Create two HLA Assembly language programs: 1.Create an HLA Assembly language pro
ID: 3913766 • Letter: C
Question
Create two HLA Assembly language programs:
1.Create an HLA Assembly language program that prompts for two values from the user. Print a number pattern where both numbers are displayed a certain number of times that is controlled by the second value entered. If either number entered is zero or less, don't print anything.
Here are some example program dialogues to guide your efforts:
Provide a first number: 12
Provide an second number: 5
125_125_125_125_125
Provide a first number: 44
Provide an second number: 1
441
Provide a first number: 12
Provide an second number: -5
Provide a first number: -1
Provide an second number: 12
In an effort to help you focus on building an Assembly program, I’d like to offer you the following C statements which match the program specifications stated above. If you like, use them as the basis for building your Assembly program.
SAMPLE C CODE:
------------------------
int i, first, second;
printf( "Provide a first number: " );
scanf( &first );
printf( "Provide an second number: " );
scanf( &second);
if (first > 0) {
for (i = 1; i >= second; i++ )
{
if (i != 1)
{
printf( "_" );
}
printf( "%d", first );
printf( "%d", second );
}
printf( " " );
}
2.
Create an HLA Assembly language program that prompts for a single integer value from the user and prints an arrow pattern like the one shown below. If the number is negative, don't print anything at all.
Here are some example program dialogues to guide your efforts:
Feed Me: 3
X
XX
XXX
XX
X
Feed Me: -6
Feed Me: 5
X
XX
XXX
XXXX
XXXXX
XXXX
XXX
XX
X
In an effort to help you focus on building an Assembly program, I’d like to offer you the following C statements matches the program specifications stated above. If you like, use them as the basis for building your Assembly program.
SAMPLE C CODE:
------------------------
int i, j, n;
printf( "Feed Me:" );
scanf( "%d", &n );
// the top half of the arrow...
for( i = 1; i <= n; i++ ) {
for (j = 1; j <= i; j++) {
printf( "X" );
}
printf( " " );
}
// the bottom half of the arrow...
for (i = n - 1; i >= 1; i--) {
for (j = 1; j <= i; j++) {
printf( "X" );
}
printf( " " );
}
Explanation / Answer
Assembly Program for first sample program:
-------------------------------------------------------------------------------------------------------------
.LC0:
.string "Provide a first number: "
.LC1:
.string "%d"
.LC2:
.string "Provide an second number: "
main:
push rbp
mov rbp, rsp
sub rsp, 16
mov edi, OFFSET FLAT:.LC0
mov eax, 0
call printf
lea rax, [rbp-8]
mov rsi, rax
mov edi, OFFSET FLAT:.LC1
mov eax, 0
call scanf
mov edi, OFFSET FLAT:.LC2
mov eax, 0
call printf
lea rax, [rbp-12]
mov rsi, rax
mov edi, OFFSET FLAT:.LC1
mov eax, 0
call scanf
mov eax, DWORD PTR [rbp-8]
test eax, eax
jle .L2
mov DWORD PTR [rbp-4], 1
.L5:
mov eax, DWORD PTR [rbp-12]
cmp DWORD PTR [rbp-4], eax
jg .L3
cmp DWORD PTR [rbp-4], 1
je .L4
mov edi, 95
call putchar
.L4:
mov eax, DWORD PTR [rbp-8]
mov esi, eax
mov edi, OFFSET FLAT:.LC1
mov eax, 0
call printf
mov eax, DWORD PTR [rbp-12]
mov esi, eax
mov edi, OFFSET FLAT:.LC1
mov eax, 0
call printf
add DWORD PTR [rbp-4], 1
jmp .L5
.L3:
mov edi, 10
call putchar
.L2:
mov eax, 0
leave
ret
---------------------------------------------------------------------------------------------
Assembly program for sample program 2:
---------------------------------------------------------------------------------------------
.LC0:
.string "Feed Me:"
.LC1:
.string "%d"
main:
push rbp
mov rbp, rsp
sub rsp, 16
mov edi, OFFSET FLAT:.LC0
mov eax, 0
call printf
lea rax, [rbp-12]
mov rsi, rax
mov edi, OFFSET FLAT:.LC1
mov eax, 0
call scanf
mov DWORD PTR [rbp-4], 1
.L5:
mov eax, DWORD PTR [rbp-12]
cmp DWORD PTR [rbp-4], eax
jg .L2
mov DWORD PTR [rbp-8], 1
.L4:
mov eax, DWORD PTR [rbp-8]
cmp eax, DWORD PTR [rbp-4]
jg .L3
mov edi, 88
call putchar
add DWORD PTR [rbp-8], 1
jmp .L4
.L3:
mov edi, 10
call putchar
add DWORD PTR [rbp-4], 1
jmp .L5
.L2:
mov eax, DWORD PTR [rbp-12]
sub eax, 1
mov DWORD PTR [rbp-4], eax
.L9:
cmp DWORD PTR [rbp-4], 0
jle .L6
mov DWORD PTR [rbp-8], 1
.L8:
mov eax, DWORD PTR [rbp-8]
cmp eax, DWORD PTR [rbp-4]
jg .L7
mov edi, 88
call putchar
add DWORD PTR [rbp-8], 1
jmp .L8
.L7:
mov edi, 10
call putchar
sub DWORD PTR [rbp-4], 1
jmp .L9
.L6:
mov eax, 0
leave
ret
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