# Week 10: Assembly Programming

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1 Week 10: Assembly Programming

2 Arithmetic instructions Instruction Opcode/Function Syntax Operation add \$d, \$s, \$t \$d = \$s + \$t addu \$d, \$s, \$t \$d = \$s + \$t addi \$t, \$s, i \$t = \$s + SE(i) addiu \$t, \$s, i \$t = \$s + SE(i) div \$s, \$t lo = \$s / \$t; hi = \$s % \$t divu \$s, \$t lo = \$s / \$t; hi = \$s % \$t mult \$s, \$t hi:lo = \$s * \$t multu \$s, \$t hi:lo = \$s * \$t sub \$d, \$s, \$t \$d = \$s - \$t subu \$d, \$s, \$t \$d = \$s - \$t Note: hi and lo refer to the high and low bits referred to in the register slide. SE = sign extend.

3 Logical instructions Instruction Opcode/Function Syntax Operation and \$d, \$s, \$t \$d = \$s & \$t andi \$t, \$s, i \$t = \$s & ZE(i) nor \$d, \$s, \$t \$d = ~(\$s \$t) or \$d, \$s, \$t \$d = \$s \$t ori \$t, \$s, i \$t = \$s ZE(i) xor \$d, \$s, \$t \$d = \$s ^ \$t xori \$t, \$s, i \$t = \$s ^ ZE(i) Note: ZE = zero extend (pad upper bits with 0 value).

4 Shift instructions Instruction Opcode/Function Syntax Operation sll \$d, \$t, a \$d = \$t << a sllv \$d, \$t, \$s \$d = \$t << \$s sra \$d, \$t, a \$d = \$t >> a srav \$d, \$t, \$s \$d = \$t >> \$s srl \$d, \$t, a \$d = \$t >>> a srlv \$d, \$t, \$s \$d = \$t >>> \$s Note: srl = shift right logical, and sra = shift right arithmetic. The v denotes a variable number of bits, specified by \$s.

5 Data movement instructions Instruction Opcode/Function Syntax Operation mfhi \$d \$d = hi mflo \$d \$d = lo mthi \$s hi = \$s mtlo \$s lo = \$s These are instructions for operating on the HI and LO registers described earlier.

6 QtSpim MIPS Simulator ú Allow us to run & test assembly code ú Can run program or step-by-step ú Can view registers in binary/decimal/hex ú Not particularly well documented May need to fiddle with settings to get it working properly Main thing: want a simple (not a bare) machine

7 Formatting Assembly Code 3 columns ú Labels ú Code ú Comments Start with.text (we ll see other options later) First line of code to run = label: main

8 Time to write our first assembly program

9 Time for more instructions!

10 Control flow in assembly Not all programs follow a linear set of instructions. ú Some operations require the code to branch to one section of code or another (if/else). ú Some require the code to jump back and repeat a section of code again (for/while). For this, we have labels on the left-hand side that indicate the points that the program flow might need to jump to. ú References to these points in the assembly code are resolved at compile time to offset values for the program counter.

11 Branch instructions Instruction Opcode/Function Syntax Operation beq \$s, \$t, label if (\$s == \$t) pc ß label bgtz \$s, label if (\$s > 0) pc ß label blez \$s, label if (\$s <= 0) pc ß label bne \$s, \$t, label if (\$s!= \$t) pc ß label Branch operations are key when implementing if statements and while loops. The labels are memory locations, assigned to each label at compile time.

12 Jump instructions Instruction Opcode/Function Syntax Operation j label pc ß label jal label \$ra = pc; pc ß label jalr \$s \$ra = pc; pc = \$s jr \$s pc = \$s jal = jump and link. ú Register \$31 (aka \$ra) stores the address that s used when returning from a subroutine. Note: jr and jalr are not j-type instructions.

13 Comparison instructions Instruction Opcode/Function Syntax Operation slt \$d, \$s, \$t \$d = (\$s < \$t) sltu \$d, \$s, \$t \$d = (\$s < \$t) slti \$t, \$s, i \$t = (\$s < SE(i)) sltiu \$t, \$s, i \$t = (\$s < SE(i)) Note: Comparison operation stores a one in the destination register if the less-than comparison is true, and stores a zero in that location otherwise.

14 If/Else statements in MIPS if ( i == j ) i++; else j--; j += i; Strategy for if/else statements: ú Test condition, and jump to if logic block whenever condition is true. ú Otherwise, perform else logic block, and jump to first line after if logic block.

15 Translated if/else statements # \$t1 = i, \$t2 = j main: beq \$t1, \$t2, IF # branch if ( i == j ) addi \$t2, \$t2, -1 # j-- j END # jump over IF IF: addi \$t1, \$t1, 1 # i++ END: add \$t2, \$t2, \$t1 # j += i Alternately, you can branch on the else condition first: # \$t1 = i, \$t2 = j main: bne \$t1, \$t2, ELSE # branch if! ( i == j ) addi \$t1, \$t1, 1 # i++ j END # jump over ELSE ELSE: addi \$t2, \$t2, -1 # j-- END: add \$t2, \$t2, \$t1 # j += i

16 A trick with if statements Use flow charts to help you sort out the control flow of the code: if ( i == j ) i++; else j--; j += i; false beq else block true # \$t1 = i, \$t2 = j main: beq \$t1, \$t2, IF addi \$t2, \$t2, -1 j END IF: addi \$t1, \$t1, 1 END: add \$t2, \$t2, \$t1 jump if block end

17 Break

18 Multiple if conditions if ( i == j or i == k ) i++ ; // if-body else j-- ; // else-body j = i + k ;

19 Multiple if conditions if ( i == j or i == k ) i++ ; // if-body else j-- ; // else-body j = i + k ; Branch statement for each condition: # \$t1 = i, \$t2 = j, \$t3 = k main: beq \$t1, \$t2, IF # cond1: branch if ( i == j ) bne \$t1, \$t3, ELSE # cond2: branch if ( i!= k ) IF: addi \$t1, \$t1, 1 # if (i==j i==k) à i++ j END # jump over else ELSE: addi \$t2, \$t2, -1 # else-body: j-- END: add \$t2, \$t1, \$t3 # j = i + k

20 main: START: END: add \$t0, \$zero, \$zero addi \$t1, \$zero, 100 beq \$t0, \$t1, END addi \$t0, \$t0, 1 j START

21 Loops in MIPS Example of a simple loop, in assembly: main: START: END: add \$t0, \$zero, \$zero addi \$t1, \$zero, 100 beq \$t0, \$t1, END addi \$t0, \$t0, 1 j START which is the same as saying (in C): int i = 0; while (i < 100) { i++; }

22 Loops in MIPS for ( <init> ; <cond> ; <update> ) { <for body> } For loops (such as above) are usually implemented with the following structure: main: START: <init> if (!<cond>) branch to END <for-body> UPDATE: <update> jump to START END:

23 Loop example in MIPS for ( i=0 ; i<100 ; i++ ) { j = j + i; } This translates to: # \$t0 = i, \$t1 = j main: add \$t0, \$zero, \$zero # set i to 0 add \$t1, \$zero, \$zero # set j to 0 addi \$t9, \$zero, 100 # set \$t9 to 100 START: beq \$t0, \$t9, EXIT # branch if i==100 add \$t1, \$t1, \$t0 # j = j + i UPDATE: addi \$t0, \$t0, 1 # i++ j START EXIT: while loops are the same, without the initialization and update sections.

24 Homework Fibonacci sequence: ú How would you convert this into assembly? int n = 10; int f1 = 1, f2 = 1; while (n!= 0) { f1 = f1 + f2; f2 = f1 f2; n = n 1; } # result is f1

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