li $t0 , 0xABCD9876
sw $t0 , 100($0)
lb $s5 , 101($0)
New to MIPS; So my understanding is,
li loads the value 0xABCD9876 into register $t0
This value is then stored into memory at address ($0+100)
lb then copies the byte at address ($0+101) into register $s5
But there's nothing in register ($0+101) is there? 0xABCD9876 was stored in ($0+100), not ($0+$101). Lost at this point.
Memory is byte-addressed. Hence, ($0+100) points at a single byte. When it is used with a sw or lw instruction, you are actually accessing not only ($0+100), but also ($0+101), ($0+102), and ($0+103) (in other words, you’re accessing four bytes (one word) beginning at that address). By storing a word and then accessing a particular byte of it, you can determine which order the word’s bytes were stored in memory, and hence determine the processor’s endianness.
Related
Supposing that f, g, h, i are stored in $s0~$s4 respectively and the base addresses of arrays A and B are in $S6 and $S7.
sll $t0, $s0, 2
add $t0, $s6, $t0
sll $tl, $sl, 2
add $tl, $s7, $tl
lw $s0, 0($t0)
addi $t2 , $t0, 4
lw $t0, 0($t2)
add $t0, $t0, $s0
SW $t0, 0($tl)
I'm not familiar with MIPS so I Wonder how to translate MIPS into C and how to minimize these MIPS instructions？
how to translate MIPS into C
You recognize the patterns, here for array indexing / array element access.
On a byte addressable machine (all modern hardware), a 4-byte integer occupies 4 bytes in memory, and each of those bytes has a unique memory address. Because of the way the hardware works, we only use one of those 4 addresses to refer to the whole 4-byte integer, namely we use the lowest address among the 4. The hardware can load a 4-byte integer from memory given that one address (the lowest).
Since each 4-byte integer in memory occupies 4 addresses, in an array of 4-byte integers, the memory address of the first element and the memory address of the second element are 4 addresses apart even though are sequential index positions (i.e. they are only 1 index position apart).
The formula for indexing a 4-byte integer array, then is to convert the index into a byte offset, then add the byte offset to the base address of the array. The first part of that: converting an index to a byte offset, is sometimes referred to as "scaling". Scaling is conceptually done by multiplication, so in A[i], i needs to be scaled by the size of the array elements of A. If 4-byte integers that means scaling (multiplying) the index by 4. A quick way of doing that is shifting by 2 bit positions, which has the same effect as multiplying by 4.
The C language automatically scales when doing array references, whereas assembly language requires explicit scaling. C can do this because it knows the type of the array, whereas assembly language does not.
In C we can do expressions like A[i]. The C language allows us to break that down somewhat into *(A+i), which separates the pointer arithmetic addition A+i from the dereferencing of that sum, dereferencing with the unary indirection operator, *. As previously mentioned, C automatically scales, so A+i becomes the equivalent of A+i*4, in which we can substitute shifting for multiplication: A+(i<<2).
Next, we need to know if the dereference is for read or for write. When A[i] is accessed for its value, we will see it on what we call the "right hand side" of an assignment operator, as in ... = A[i]. When A[i] is access to update/store a value, we will see it on what we call the left hand side of an assignment operator, as in A[i] = ....
So, the sequence for doing A[i] for read (right hand side) in C is the following in assembly:
sll $temp1, $i, 2
addu $temp2, $A, $temp1
lw $temp3, 0($temp1)
Where $tempN is some register (usually a designated temporary) chosen to hold an intermediate value. Since multiple instructions are needed to accomplish anything, sequences of instructions are interconnected with registers that hold the intermediate states. And also, in assembly we name registers, not variables, so in my above $i and $A should be a registers names representing those variables rather than variable names directly used.
The pattern for write/store array access is similar but ends with a sw instruction instead, to store some value into memory at the index position.
These instruction sequence are interconnected by the use of these registers, and the sequences can be interrupted or interspersed with other instructions — what we have to follow then is the above pattern by paying attention to to the register usages that interconnect them rather than the specific sequences.
In your sample code:
sll $t0, $s0, 2 # sourcing an index in $s0, scaling it into temp $t0
add $t0, $s6, $t0 # adding a base array in $s6, putting back into $t0
sll $tl, $sl, 2
add $tl, $s7, $tl
lw $s0, 0($t0) # accessing the value of $s6[$s0*4], aka A[f]
addi $t2 , $t0, 4
lw $t0, 0($t2)
add $t0, $t0, $s0
SW $t0, 0($tl)
We can see the pattern for a read access to an index in $s0, and an array in $s6, these, we are told, map to f and A, so those three instructions comprise A[f] to read a value from A at index f.
The rest are done similarly. Your job is to use this knowledge to find the other array indexing patterns in the above sequence. Find out how the results of the array indexing operations are used and you'll have the complete C code.
NOTE that the sample you've been given incorrectly uses add and addi when pointer arithmetic should use addu and addiu — we don't want signed integer overflow checking on pointer arithmetic, as pointers are unsigned.
One of the add instructions is not for pointer arithmetic, but should probably still have used addu if this is intended to be replicated in C, because the C language does not have a built in operator to trap on overflow.
How does MIPS's assembler labels and J type instruction work?
I am currently making a MIPS simulator using C++ and came into a big question. How exactly does MIPS assembler manage label's and their address while on a J type instruction?
Let's assume that we have a following code. Also let's assume that start: starts at 0x00400000. Comments after code represent where the machine codes will be stored in memory.
start:
andi $t0, $t0, 0 # 0x0040 0000
andi $t1, $t1, 0 # 0x0040 0004
andi $t2, $t2, 0 # 0x0040 0008
addi $t3, $t3, 4 # 0x0040 000C
loop:
addi $t2, $t2, 1 # 0x0040 0010
beq $t2, $t3, exit # 0x0040 0014
j loop # 0x0040 0018
exit:
addi $t0, $t0, 1000 # 0x0040 002C
As I am understanding right at the moment, j loop expression will set PC as 0x0040 0010.
When J type instruction uses 32 bits and with MSB 6 bits as its opcode, it only has 26 bits left to represent address of instruction. Then how is it possible to represent 32 bit address system using only 26 bits?
With the example above, it can represent 0x00400010 with only 24bits. However, in references, text segment is located from 0x00400000 to 0x10000000 which needs 32bit to represent.
I have tried to understand this using MARS simulator, however it just represents j loop as j 0x00400010 which seems nonsense to me since 0x00400010 is 32 bits.
My current guess
One of my current guesses is following.
Assembler saves the loop: label's address into some memory address that is reachable by 26 bits. Then when expression j loop is called, label loop is translated to the memory address that contains 0x00400010 For example, 0x00400010 is saved in some address like 0x00300000 and when j loop is called, loop is translated into 0x00300000 and it is able to get value from 0x00300000 and reach out 0x00400010. (This is just one of my guess)
You have a number of questions here.
First, let's try to differentiate between the assembler's operation and the MIPS machine code that it generates and the processor executes.
The assembler manages labels and address in two ways. First, it has a symbol table, which is like a dictionary, a data structure of key-value pairs where the names are keys and the addresses (that those names will refer to when the program is running) are the values in the pairs.
Second, the assembler manages the code and data sections with a location counter. That location counter advances each time the program provides some code or data. When new label is defined, the current location counter is then used as the address value in a new key-value pair.
The processor never sees the labels: they do not execute and they do not occupy any space in the code or data. The processor sees only machine code instructions, which on MIPS are all 32-bits wide. Each machine code instruction is divided into fields. There are instruction types or formats, which on MIPS are straightforward: I-Type, J-Type, and R-Type. These formats then define the instruction fields, and the assembler follows these encodings. All the instruction formats share the 6-bit opcode field, and this opcode field tells the processor what format the instruction is, which fields it therefore has, and thus how to interpret and execute the rest of the instruction.
The assembler removes labels from the assembly — labels and their names do not exist in the program binary. The label definitions themselves (label:) are omitted from the program binary but usages of labels are translated into numbers, so a machine code instruction that uses a label will have some instruction field that is numeric, and the assembler will provide a proper value for that numeric field so that the effect of the reaching or otherwise accessing what the label referred to is accomplished. (The label is no longer in the program binary, but the code or data memory that the label referred does remain).
The assembler sets up branch instructions, j instructions, and la/lw instructions, using numbers that tell the processor how far forward or backward to move the program counter, or, what address some data of interest is at. The lw/la instructions access data, and these use 2 x 32-bit instructions each holding 16 bits of the address of interest. Between the two instructions, they put together a full 32-bit address for data access. For branches to fully reach any 32-bit address, they would have to put together the 32-bit address in a similar manner (two instruction pair) and use an indirect/register branch.
Lets say i have the following command in MIPS:
sw $t0, 0($sp) # $sp=-4
Does that mean that the register $t0 is saved from 0 to -3 byte or from -4 to -7 byte ?
Stack growing downward in memory to make space for new data to be saved. The stack pointer $sp point to the top of stack.
The stack pointer $sp starts at a high memory address and decrements to expand as needed. Figure (b) shows the stack expanding
to allow two more data words of temporary storage. To do so, $sp decrements by 8 to become 0x7FFFFFF4. Two additional data words,
0xAABBCCDD and 0x11223344, are temporarily stored on the stack.
So in your case if I understand your question well the sw is word- addressable and the word will be stored on that location in memory which $sp point to. In case you store the next word it must have an offset of 4.
[Harris&Harris]
UPDATE
Take this example when you use lb
Say you have this word 0x23456789
when using lb $s0,1($0) After the load byte instruction, lb $s0, 1($0),
$s0 would contain 0x00000045 on a big-endian system and 0x00000067 on a
little-endian system.[Harris&Harris]
What is the use of a $zero register in MIPS?
What does it mean?
lw $t0, myInteger($zero)
The zero register always holds the constant 0. There's not really anything special about it except for the fact that 0 happens to be a very useful constant. So useful that the MIPS designers dedicated a register to holding its value. (This way you don't have to waste another register, or any memory, holding the value.)
EDIT:
As for the question of what that line of code means, it loads the word from MEMORY[myInteger + 0] into the $t0 register. The lw command takes both a constant (myInteger) and a register ($zero). Not sure why that is, but that's just how the instructions work. Since myInteger was used as the constant, a register had to be provided, so $zero was used.
How does 'alignment of memory operands' help MIPS to be pipelined?
The book says:
Fourth, as discussed in Chapter 2, operands must be aligned in memory. Hence,
we need not worry about a single data transfer instruction requiring two data
memory accesses; the requested data can be transferred between processor and
memory in a single pipeline stage.
I think I understand that one data transfer instruction does not require two or more data memory aaccesses.
However, I am not sure what does it have to do with the alignment of memory operands.
Thanks, in advance!
The lw instruction requires that the memory address be word aligned.
Therefore, to access an unaligned word, one would need to access the two word boundaries that the required word intersects and mask out the necessary bytes.
For example, suppose you desire to load a word stored at address 0x2. 0x2 is not word aligned, so you would need to load the half word stored at 0x2 and the half-word stored at 0x4.
To do so, one might write:
lh $t0 2($zero)
lh $t1 4($zero)
sll $t1 $t1 16
or $t2 $t0 $t1
This only gets more complicated if you want to load for example a word stored at address 0x3:
# load first byte
lb $t0 3($zero)
# load second word, mask out first 3 bytes
lw $t1 4($zero)
lui $t2 0x0000FFFF
ori $t2 $t2 0xFFFFFFFF
or $t1 $t1 $t2
# combine
sll $t1 $t1 8
or $t2 $t0 $t1
So, it can be seen that the requirement for word alignment doesn't help MIPS to be pipelined, but rather that access to unaligned words requires excess memory accesses — this is a limitation of the ISA.