Arm to mips: Difference between revisions

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Created page with "==source code == example using python def fib(n): a,b = 1,1 for i in range(n-1): a,b = b,a+b return a"
 
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==source code ==
==Introduction==
example using python


def fib(n):
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 [http://logos.cs.uic.edu/366/notes/mips%20quick%20tutorial.htm MIPS Architecture and Assembly Language]
     a,b = 1,1
 
     for i in range(n-1):
'''MIPS'''
         a,b = b,a+b
 
     return a
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 ==
<pre>
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)));
}
</pre>
 
==Example1 Mips ==
 
<pre>
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
</pre>
 
 
==Example1 ARM ==
 
<pre>
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}
 
</pre>
 
==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==
<pre>
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}
 
</pre>
 
 
==Example2 MIPs gcc 5.4==
 
<pre>
$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
 
 
</pre>
 
==Example2 MIPs Stack==
 
<pre> 
  .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"
</pre>

Latest revision as of 04:54, 9 September 2016

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"