $s6 be the base address of an array A and $t0 contains some value. When you add:

 

add $t0, $s6, $t0

 

Does register $s6 refer to a value or an address? More importantly, what does $t0 contain? Does $t0 contain a value or an address? How does the computer know whether the register refers to values or addresses?

You're just calculating new address.

 

In C/C++ it would be:

char *address;

int offset;

char *new_address = address + offset;

 

Actually it's meaningless whether it's address or integer. Because result will be the same. Just adding two together.

 

On many CPUs if you want to increment/change some memory location, you need to load that memory location to register, increment/change register, and store register in the same place.

 

If it is a value, then how is it possible to do this afterward:

 

lw $s0, 0($t0)

You calculated register $t0, and now want to read word from memory address at $t0+0, and store it in register $s0

 

Equivalent to C/C++:

unsigned long data = *( (unsigned long *) new_address ); // if word = 32 bits.

 

If it would be f.e.

 

lw $s0, 100($t0)

 

Equivalent to C/C++:

unsigned long data = *( (unsigned long *) ( &new_address[ 100 ] ) ); // if word = 32 bits.

 

How can a register have a set of offset addresses?

In this case offset 0 is hard-coded in instruction.

See Encoding of LW instruction in link below.

 

If so where are they stored, because obviously $t0 is used to store the data.

Data from memory location at $t0+0 will be stored in $s0

 

I thought offsets only applied to arrays.

Offsets are essential for structures and class object members.

 

Arrays have indexes. Offset in array is equivalent to sizeof( struct ) * index

Full multiply of size of single element of array.

 

f.e.

struct Data

{

char Name[ 16 ];

long Age;

long Sex;

};

 

will have Name at offset 0

Age at offset 16

Sex at offset 16+4=20

and sizeof( Data ) = 16+4+4=24

 

So does $s6 and $t0 both refer to an address and a value?

 

Is $s6 by default equivalent to 0($s6)?

I don't think so

$s6 is most probably operation on register

while

0($s6) is read/write of memory location specified by register.

 

See this

http://www.mrc.uidaho.edu/mrc/people/jff/digital/MIPSir.html

You have description of operation in 2nd row of table of each instruction.

It has even Encoding supplied, so you can see which registers/data/offsets are in which bits of instruction.