How can I multiply and divide by odd constants in arm Assembly?
0
I don’t know how to multiply or divide by an odd constant. In the image (Link to image ), I know how to solve the first two problem (with the left and the right shift) but not the last two.
How can I do the exercise number 3 and number 4?
I resolved the first like this:
ADD r1, r2, lsl #3
So I’d Like something like that for the last two.
assembly arm
edited Nov 23 '18 at 16:02
Stoogy
735724
asked Nov 23 '18 at 12:47
Noemi PecorariNoemi Pecorari
82
add a comment |
0
I don’t know how to multiply or divide by an odd constant. In the image (Link to image ), I know how to solve the first two problem (with the left and the right shift) but not the last two.
How can I do the exercise number 3 and number 4?
I resolved the first like this:
ADD r1, r2, lsl #3
So I’d Like something like that for the last two.
assembly arm
edited Nov 23 '18 at 16:02
Stoogy
735724
asked Nov 23 '18 at 12:47
Noemi PecorariNoemi Pecorari
82
4
You know thatx*5=x*4+xright? Also(3/4)*x=(3*x)/4=(2*x+x)/4. Any problems doing these? PS: these will take more than a single instruction.
– Jester
Nov 23 '18 at 12:53
2
Welcome to StackOverflow! Please be aware that this isn't a homework service - asking a good SO question is about documenting your attempts to solve a problem and then asking for help with something specific that you don't understand, rather than asking for someone to write code for you.
– cooperised
Nov 23 '18 at 13:18
You're only dividing by even constants, powers of 2 in fact, so it's just shifts. Is(3/4) * xreally supposed to be evaluated with integer math? If so, the answer is zero because(3/4)truncates to 0. Otherwise, are you supposed to be keeping some fraction bits in the other cases where you divide by4?
– Peter Cordes
Nov 23 '18 at 18:12
I don’t know I only have to put the value from 3/4 * r2 in r1
– Noemi Pecorari
Nov 25 '18 at 19:39
add a comment |
0
0
0
I don’t know how to multiply or divide by an odd constant. In the image (Link to image ), I know how to solve the first two problem (with the left and the right shift) but not the last two.
How can I do the exercise number 3 and number 4?
I resolved the first like this:
ADD r1, r2, lsl #3
So I’d Like something like that for the last two.
assembly arm
edited Nov 23 '18 at 16:02
Stoogy
735724
asked Nov 23 '18 at 12:47
Noemi PecorariNoemi Pecorari
82
I don’t know how to multiply or divide by an odd constant. In the image (Link to image ), I know how to solve the first two problem (with the left and the right shift) but not the last two.
How can I do the exercise number 3 and number 4?
I resolved the first like this:
ADD r1, r2, lsl #3
So I’d Like something like that for the last two.
assembly arm
assembly arm
edited Nov 23 '18 at 16:02
Stoogy
735724
asked Nov 23 '18 at 12:47
Noemi PecorariNoemi Pecorari
82
edited Nov 23 '18 at 16:02
Stoogy
735724
asked Nov 23 '18 at 12:47
Noemi PecorariNoemi Pecorari
82
edited Nov 23 '18 at 16:02
Stoogy
735724
edited Nov 23 '18 at 16:02
Stoogy
735724
edited Nov 23 '18 at 16:02
Stoogy
735724
735724
asked Nov 23 '18 at 12:47
Noemi PecorariNoemi Pecorari
82
asked Nov 23 '18 at 12:47
Noemi PecorariNoemi Pecorari
82
asked Nov 23 '18 at 12:47
Noemi PecorariNoemi Pecorari
82
82
4
You know thatx*5=x*4+xright? Also(3/4)*x=(3*x)/4=(2*x+x)/4. Any problems doing these? PS: these will take more than a single instruction.
– Jester
Nov 23 '18 at 12:53
2
Welcome to StackOverflow! Please be aware that this isn't a homework service - asking a good SO question is about documenting your attempts to solve a problem and then asking for help with something specific that you don't understand, rather than asking for someone to write code for you.
– cooperised
Nov 23 '18 at 13:18
You're only dividing by even constants, powers of 2 in fact, so it's just shifts. Is(3/4) * xreally supposed to be evaluated with integer math? If so, the answer is zero because(3/4)truncates to 0. Otherwise, are you supposed to be keeping some fraction bits in the other cases where you divide by4?
– Peter Cordes
Nov 23 '18 at 18:12
I don’t know I only have to put the value from 3/4 * r2 in r1
– Noemi Pecorari
Nov 25 '18 at 19:39
add a comment |
4
You know thatx*5=x*4+xright? Also(3/4)*x=(3*x)/4=(2*x+x)/4. Any problems doing these? PS: these will take more than a single instruction.
– Jester
Nov 23 '18 at 12:53
2
Welcome to StackOverflow! Please be aware that this isn't a homework service - asking a good SO question is about documenting your attempts to solve a problem and then asking for help with something specific that you don't understand, rather than asking for someone to write code for you.
– cooperised
Nov 23 '18 at 13:18
You're only dividing by even constants, powers of 2 in fact, so it's just shifts. Is(3/4) * xreally supposed to be evaluated with integer math? If so, the answer is zero because(3/4)truncates to 0. Otherwise, are you supposed to be keeping some fraction bits in the other cases where you divide by4?
– Peter Cordes
Nov 23 '18 at 18:12
I don’t know I only have to put the value from 3/4 * r2 in r1
– Noemi Pecorari
Nov 25 '18 at 19:39
4
4
You know that
x*5=x*4+x right? Also (3/4)*x=(3*x)/4=(2*x+x)/4. Any problems doing these? PS: these will take more than a single instruction.– Jester
Nov 23 '18 at 12:53
You know that
x*5=x*4+x right? Also (3/4)*x=(3*x)/4=(2*x+x)/4. Any problems doing these? PS: these will take more than a single instruction.– Jester
Nov 23 '18 at 12:53
2
2
Welcome to StackOverflow! Please be aware that this isn't a homework service - asking a good SO question is about documenting your attempts to solve a problem and then asking for help with something specific that you don't understand, rather than asking for someone to write code for you.
– cooperised
Nov 23 '18 at 13:18
Welcome to StackOverflow! Please be aware that this isn't a homework service - asking a good SO question is about documenting your attempts to solve a problem and then asking for help with something specific that you don't understand, rather than asking for someone to write code for you.
– cooperised
Nov 23 '18 at 13:18
You're only dividing by even constants, powers of 2 in fact, so it's just shifts. Is
(3/4) * x really supposed to be evaluated with integer math? If so, the answer is zero because (3/4) truncates to 0. Otherwise, are you supposed to be keeping some fraction bits in the other cases where you divide by 4?– Peter Cordes
Nov 23 '18 at 18:12
You're only dividing by even constants, powers of 2 in fact, so it's just shifts. Is
(3/4) * x really supposed to be evaluated with integer math? If so, the answer is zero because (3/4) truncates to 0. Otherwise, are you supposed to be keeping some fraction bits in the other cases where you divide by 4?– Peter Cordes
Nov 23 '18 at 18:12
I don’t know I only have to put the value from 3/4 * r2 in r1
– Noemi Pecorari
Nov 25 '18 at 19:39
I don’t know I only have to put the value from 3/4 * r2 in r1
– Noemi Pecorari
Nov 25 '18 at 19:39
add a comment |
1 Answer
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1) MUL r1, r2, lsl #2 → ADD r1, r1, r2
2) MUL r1, r2, lsl #1 → ADD r1, r1, r2 → MUL r1, r1, lsr #2
answered Nov 25 '18 at 19:47
Andrea CulotAndrea Culot
4918
add a comment |
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1 Answer
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1) MUL r1, r2, lsl #2 → ADD r1, r1, r2
2) MUL r1, r2, lsl #1 → ADD r1, r1, r2 → MUL r1, r1, lsr #2
answered Nov 25 '18 at 19:47
Andrea CulotAndrea Culot
4918
add a comment |
0
1) MUL r1, r2, lsl #2 → ADD r1, r1, r2
2) MUL r1, r2, lsl #1 → ADD r1, r1, r2 → MUL r1, r1, lsr #2
answered Nov 25 '18 at 19:47
Andrea CulotAndrea Culot
4918
add a comment |
0
0
0
1) MUL r1, r2, lsl #2 → ADD r1, r1, r2
2) MUL r1, r2, lsl #1 → ADD r1, r1, r2 → MUL r1, r1, lsr #2
answered Nov 25 '18 at 19:47
Andrea CulotAndrea Culot
4918
1) MUL r1, r2, lsl #2 → ADD r1, r1, r2
2) MUL r1, r2, lsl #1 → ADD r1, r1, r2 → MUL r1, r1, lsr #2
answered Nov 25 '18 at 19:47
Andrea CulotAndrea Culot
4918
answered Nov 25 '18 at 19:47
Andrea CulotAndrea Culot
4918
answered Nov 25 '18 at 19:47
Andrea CulotAndrea Culot
4918
answered Nov 25 '18 at 19:47
Andrea CulotAndrea Culot
4918
4918
add a comment |
add a comment |
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4
You know that
x*5=x*4+xright? Also(3/4)*x=(3*x)/4=(2*x+x)/4. Any problems doing these? PS: these will take more than a single instruction.– Jester
Nov 23 '18 at 12:53
2
Welcome to StackOverflow! Please be aware that this isn't a homework service - asking a good SO question is about documenting your attempts to solve a problem and then asking for help with something specific that you don't understand, rather than asking for someone to write code for you.
– cooperised
Nov 23 '18 at 13:18
You're only dividing by even constants, powers of 2 in fact, so it's just shifts. Is
(3/4) * xreally supposed to be evaluated with integer math? If so, the answer is zero because(3/4)truncates to 0. Otherwise, are you supposed to be keeping some fraction bits in the other cases where you divide by4?– Peter Cordes
Nov 23 '18 at 18:12
I don’t know I only have to put the value from 3/4 * r2 in r1
– Noemi Pecorari
Nov 25 '18 at 19:39