Arm to mips
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Introduction
In this page there two code example for ARM and MIPS architecture.Main idea is to see the difference between ARM and MIPS. Here only the MIPs assembly is commented. A good read about MIPS Architecture and Assembly Language
MIPS
The MIPS instruction set acknowledges 32 general-purpose registers in the register file. For most processors implementing the MIPS instruction set architecture, each register is 32 bits in size. Registers are designated using the “$” symbol. For all practical purposes, three of these registers have special functionality ($0,29,$31). Also, it should be noted that these registers can be accessed via standardized naming conventions in software. The C/C++ library file “regdef.h” is an implementation of one such naming convention.
Example1 code
int binary(int a, int b)
{
return a + b;
}
void stk(void)
{
binary(binary(binary(1, 2), binary(3, 4)), binary(binary(5, 6), binary(7, 8)));
}
Example1 Mips
binary:
j $31 # return
addu $2,$4,$5 # r2 = a + b
stk:
subu $sp,$sp,32 # allocate space for local vars & 4 slots
li $4,0x00000001 # 1
li $5,0x00000002 # 2
sw $31,24($sp) # store return address on stack
sw $17,20($sp) # preserve r17 on stack
jal binary # call binary(1,2)
sw $16,16($sp) # preserve r16 on stack
li $4,0x00000003 # 3
li $5,0x00000004 # 4
jal binary # call binary(3,4)
move $16,$2 # r16 = binary(1,2)
move $4,$16 # r4 = binary(1,2)
jal binary # call binary(binary(1,2), binary(3,4))
move $5,$2 # r5 = binary(3,4)
li $4,0x00000005 # 5
li $5,0x00000006 # 6
jal binary # call binary(5,6)
move $17,$2 # r17 = binary(binary(1,2), binary(3,4))
li $4,0x00000007 # 7
li $5,0x00000008 # 8
jal binary # call binary(7,8)
move $16,$2 # r16 = binary(5,6)
move $4,$16 # r4 = binary(5,6)
jal binary # call binary(binary(5,6), binary(7,8))
move $5,$2 # r5 = binary(7,8)
move $4,$17 # r4 = binary(binary(1,2), binary(3,4))
jal binary # call binary(binary(binary(1,2), binary(3,4)), binary(binary(5,6), binary(7,8)))
move $5,$2 # r5 = binary(binary(5,6), binary(7,8))
lw $31,24($sp) # restore return address from stack
lw $17,20($sp) # restore r17 from stack
lw $16,16($sp) # restore r16 from stack
addu $sp,$sp,32 # remove local vars and 4 slots
j $31 # return
nop
Example1 ARM
binary(int, int):
push {r7}
sub sp, sp, #12
add r7, sp, #0
str r0, [r7, #4]
str r1, [r7, #0]
ldr r2, [r7, #4]
ldr r3, [r7, #0]
adds r3, r2, r3
mov r0, r3
add r7, r7, #12
mov sp, r7
pop {r7}
bx lr
stk():
push {r4, r5, r7, lr}
add r7, sp, #0
mov r0, #1
mov r1, #2
bl binary(int, int)
mov r4, r0
mov r0, #3
mov r1, #4
bl binary(int, int)
mov r3, r0
mov r0, r4
mov r1, r3
bl binary(int, int)
mov r4, r0
mov r0, #5
mov r1, #6
bl binary(int, int)
mov r5, r0
mov r0, #7
mov r1, #8
bl binary(int, int)
mov r3, r0
mov r0, r5
mov r1, r3
bl binary(int, int)
mov r3, r0
mov r0, r4
mov r1, r3
bl binary(int, int)
pop {r4, r5, r7, pc}
Example2 code
int fib(int n) {
if (n == 0)
return 0;
else if (n == 1)
return 1;
return fib(n - 1) + fib(n - 2);
}
Example2 ARM gcc 4.5.3
fib(int):
push {r4, r7, lr}
sub sp, sp, #12
add r7, sp, #0
str r0, [r7, #4]
ldr r3, [r7, #4]
cmp r3, #0
bne .L2
mov r3, #0
b .L3
.L2:
ldr r3, [r7, #4]
cmp r3, #1
bne .L4
mov r3, #1
b .L3
.L4:
ldr r3, [r7, #4]
add r3, r3, #-1
mov r0, r3
bl fib(int)
mov r3, r0
mov r4, r3
ldr r3, [r7, #4]
sub r3, r3, #2
mov r0, r3
bl fib(int)
mov r3, r0
adds r3, r4, r3
.L3:
mov r0, r3
add r7, r7, #12
mov sp, r7
pop {r4, r7, pc}
Example2 MIPs gcc 5.4
$LFB0 = .
fib(int):
addiu $sp,$sp,-40
sw $31,36($sp)
sw $fp,32($sp)
sw $16,28($sp)
move $fp,$sp
sw $4,40($fp)
lw $2,40($fp)
bne $2,$0,$L2
nop
move $2,$0
b $L3
nop
lw $3,40($fp)
li $2,1 # 0x1
bne $3,$2,$L4
nop
li $2,1 # 0x1
b $L3
nop
lw $2,40($fp)
addiu $2,$2,-1
move $4,$2
jal fib(int)
nop
move $16,$2
lw $2,40($fp)
addiu $2,$2,-2
move $4,$2
jal fib(int)
nop
addu $2,$16,$2
move $sp,$fp
lw $31,36($sp)
lw $fp,32($sp)
lw $16,28($sp)
addiu $sp,$sp,40
j $31
nop
Example2 MIPs Stack
.text main: # Prompt user to input non-negative number la $a0,prompt li $v0,4 syscall li $v0,5 #Read the number(n) syscall move $t2,$v0 # n to $t2 # Call function to get fibonnacci #n move $a0,$t2 move $v0,$t2 jal fib #call fib (n) move $t3,$v0 #result is in $t3 # Output message and n la $a0,result #Print F_ li $v0,4 syscall move $a0,$t2 #Print n li $v0,1 syscall la $a0,result2 #Print = li $v0,4 syscall move $a0,$t3 #Print the answer li $v0,1 syscall la $a0,endl #Print '\n' li $v0,4 syscall # End program li $v0,10 syscall fib: # Compute and return fibonacci number beqz $a0,zero #if n=0 return 0 beq $a0,1,one #if n=1 return 1 #Calling fib(n-1) sub $sp,$sp,4 #storing return address on stack sw $ra,0($sp) sub $a0,$a0,1 #n-1 jal fib #fib(n-1) add $a0,$a0,1 lw $ra,0($sp) #restoring return address from stack add $sp,$sp,4 sub $sp,$sp,4 #Push return value to stack sw $v0,0($sp) #Calling fib(n-2) sub $sp,$sp,4 #storing return address on stack sw $ra,0($sp) sub $a0,$a0,2 #n-2 jal fib #fib(n-2) add $a0,$a0,2 lw $ra,0($sp) #restoring return address from stack add $sp,$sp,4 #--------------- lw $s7,0($sp) #Pop return value from stack add $sp,$sp,4 add $v0,$v0,$s7 # f(n - 2)+fib(n-1) jr $ra # decrement/next in stack zero: li $v0,0 jr $ra one: li $v0,1 jr $ra .data prompt: .asciiz "This program calculates Fibonacci sequence with recursive functions.\nEnter a non-negative number: " result: .asciiz "F_" result2: .asciiz " = " endl: .asciiz "\n"
