1 / 31

MIPS Assembly Tutorial

MIPS Assembly Tutorial. Types of Instructions. There are 3 main types of assembly instructions Arithmetic - add, sub, mul, shifts, and, or, etc. Load/store Conditional - branches. Arithmetic Instructions. add a, b, c a = b+c add a, a, d a = d+a = d+b+c add a, a, e a = e+a = e+d+b+c.

vallari-karn
Download Presentation

MIPS Assembly Tutorial

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. MIPS Assembly Tutorial

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

  3. Arithmetic Instructions add a, b, c a = b+c add a, a, d a = d+a = d+b+c add a, a, e a = 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

  4. 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

  5. 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

  6. 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

  7. 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)

  8. 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

  9. 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: Offset This is simplified, more details later… lw r4, 8(r3) add r1, r2, r4 Base Address

  10. 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

  11. 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

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

  13. 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)

  14. 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]

  15. 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.

  16. 000000 0 00000 0 00001 1 00010 2 00000 0 32 100000 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 binary 6 bits 5 bits 5 bits 5 bits 5 bits 6 bits Each segment is referred to as a field. Details to come….

  17. op rs rt rd shamt funct MIPS Fields • MIPS fields are giving names to make them easier to discuss 6 bits 5 bits 5 bits 5 bits 5 bits 6 bits • 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

  18. 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

  19. op rs rt address 6 bits 5 bits 5 bits 16 bits MIPS Fields • I-type (I-format) • I=immediate • Now LW can specify an address up to 16-bits • Opcode determines the format

  20. MIPS Instruction Encoding

  21. 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)

  22. 101011 100011 43 35 00000 00000 0 0 00001 1 1 00001 0000 0100 1011 0000 1200 1200 0000 0100 1011 0000 0 000000 00000 0 00010 2 0 00000 00000 0 32 32 MIPS Asm -> Machine Language lw r0, 1200(r1) add r0, r2, r0 sw r0, 1200(r1) Solution: decimal Address/shamt op rs rt rd funct binary

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

  24. 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

  25. 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:

  26. 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

  27. 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

  28. 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:

  29. 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

  30. 001000 8 00100 4 4 00100 0000 0000 0000 0100 4 Immediate Operands Example: What is the machine code for the following? (Remember the I-format instruction) addi r4, r4, 4 Solution: decimal op rs rt Immediate binary

  31. opcode Target address 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

More Related