Memory Operations and Arrays in ARM Assembly

Question

Open the find_min_array.s file, and open it up in a text editor of your choice. Note that word processors (e.g., Microsoft Word, Pages, Google Docs) will probably not work for this purpose, as you must save your file as plain text. You must write ARM assembly code which will find and print out the smallest element of the array, where the array is specified with the array label and the array length is specified with the array_length label. Example output of this code is shown below, based on the provided array and array length in find_min_array.s:

-5

Multiple different implementation approaches are possible. One such implementation approach is shown below, implemented in pseudocode:

min = array[0];

for each element of the array:

if element < min:

min = element;

endif

endfor

Open the add_amount_array.s file, and open it up in a text editor of your choice. Note that word processors (e.g., Microsoft Word, Pages, Google Docs) will probably not work for this purpose, as you must save your file as plain text. This program will read values from a source array, add a specified amount to each value, and put the result in another (sink) array. A number of definitions are provided in the file, summarized below:

1. array_source: The source array to read from
2. array_sink: The destination array to write results to
3. array_length: The length of the arrays. It is assumed that the array_source and array_sink arrays have the same length.
4. add_amount: The amount to add to each element

Multiple different implementation approaches are possible. One such implementation approach is shown below, implemented in pseudocode:

counter = 0;

while counter < array_length:

array_sink[counter] = array_source[counter] + add_amount;

counter++;

endwhile

Answer 1

#include <stdio.h>

#include <stdlib.h>

void Go(double * source, double * sink, int nSize, double incrAmt, double * minVal)

{

int iLoop;

*minVal = source[0];

for (iLoop=0; iLoop<nSize; iLoop++)

{

sink[iLoop] = source[iLoop]+incrAmt;

if (source[iLoop]<*minVal)

{

*minVal = source[iLoop];

}

}

}

int main()

{

double A[] = {35.42, 2.723, 42.35, 3.1415926, 7.11, 7.24, 7.25, 13.59, 24.36, 36.32};

double* a;

int N=10;

double deltaX=5.0f;

double minVal;

int iLoop;

a = (double*)malloc(N*sizeof(double));

Go(A,a,N,deltaX,&minVal);

for (iLoop=0; iLoop<N; iLoop++)

{

printf("%8.6lf \n",a[iLoop]);

}

printf(" the min is %6.2f \n",minVal);

free(a);

}

/*******************************************************************************/

'*****************************************************************************

Go:

push {r11, lr}

mov r11, sp

sub sp, sp, #40

ldr r3, [r11, #12]

ldr r12, [r11, #8]

ldr lr, [r11, #16]

str r0, [r11, #-4]

str r1, [r11, #-8]

str r2, [r11, #-12]

str r3, [sp, #20]

str r12, [sp, #16]

ldr r0, [r11, #-4]

ldr r1, [r0]

ldr r0, [r0, #4]

ldr r2, [r11, #16]

str r0, [r2, #4]

str r1, [r2]

mov r0, #0

str r0, [sp, #12]

b .LBB0_1

.LBB0_1: @ =>This Inner Loop Header: Depth=1

ldr r0, [sp, #12]

ldr r1, [r11, #-12]

cmp r0, r1

bge .LBB0_6

b .LBB0_2

.LBB0_2: @ in Loop: Header=BB0_1 Depth=1

ldr r0, [r11, #-4]

ldr r1, [sp, #12]

ldr r2, [r0, r1, lsl #3]!

ldr r0, [r0, #4]

ldr r3, [sp, #16]

ldr r12, [sp, #20]

str r0, [sp, #8] @ 4-byte Spill

mov r0, r2

ldr r2, [sp, #8] @ 4-byte Reload

str r1, [sp, #4] @ 4-byte Spill

mov r1, r2

mov r2, r3

mov r3, r12

bl __aeabi_dadd

ldr r2, [r11, #-8]

ldr r3, [sp, #4] @ 4-byte Reload

str r0, [r2, r3, lsl #3]!

str r1, [r2, #4]

ldr r0, [r11, #-4]

ldr r1, [sp, #12]

ldr r1, [r0, r1, lsl #3]!

ldr r0, [r0, #4]

ldr r2, [r11, #16]

ldr r12, [r2]

ldr r3, [r2, #4]

str r0, [sp] @ 4-byte Spill

mov r0, r1

ldr r1, [sp] @ 4-byte Reload

mov r2, r12

bl __aeabi_dcmplt

cmp r0, #0

beq .LBB0_4

b .LBB0_3

.LBB0_3: @ in Loop: Header=BB0_1 Depth=1

ldr r0, [r11, #-4]

ldr r1, [sp, #12]

ldr r1, [r0, r1, lsl #3]!

ldr r0, [r0, #4]

ldr r2, [r11, #16]

str r0, [r2, #4]

str r1, [r2]

b .LBB0_4

.LBB0_4: @ in Loop: Header=BB0_1 Depth=1

b .LBB0_5

.LBB0_5: @ in Loop: Header=BB0_1 Depth=1

ldr r0, [sp, #12]

add r0, r0, #1

str r0, [sp, #12]

b .LBB0_1

.LBB0_6:

mov sp, r11

pop {r11, lr}

bx lr

main:

push {r11, lr}

mov r11, sp

sub sp, sp, #144

mov r0, #0

str r0, [r11, #-4]

ldr r1, .LCPI1_0

add r2, sp, #56

mov r3, #80

str r0, [sp, #24] @ 4-byte Spill

mov r0, r2

str r2, [sp, #20] @ 4-byte Spill

mov r2, r3

bl memcpy

mov r1, #10

str r1, [sp, #48]

mov r1, #1310720

orr r1, r1, #1073741824

str r1, [sp, #44]

ldr r1, [sp, #24] @ 4-byte Reload

str r1, [sp, #40]

ldr r2, [sp, #48]

lsl r2, r2, #3

str r0, [sp, #16] @ 4-byte Spill

mov r0, r2

bl malloc

str r0, [sp, #52]

ldr r1, [sp, #52]

ldr r2, [sp, #48]

ldr r0, [sp, #40]

ldr r3, [sp, #44]

mov r12, sp

add lr, sp, #32

str lr, [r12, #8]

str r3, [r12, #4]

str r0, [r12]

ldr r0, [sp, #20] @ 4-byte Reload

bl Go

ldr r0, [sp, #24] @ 4-byte Reload

str r0, [sp, #28]

b .LBB1_1

.LBB1_1: @ =>This Inner Loop Header: Depth=1

ldr r0, [sp, #28]

ldr r1, [sp, #48]

cmp r0, r1

bge .LBB1_4

b .LBB1_2

.LBB1_2: @ in Loop: Header=BB1_1 Depth=1

ldr r0, [sp, #52]

ldr r1, [sp, #28]

ldr r2, [r0, r1, lsl #3]!

ldr r3, [r0, #4]

ldr r0, .LCPI1_2

bl printf

b .LBB1_3

.LBB1_3: @ in Loop: Header=BB1_1 Depth=1

ldr r0, [sp, #28]

add r0, r0, #1

str r0, [sp, #28]

b .LBB1_1

.LBB1_4:

ldr r2, [sp, #32]

ldr r3, [sp, #36]

ldr r0, .LCPI1_1

bl printf

ldr r1, [sp, #52]

str r0, [sp, #12] @ 4-byte Spill

mov r0, r1

bl free

ldr r0, [r11, #-4]

mov sp, r11

pop {r11, lr}

bx lr

.LCPI1_0:

.long .L__const.main.A

.LCPI1_1:

.long .L.str.1

.LCPI1_2:

.long .L.str

.L__const.main.A:

.long 2405181686 @ double 35.420000000000002

.long 1078048194

.long 962072674 @ double 2.7229999999999999

.long 1074120884

.long 3435973837 @ double 42.350000000000001

.long 1078275276

.long 1293080650 @ double 3.1415926000000001

.long 1074340347

.long 3607772529 @ double 7.1100000000000003

.long 1075605667

.long 2405181686 @ double 7.2400000000000002

.long 1075639746

.long 0 @ double 7.25

.long 1075642368

.long 2061584302 @ double 13.59

.long 1076571668

.long 4123168604 @ double 24.359999999999999

.long 1077435432

.long 3264175145 @ double 36.32

.long 1078077685

.L.str:

.asciz "%8.6lf \n"

.L.str.1:

.asciz " the min is %6.2f \n"