1. 24/09/13 www.eej.ulster.ac.uk/~ian/modules/EEE527/files 1 Embedded Systems Lecture 1: The MIPS 32 bit microprocessor Ian McCrumRoom 5B18, Tel: 90 366364 voice mail on 6 th ring Email: IJ.McCrum@Ulster.ac.uk Web site: http://www.eej.ulst.ac.ukIJ.McCrum@Ulster.ac.ukhttp://www.eej.ulst.ac.uk

10. Types of Instructions There are 3 main types of assembly instructions –Arithmetic - add, sub, mul, shifts, and, or, etc. –Load/store –Conditional - branches

11. Arithmetic Instructions add a, b, ca = b+c add a, a, da = d+a = d+b+c add a, a, ea = e+a = e+d+b+c Example: Translate the following instructions to assembly code a = b+c d = a-e Solution: add a, b, c sub d, a, e

12. Arithmetic Instructions Example: Translate the following instructions to assembly code. Remember with RISC, only 1 operation per instruction! HINT - you may need temporary variables f = (g+h) - (i+j) Solution: add t0, g, h add t1, i, j sub f, t0, t1

13. Operands In assembly code, you can’t use variables such as a, b, c, etc In RISC instruction sets, operands must be registers such as r1, r2, r3, etc –r0 is typically reserved to hold the immediate value of 0 –There is a limited number of registers MIPS has 32

14. Arithmetic Instructions Using Registers Example: Translate the following instructions to assembly code. Assume that g, h, i, and j are already stored in r1, r2, r3, and r4. Store f in r5 f = (g+h) - (i+j) Solution: add r6, r1, r2 add r7, r3, r4 sub r5, r6, r7

15. What about more data?? With only a limited number of registers, not all data can be stored in registers at the same time. –Registers only store data that is currently being operated on Variables are stored in memory and then loaded into registers when needed using data transfer instructions –Load word (lw) and store word (sw)

16. Load and store word Load word format –lw destination register, memory location Store word format –sw source register, memory location Memory location format –Offset(base address) Base address = starting location of data in memory Offset = how far away is location to access from base address –Values are added together

17. LW Example Example: Assume that A is an array of 100 words. Assume the base address is stored in r3, g is in r1 and h is in r2 g = h + A[8] Solution: lw r4, 8(r3) add r1, r2, r4 Offset Base Address This is simplified, more details later…

18. Data in Memory All variables/data are stored in memory –You will need to do this in your assembler –Your ISS will need a data structure to hold main memory Array is fine

19. Addressing Data Architecture usually addresses data in bytes (byte addressable) –32-bit architecture = 4 bytes = 1 word lw/sw load/store 1 word or 4 bytes –Thus, data/inst addresses are multiples of 4 Data is word aligned to be more efficient

20. Data in Memory...... 100 10 101 1 12 8 4 0 AddressData......

21. LW/SW Example Example: Assume that A is an array of 100 words. Assume the base address is stored in r3 and h is stored in r2. You may directly calculate the offset. Remember, each data piece is 4 bytes when calculating the offset A[12] = h+A[8] Solution: lw r1, 32(r3) add r4, r2, r1 sw r4, 48(r3)

22. LW/SW Example Example: Assume that A is an array of 100 words. Assume the base address is stored in r3 and g, h, and i are in r1, r2, and r4 respectively. Calculate the offset using assembly instructions but don’t use multiplication yet (mult instruction is different) g = h + A[i] Solution: add r5, r4, r4# Temp reg r5=2*i add r5, r5, r5# Temp reg r5=4*i add r5, r5, r3# t1 = addr of A[i] (4*i+r3) lw r6, 0(r5)# Temp reg r0=a[i] add r1, r6, r2# g=h+a[i]

23. Translating MIPS Assm Language to Machine Language Translate human readable assembly code to machine readable code (binary) –I will show examples in decimal for readability –This is what you assembler will do but it will output in binary.

24. MIPS -> Machine Language Example: Show the real MIPS language version of the following instruction in both decimal and binary add r0, r1, r2 Solution:decimal Each segment is referred to as a field. Details to come…. binary 0012032 00000000000000010001000000100000 6 bits5 bits 6 bits

25. MIPS Fields MIPS fields are giving names to make them easier to discuss op: Basic operation of the instruction, typically called the opcode rs: The first register source operand rt: The second register source operand rd: The register destination operand, it gets the result of the operation shamt: Shift amount (0 if not shift instruction) funct: Function. This field selects the specific variant of the operation in the op field, and is sometimes called the function code oprsrtrdshamtfunct 6 bits5 bits 6 bits

26. MIPS Fields Problem occurs with an instruction needs a longer field than that showed on the previous slide –I.e. LW must specify 2 registers and a constant. Limited to 5-bit constant if use previous format. Solution: There are different formats for different types of instructions –Previous slide is R-type (R-format): R=register

27. MIPS Fields I-type (I-format) –I=immediate –Now LW can specify an address up to 16- bits Opcode determines the format oprsrtaddress 6 bits5 bits 16 bits

28. MIPS Instruction Encoding

29. MIPS Asm -> Machine Language Example: Assume r1 is the base of A and r2 corresponds to h, the C statement: is compiled to: What is the MIPS machine code for these three instructions? (Use figure 3.5) A[300] = h + A[300] lw r0, 1200(r1) add r0, r2, r0 sw r0, 1200(r1)

30. MIPS Asm -> Machine Language decimal binary oprsrtrd Address /shamt funct Solution: lw r0, 1200(r1) add r0, r2, r0 sw r0, 1200(r1) 0020032 35011200 43011200 000000000000001000000 32 10001100000000010000 0100 1011 0000 10101100000000010000 0100 1011 0000

31. Decision Instructions Branch/jump instructions –Conditional branches beq register1, register2, Label bne register1, register2, Label –Unconditional branches j Label

32. Decision Instructions Example: Assume f->r0, g->r1, h->r2, i->r3, j->r4 if ( i==j ) goto L1 f = g+h L1:f = f-i Solution: beq r3, r4, L1 add r0, r1, r2 L1:sub r0, r0, r3 Labels will need to be translated to instruction address in your assembler

33. Decision Instructions Example: Assume f->r0, g->r1, h->r2, i->r3, j->r4 if ( i==j ) f = g+h L1:else f = g-h L2: Solution: bne r3, r4, L1 add r0, r1, r2 j L2 L1:sub r0, r1, r2 L2:

34. Decision Instructions Example: A is 100 elements with the base address in r5. g->r1, h->r2, i->r3, j->r4 Loop:g = g+A[i] i = i+j if ( i!=h ) goto Loop Solution: Loop:add r6, r3, r3 add r6, r6, r6 add r6, r6, r5 lw r7, 0(r6) add r1, r1, r7 add r3, r3, r4 bne r3, r2, Loop

35. While Loop Goto statements are bad, just used them as an example. You will want to use while loops –Or for loops but I am just showing you while loops

36. While Loop Example: Base address of save is in r6. i->r3, j->r4, k->r5 while ( save[i] == k ) i = i+j Solution: Loop:add r1, r3, r4 add r1, r1, r1 add r1, r1, r6 lw r0, 0(r1) bne r0, r5, Exit add r3, r3, r4 j Loop Exit:

37. Other Styles of MIPS Addressing Constant or immediate operands –Programs often use constant values –I.e. incrementing to the next data element while scanning an array addi instruction - adds an immediate value to a register

38. Immediate Operands Example: What is the machine code for the following? (Remember the I-format instruction) addi r4, r4, 4 Solution: decimal binary oprsrtImmediate 8444 00100000100 0000 0000 0000 0100

39. Addressing in Branches and Jumps Last instruction format - J-type (J- format) Branches do not use J-type. –Must specify 2 registers to compare –Use I-type opcodeTarget address

40. Implementing Conditional Statements We're going to translate some easy conditional statements. if ( i == j ) i++ ; j-- ; Translating conditional statements is interesting. In C, for example, when the condition is true, you execute the body. This is the fall-through case. That is, you execute the next statement. When the condition is false, you don't execute the body, you jump over it. This is the jump case. Therefore, you jump when the condition is false. In ISA programming, you jump when the condition is true. Thus, we often need to negate the condition.Here‘s the translation of the above if-statement, assuming $r1 stores i and $r2 stores j. bne $r1, $r2, L1 # branch if ! ( i == j ) addi $r1, $r1, 1 # i++ L1: addi $r2, $r2, -1 # j– The label L1 has the same address as the instruction immediately following the colon. Thus, the above code is the same as: bne $r1, $r2, L1 # branch if ! ( i == j ) addi $r1, $r1, 1 # i++ L1: addi $r2, $r2, -1 # j-- Even though it appears that label L1 has an empty instruction, it doesn't. It is still associated with the second addi instruction. Taken from http://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htmhttp://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htm

41. Translating if-else if ( i == j ) i++ ; else j-- ; j += i ; Let's think about what happens. As before, if the condition is false, we want to jump. This time, we want to jump to the else. Thus, we write the code like: bne $r1, $r2, ELSE # branch if ! ( i == j ) addi $r1, $r1, 1 # i++ j L1 # jump over else ELSE: addi $r2, $r2, -1 # j– L1: add $r2, $r2, $r1 # j += i Taken from http://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htmhttp://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htm

42. Translating if-else with && Translating && is interesting because there's short-circuiting involved. To see how this works, let's translate: if ( i == j && i == k ) i++ ; // if-body else j-- ; // else-body j = i + k ; let stand for i == j and stand for i == k. if ( && ) i++ ; // if-body else j-- ; // else-body j = i + k ; Short-circuiting occurs when evaluates to false. The control-flow then jumps over (that is, is not evaluated), and continues executing in the else-body.If evaluates to true, we want to fall-through and check. If evaluates false, we again jump, this time over the if- body, and to the else-body.If is true, we fall-through to the if-body. Notice that we jump when the condition evaluates to false for both cases, so we'll be interested in jumping on negations of conditions. Taken from http://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htmhttp://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htm

43. Here's the translated code, assuming $r3 stores k. bne $r1, $r2, ELSE # cond1: branch if ! ( i == j ) bne $r1, $r3, ELSE # cond2: branch if ! ( i == k ) addi $r1, $r1, 1 # if-body: i++ j L1 # jump over else ELSE: addi $r2, $r2, -1 # else-body: j– L1: add $r2, $r1, $r3 # j = i + k From the C Code if ( i == j && i == k ) i++ ; // if-body else j-- ; // else-body j = i + k ; Taken from http://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htmhttp://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htm

44. Translating if-else with || if ( i == j || i == k ) i++ ; // if-body else j-- ; // else-body j = i + k ; Again, let's use to stand for i == j and to stand for i == k. if ( || ) i++ ; // if-body else j-- ; // else-body j = i + k ; Short-circuiting occurs when evaluates to true. That is, we want to jump over checking the second condition and into the if-body. Notice that we go to the if-body when the condition evaluates to true. If is false, we want to fall-through and check. If is false, we now jump to the else-body. If is true, we fall-through to the if- body. Notice that we jump when evaluates to true (to the if-body) and when evaluates to false (to the else-body). Taken from http://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htmhttp://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htm

45. Translating if-else with || /cont if ( i == j || i == k ) i++ ; // if-body else j-- ; // else-body j = i + k ; Here's the translated code: beq $r1, $r2, IF # cond1: branch if (i==j) bne $r1, $r3, ELSE # cond2: branch if !(i==k) IF: addi $r1, $r1, 1 # if-body: i++ j L1 # jump over else ELSE: addi $r2, $r2, -1 # else-body: j– L1: add $r2, $r1, $r3 # j = i + k Taken from http://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htmhttp://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htm

46. switch statements Switch statements are interesting. Note switch works on a very limited number of types (int and char, primarily). It doesn't work on strings (even if students wish they did). Switch evaluates case-by-case. When one condition fails, the next is checked. When a condition is true, the code associated with that condition is run. However, if you don't put break the code for the next condition will also run. Unfortunately, most people expect a break to occur when the condition is done. They don't expect a fall- through case. Here's an example of a switch. switch( i ) { case 1: i++ ; // falls through case 2: i += 2 ; break; case 3: i += 3 ; } Taken from http://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htmhttp://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htm

47. Here's the translated code addi $r4, $r0, 1 # set temp to 1 bne $r1, $r4, C2_COND # case 1 false: branch to case 2 cond j C1_BODY # case 1 true: branch to case 1 C2_COND: addi $r4, $r0, 2 # set temp to 2 bne $r1, $r4, C3_COND # case 2 false: branch to case 2 cond j C2_BODY # case 2 true: branch to case 2 body C3_COND: addi $r4, $r0, 3 # set temp to 3 bne $r1, $r4, EXIT # case 3 false: branch to exit j C3_BODY # case 3 true: branch to case 3 body C1_BODY: addi $r1, $r1, 1 # case 1 body: i++ C2_BODY: addi $r1, $r1, 2 # case 2 body: i += 2 j EXIT # break C3_BODY: addi $r1, $r1, 3 # case 3 body: i += 3 EXIT: switch( i ) { case 1: i++ ; // falls through case 2: i += 2 ; break; case 3: i += 3 ; } When case 1 is true, you want to run i++, then i += 2. You don't want to test for case 2, because that's now how the semantics of switch works.

48. bge, bgt, blt, ble bge, bgt, blt, ble are all pseudo-instructions. That is, there is no corresponding machine code to these instructions. Let's see an example of how the assembler might translate bge. The key is to use slt which means "set on less than". Here is the syntax of slt. slt $r1, $r2, $r3 # R[1] = R[2] < R[3] ? 1 : 0 The semantics are shown in the comments: if R[2] < R[3] false, we now jump to the else- bodythen R[1] = 1, otherwise it's assigned to 0. Here is the syntax and semantics of bge: bge $r1, $r2, LABEL # jump to LABEL if R[1] >= R[2] If R[1] >= R[2] we know that this is equivalent to !( R[1] < R[2]). Thus, if we check R[1] < R[2] using slt, we expect it to be false. Here's the translation of bge. slt $r3, $r1, $r2 # check if R[1] < R[2] beq $r3, $r0, LABEL # branch if previous condition is false As an exercise, you should translate the other three pseudo-instructions

49. Implementing Loops Unlike conditional statements, which have assembly instructions that support them more-or- less directly (i.e., beq), loops do not have similar support.To translate loops, it's easier to convert the loops to if-statements with goto statements, prior to translation. Doing so also gives you insight into how a loop really behaves. We'll translate a while-loop, then a for-loop. You can do the do-while loop as an exercise (and you should!). Here's a generic while-loop : while ( ) { } This is how it's translated to an if-statement with goto's. L1: if ( ) { goto L1 ; } Taken from http://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htmhttp://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htm

50. Here's an example with real code. while ( i < j ) { k++ ; i = i * 2 ; } Then translate it to if-statements with goto's. L1: if ( i < j ) { k++ ; i = i * 2 ; goto L1 ; } This should be easy to convert to MIPS. Assume $r1 stores i, $r2 stores j, and $r3 stores k. L1: bge $r1, $r2, EXIT # branch if ! ( i < k ) addi $r3, $r3, 1 # k++ add $r1, $r1, $r1 # i = i * 2 j L1 # jump back to top of loop EXIT: Taken from http://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htmhttp://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htm

51. Translating for-loops To translate a for-loop, we'll only go part way, and translate it to if-statements and goto's. You can do the rest on your own. Here's a generic for-loop: for ( ; ; ) { } This is how it's translated to an if-statement with goto's. L1: if ( ) { UPDATE: // should affect condition! goto L1 ; } EXIT: There's only one special case. Suppose the has a continue statement. Then, you need to jump to the code, which is at the UPDATE label. break should still work fine, because you can jump to the EXIT label. Taken from http://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htmhttp://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htm

52. Accessing Array Data in MIPS Since arrays can store LOTS of data, and since we have only a small (~32) number of registers, it is infeasible to use the registers for long-term storage of the array data. Hence, arrays are stored in the Data Segment of a MIPS program. Fundamentally, there are three operations which one can perform on an array: –Getting the data from an array cell, e.g, x = list[i]; –Storing data into an array cell, e.g. list[i] = x; –Determining the length of an array, i.e. list.length. To access the data in the array requires that we know the address of the data and then use the load word (lw) or store word (sw) instructions. Words (which is how integers are stored) in MIPS take up 32 bits or 4 bytes. Therefore, if we have a declaration in the.data segment such as: list:.word 3, 0, 1, 2, 6, -2, 4, 7, 3, 7 the address that is loaded by the instruction la $t3, list is the address of the first '3' in the list. The address of the '0' is 4 greater than that number, and the address of the '6' is 16 greater than that number. Taken from http://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htmhttp://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htm

53. Accessing Array Data in MIPS The following snippet of code will place the value of list[6] into the $t4: la $t3, list # put address of list into $t3 li $t2, 6 # put the index into $t2 add $t2, $t2, $t2 # double the index add $t2, $t2, $t2 # double the index again (now 4x) add $t1, $t2, $t3 # combine both parts of the address lw $t4, 0($t1) # get the value from the array cell If we wish to assign to the contents of $t4 to list[6] instead, the last line would simply be: sw $t4, 0($t1) # store value in the array cell Taken from http://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htmhttp://www.cs.pitt.edu/~xujie/cs447/AccessingArray.htm

54. Some notes on arrays Most assembly languages, like MIPS, do not have built-in capabilities for sophisticated data structures. Even the most commonly available structure, the array, is not available in MIPS. As we shall see shortly, we can set aside a block of memory in MIPS in assembly language and treat it similar to an array. What is an array anyway? One way to think about an array is a set of values that can simultaneously be considered a single collective entity and many individual elements. In high level languages, the array is a set of references using an identifier like any other variable. To access each individual element, one uses an integer to specify a particular element. In most high-level languages, arrays are stored as a contiguous block of n memory cells starting with a base address. In java, we might declare and use an array as follows. int a[] = new int[10]; int sum = 0;... for (int i = 0; i < 10; i++) sum += a[i]; In this example, the elements in the array are added and the result is stored in the variable sum. Note a is the identifier that refers to the array and [i] refers to the ith element of the array. Here a is the base location of the array and i indicates the offset into memory from the base address. From http://zeta.albion.edu/~dreimann/Spring2012/courses/cs354/projects/primes.phphttp://zeta.albion.edu/~dreimann/Spring2012/courses/cs354/projects/primes.php

55. In MIPS, there is no formal array construct. The first issue to resolve is the association of a block of memory with a particular identifier. This can be done using a label and a.space directive in the.data section. For example.data a:.space 40 reserves 40 bytes (10 words) of memory of data associated with the label a. The memory location where this label is stored (by the assembler) becomes the base address of the array. To access the ith element from the array, we need to determine the memory offset from the beginning address and add the number of bytes per element. For simplicity, let's assume the array stores elements that each require on word of storage and $t0 is the register that represents i. Then la $t1, a muli $t2, $t0, 4 # or use two adds... add $t2, $t2, $t1 lw $t3, 0($t2) will load the ith element from the array into register $t3. Here $t1 contains the base address of the array and $t2 is the address of the ith element of the array. From http://zeta.albion.edu/~dreimann/Spring2012/courses/cs354/projects/primes.php

56. While this will work, it is not the only way to access the elements of the array. Consider storing the integers 0-9 in the first ten elements of an array then reading the elements and adding them together. The program sum10.s shown below illustrates one way to do this.sum10.s.text main: # Fill the array la $t4, n # address of n lw $t4, 0($t4) # t4 = n and $t0, $0, $0 # i = 0 la $t1, a # address of a loop1: sll $t2, $t0, 2 # byte offset of ith element add $t2, $t2, $t1 # address of a[i] sw $t0, 0($t2) # put i into a[i] addi $t0, $t0, 1 # increment i slt $t5, $t0, $t4 # is $t0 < $t4 ? bne $t5, $0, loop1 # branch if so #/CONT (note sll is the same as multiplying by 2) From http://zeta.albion.edu/~dreimann/Spring2012/courses/cs354/projects/primes.phphttp://zeta.albion.edu/~dreimann/Spring2012/courses/cs354/projects/primes.php

57. # Sum the array values and $s0, $0, $0 # sum = 0 and $t0, $0, $0 # i = 0 add $t2, $t1, $0 # address of a[i] loop2: lw $t3, 0($t2) # load a[i] add $s0, $s0, $t3 # increment sum addi $t0, $t0, 1 # increment i addi $t2, $t2, 4 # increment address of a[i] slt $t5, $t0, $t4 # is $t0 < $t4 ? bne $t5, $0, loop2 # branch if so # Output Sum li $v0, 1 # Load 1=print_int into $v0 add $a0, $s0, $zero # Load first number into $a0 syscall # Output prompt via syscall li $v0, 10 syscall# exit. data n:.word 10 # n = 10.align 4 a:.space 40 # Allocate 10 words (40 bytes) From http://zeta.albion.edu/~dreimann/Spring2012/courses/cs354/projects/primes.phphttp://zeta.albion.edu/~dreimann/Spring2012/courses/cs354/projects/primes.php