Which arithmetic operations are the same on unsigned and two's complement signed numbers?Assembly: is numeric operations for signed different from that of unsigned?Is there a single instruction to add numbers in U2 format in assembly?Why doesn't C have unsigned floats?Why are signed and unsigned multiplication different instructions on x86(-64)?how does c compiler handle unsigned and signed integer? Why the assembly code for unsigned and signed arithmetic operation are the same?Difference between signed and unsigned on bitwise operationsWhy would we use addiu instead of addi?Which 2's complement integer operations can be used without zeroing high bits in the inputs, if only the low part of the result is wanted?Why does ARM distinguish between SDIV and UDIV but not with ADD, SUB and MUL?How to handle signed integers in ARM?What is “2's Complement”?Why prefer two's complement over sign-and-magnitude for signed numbers?two's complement, why the name “two”Performing arithmetic operations in binary using only bitwise operatorsHow to divide two negative numbers using the two's complement method?How to get the signed complement of a number?Is the two's complement always a signed number?How to add hex numbers using two's complement?Why does Javascript treat a number as a two's complementAre the bytes stored in 6502 memory signed or unsigned?
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Which arithmetic operations are the same on unsigned and two's complement signed numbers?
Assembly: is numeric operations for signed different from that of unsigned?Is there a single instruction to add numbers in U2 format in assembly?Why doesn't C have unsigned floats?Why are signed and unsigned multiplication different instructions on x86(-64)?how does c compiler handle unsigned and signed integer? Why the assembly code for unsigned and signed arithmetic operation are the same?Difference between signed and unsigned on bitwise operationsWhy would we use addiu instead of addi?Which 2's complement integer operations can be used without zeroing high bits in the inputs, if only the low part of the result is wanted?Why does ARM distinguish between SDIV and UDIV but not with ADD, SUB and MUL?How to handle signed integers in ARM?What is “2's Complement”?Why prefer two's complement over sign-and-magnitude for signed numbers?two's complement, why the name “two”Performing arithmetic operations in binary using only bitwise operatorsHow to divide two negative numbers using the two's complement method?How to get the signed complement of a number?Is the two's complement always a signed number?How to add hex numbers using two's complement?Why does Javascript treat a number as a two's complementAre the bytes stored in 6502 memory signed or unsigned?
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I'm designing a simple toy instruction set and accompanying emulator, and I'm trying to figure out what instructions to support. In the way of arithmetic, I currently have unsigned add, subtract, multiply, and divide. However, I can't seem to find a definitive answer to the following question: Which of the arithmetic operators need signed versions, and for which are the unsigned and two's complement signed versions equivalent?
So, for example, 1111 in two's complement is equal to -1. If you add 1 to it and pretend that it's an unsigned number , you get 0000, which is correct even when thinking of it as -1. However, does that hold for all numbers? And what about for the other three operations (subtraction, multiplication, division)?
binary twos-complement instructions instruction-set
edited Oct 14 '18 at 13:00
phuclv
17.1k957242
asked Jan 31 '14 at 8:16
joshlfjoshlf
10.7k34763
add a comment |
I'm designing a simple toy instruction set and accompanying emulator, and I'm trying to figure out what instructions to support. In the way of arithmetic, I currently have unsigned add, subtract, multiply, and divide. However, I can't seem to find a definitive answer to the following question: Which of the arithmetic operators need signed versions, and for which are the unsigned and two's complement signed versions equivalent?
So, for example, 1111 in two's complement is equal to -1. If you add 1 to it and pretend that it's an unsigned number , you get 0000, which is correct even when thinking of it as -1. However, does that hold for all numbers? And what about for the other three operations (subtraction, multiplication, division)?
binary twos-complement instructions instruction-set
edited Oct 14 '18 at 13:00
phuclv
17.1k957242
asked Jan 31 '14 at 8:16
joshlfjoshlf
10.7k34763
Somewhat related: Which 2's complement integer operations can be used without zeroing high bits in the inputs, if only the low part of the result is wanted?
– Peter Cordes
Jul 9 '18 at 23:02
add a comment |
I'm designing a simple toy instruction set and accompanying emulator, and I'm trying to figure out what instructions to support. In the way of arithmetic, I currently have unsigned add, subtract, multiply, and divide. However, I can't seem to find a definitive answer to the following question: Which of the arithmetic operators need signed versions, and for which are the unsigned and two's complement signed versions equivalent?
So, for example, 1111 in two's complement is equal to -1. If you add 1 to it and pretend that it's an unsigned number , you get 0000, which is correct even when thinking of it as -1. However, does that hold for all numbers? And what about for the other three operations (subtraction, multiplication, division)?
binary twos-complement instructions instruction-set
edited Oct 14 '18 at 13:00
phuclv
17.1k957242
asked Jan 31 '14 at 8:16
joshlfjoshlf
10.7k34763
I'm designing a simple toy instruction set and accompanying emulator, and I'm trying to figure out what instructions to support. In the way of arithmetic, I currently have unsigned add, subtract, multiply, and divide. However, I can't seem to find a definitive answer to the following question: Which of the arithmetic operators need signed versions, and for which are the unsigned and two's complement signed versions equivalent?
So, for example, 1111 in two's complement is equal to -1. If you add 1 to it and pretend that it's an unsigned number , you get 0000, which is correct even when thinking of it as -1. However, does that hold for all numbers? And what about for the other three operations (subtraction, multiplication, division)?
binary twos-complement instructions instruction-set
binary twos-complement instructions instruction-set
edited Oct 14 '18 at 13:00
phuclv
17.1k957242
asked Jan 31 '14 at 8:16
joshlfjoshlf
10.7k34763
edited Oct 14 '18 at 13:00
phuclv
17.1k957242
asked Jan 31 '14 at 8:16
joshlfjoshlf
10.7k34763
edited Oct 14 '18 at 13:00
phuclv
17.1k957242
edited Oct 14 '18 at 13:00
phuclv
17.1k957242
edited Oct 14 '18 at 13:00
phuclv
17.1k957242
17.1k957242
asked Jan 31 '14 at 8:16
joshlfjoshlf
10.7k34763
asked Jan 31 '14 at 8:16
joshlfjoshlf
10.7k34763
asked Jan 31 '14 at 8:16
joshlfjoshlf
10.7k34763
10.7k34763
Somewhat related: Which 2's complement integer operations can be used without zeroing high bits in the inputs, if only the low part of the result is wanted?
– Peter Cordes
Jul 9 '18 at 23:02
add a comment |
Somewhat related: Which 2's complement integer operations can be used without zeroing high bits in the inputs, if only the low part of the result is wanted?
– Peter Cordes
Jul 9 '18 at 23:02
Somewhat related: Which 2's complement integer operations can be used without zeroing high bits in the inputs, if only the low part of the result is wanted?
– Peter Cordes
Jul 9 '18 at 23:02
Somewhat related: Which 2's complement integer operations can be used without zeroing high bits in the inputs, if only the low part of the result is wanted?
– Peter Cordes
Jul 9 '18 at 23:02
add a comment |
3 Answers
3
active
oldest
votes
Add and subtract are the same for signed and unsigned 2s complement, assuming you're going to handle overflow/underflow in the normal way for most CPUs, i.e. just wrap around. Multiply and divide are different. So you only need one addition routine and one subtraction routine regardless of signedness, but you need separate signed and unsigned multiply and divide.
answered Jan 31 '14 at 8:19
Paul RPaul R
179k25306467
4
Non-widening multiplication result is the same for both signed and unsigned types
– phuclv
Mar 29 '16 at 3:14
add a comment |
Addition, subtraction and multiplication are the same provided:
- Your inputs and outputs are the same size
- Your behaviour on overflow is wraparound modulo 2n
Division is different.
Many instruction sets offer multiplication operations where the output is larger than the input, again these are different for signed and unsigned.
Furthermore if you are writing your emulator in C there are some misfeatures of the language that you need to be aware of.
- Overflow of signed arithmetic in C is undefined behaviour. To get reliable modulo 2n behaviour arithmetic must be performed using unsigned types.
- C will promote types smaller than int to int. Great care is needed to avoid such promotions (adding 0u or multiplying by 1u at the start of your calculation is one way).
- Conversion from unsigned types to signed types is implementation defined, the implementations i've seen do the sensible thing but there may be some that don't.
answered Mar 2 '18 at 13:37
plugwashplugwash
4,7361327
add a comment |
All your operations need overflow checks, or they will return incorrect values in some cases. The unsigned versions of these checks are different from the signed ones, so you'll need to implement each routine separately.
answered Jan 31 '14 at 8:20
Guntram BlohmGuntram Blohm
8,50321426
1
Since this is an instruction set emulator I would expect that integer overflow/underflow would behave in the usual way, i.e. wrap around modulo 2^N, so addition and subtraction would be the same for both signed and unsigned.
– Paul R
Jan 31 '14 at 8:34
The way real instruction sets usually handle this is by having multiple flags that can be used to detect different types of overflow.
– plugwash
Mar 2 '18 at 13:23
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Add and subtract are the same for signed and unsigned 2s complement, assuming you're going to handle overflow/underflow in the normal way for most CPUs, i.e. just wrap around. Multiply and divide are different. So you only need one addition routine and one subtraction routine regardless of signedness, but you need separate signed and unsigned multiply and divide.
answered Jan 31 '14 at 8:19
Paul RPaul R
179k25306467
4
Non-widening multiplication result is the same for both signed and unsigned types
– phuclv
Mar 29 '16 at 3:14
add a comment |
Add and subtract are the same for signed and unsigned 2s complement, assuming you're going to handle overflow/underflow in the normal way for most CPUs, i.e. just wrap around. Multiply and divide are different. So you only need one addition routine and one subtraction routine regardless of signedness, but you need separate signed and unsigned multiply and divide.
answered Jan 31 '14 at 8:19
Paul RPaul R
179k25306467
4
Non-widening multiplication result is the same for both signed and unsigned types
– phuclv
Mar 29 '16 at 3:14
add a comment |
Add and subtract are the same for signed and unsigned 2s complement, assuming you're going to handle overflow/underflow in the normal way for most CPUs, i.e. just wrap around. Multiply and divide are different. So you only need one addition routine and one subtraction routine regardless of signedness, but you need separate signed and unsigned multiply and divide.
answered Jan 31 '14 at 8:19
Paul RPaul R
179k25306467
Add and subtract are the same for signed and unsigned 2s complement, assuming you're going to handle overflow/underflow in the normal way for most CPUs, i.e. just wrap around. Multiply and divide are different. So you only need one addition routine and one subtraction routine regardless of signedness, but you need separate signed and unsigned multiply and divide.
answered Jan 31 '14 at 8:19
Paul RPaul R
179k25306467
edited Jan 31 '14 at 8:33
answered Jan 31 '14 at 8:19
Paul RPaul R
179k25306467
answered Jan 31 '14 at 8:19
Paul RPaul R
179k25306467
answered Jan 31 '14 at 8:19
Paul RPaul R
179k25306467
179k25306467
4
Non-widening multiplication result is the same for both signed and unsigned types
– phuclv
Mar 29 '16 at 3:14
add a comment |
4
Non-widening multiplication result is the same for both signed and unsigned types
– phuclv
Mar 29 '16 at 3:14
4
4
Non-widening multiplication result is the same for both signed and unsigned types
– phuclv
Mar 29 '16 at 3:14
Non-widening multiplication result is the same for both signed and unsigned types
– phuclv
Mar 29 '16 at 3:14
add a comment |
Addition, subtraction and multiplication are the same provided:
- Your inputs and outputs are the same size
- Your behaviour on overflow is wraparound modulo 2n
Division is different.
Many instruction sets offer multiplication operations where the output is larger than the input, again these are different for signed and unsigned.
Furthermore if you are writing your emulator in C there are some misfeatures of the language that you need to be aware of.
- Overflow of signed arithmetic in C is undefined behaviour. To get reliable modulo 2n behaviour arithmetic must be performed using unsigned types.
- C will promote types smaller than int to int. Great care is needed to avoid such promotions (adding 0u or multiplying by 1u at the start of your calculation is one way).
- Conversion from unsigned types to signed types is implementation defined, the implementations i've seen do the sensible thing but there may be some that don't.
answered Mar 2 '18 at 13:37
plugwashplugwash
4,7361327
add a comment |
Addition, subtraction and multiplication are the same provided:
- Your inputs and outputs are the same size
- Your behaviour on overflow is wraparound modulo 2n
Division is different.
Many instruction sets offer multiplication operations where the output is larger than the input, again these are different for signed and unsigned.
Furthermore if you are writing your emulator in C there are some misfeatures of the language that you need to be aware of.
- Overflow of signed arithmetic in C is undefined behaviour. To get reliable modulo 2n behaviour arithmetic must be performed using unsigned types.
- C will promote types smaller than int to int. Great care is needed to avoid such promotions (adding 0u or multiplying by 1u at the start of your calculation is one way).
- Conversion from unsigned types to signed types is implementation defined, the implementations i've seen do the sensible thing but there may be some that don't.
answered Mar 2 '18 at 13:37
plugwashplugwash
4,7361327
add a comment |
Addition, subtraction and multiplication are the same provided:
- Your inputs and outputs are the same size
- Your behaviour on overflow is wraparound modulo 2n
Division is different.
Many instruction sets offer multiplication operations where the output is larger than the input, again these are different for signed and unsigned.
Furthermore if you are writing your emulator in C there are some misfeatures of the language that you need to be aware of.
- Overflow of signed arithmetic in C is undefined behaviour. To get reliable modulo 2n behaviour arithmetic must be performed using unsigned types.
- C will promote types smaller than int to int. Great care is needed to avoid such promotions (adding 0u or multiplying by 1u at the start of your calculation is one way).
- Conversion from unsigned types to signed types is implementation defined, the implementations i've seen do the sensible thing but there may be some that don't.
answered Mar 2 '18 at 13:37
plugwashplugwash
4,7361327
Addition, subtraction and multiplication are the same provided:
- Your inputs and outputs are the same size
- Your behaviour on overflow is wraparound modulo 2n
Division is different.
Many instruction sets offer multiplication operations where the output is larger than the input, again these are different for signed and unsigned.
Furthermore if you are writing your emulator in C there are some misfeatures of the language that you need to be aware of.
- Overflow of signed arithmetic in C is undefined behaviour. To get reliable modulo 2n behaviour arithmetic must be performed using unsigned types.
- C will promote types smaller than int to int. Great care is needed to avoid such promotions (adding 0u or multiplying by 1u at the start of your calculation is one way).
- Conversion from unsigned types to signed types is implementation defined, the implementations i've seen do the sensible thing but there may be some that don't.
answered Mar 2 '18 at 13:37
plugwashplugwash
4,7361327
answered Mar 2 '18 at 13:37
plugwashplugwash
4,7361327
answered Mar 2 '18 at 13:37
plugwashplugwash
4,7361327
answered Mar 2 '18 at 13:37
plugwashplugwash
4,7361327
4,7361327
add a comment |
add a comment |
All your operations need overflow checks, or they will return incorrect values in some cases. The unsigned versions of these checks are different from the signed ones, so you'll need to implement each routine separately.
answered Jan 31 '14 at 8:20
Guntram BlohmGuntram Blohm
8,50321426
1
Since this is an instruction set emulator I would expect that integer overflow/underflow would behave in the usual way, i.e. wrap around modulo 2^N, so addition and subtraction would be the same for both signed and unsigned.
– Paul R
Jan 31 '14 at 8:34
The way real instruction sets usually handle this is by having multiple flags that can be used to detect different types of overflow.
– plugwash
Mar 2 '18 at 13:23
add a comment |
All your operations need overflow checks, or they will return incorrect values in some cases. The unsigned versions of these checks are different from the signed ones, so you'll need to implement each routine separately.
answered Jan 31 '14 at 8:20
Guntram BlohmGuntram Blohm
8,50321426
1
Since this is an instruction set emulator I would expect that integer overflow/underflow would behave in the usual way, i.e. wrap around modulo 2^N, so addition and subtraction would be the same for both signed and unsigned.
– Paul R
Jan 31 '14 at 8:34
The way real instruction sets usually handle this is by having multiple flags that can be used to detect different types of overflow.
– plugwash
Mar 2 '18 at 13:23
add a comment |
All your operations need overflow checks, or they will return incorrect values in some cases. The unsigned versions of these checks are different from the signed ones, so you'll need to implement each routine separately.
answered Jan 31 '14 at 8:20
Guntram BlohmGuntram Blohm
8,50321426
All your operations need overflow checks, or they will return incorrect values in some cases. The unsigned versions of these checks are different from the signed ones, so you'll need to implement each routine separately.
answered Jan 31 '14 at 8:20
Guntram BlohmGuntram Blohm
8,50321426
answered Jan 31 '14 at 8:20
Guntram BlohmGuntram Blohm
8,50321426
answered Jan 31 '14 at 8:20
Guntram BlohmGuntram Blohm
8,50321426
answered Jan 31 '14 at 8:20
Guntram BlohmGuntram Blohm
8,50321426
8,50321426
1
Since this is an instruction set emulator I would expect that integer overflow/underflow would behave in the usual way, i.e. wrap around modulo 2^N, so addition and subtraction would be the same for both signed and unsigned.
– Paul R
Jan 31 '14 at 8:34
The way real instruction sets usually handle this is by having multiple flags that can be used to detect different types of overflow.
– plugwash
Mar 2 '18 at 13:23
add a comment |
1
Since this is an instruction set emulator I would expect that integer overflow/underflow would behave in the usual way, i.e. wrap around modulo 2^N, so addition and subtraction would be the same for both signed and unsigned.
– Paul R
Jan 31 '14 at 8:34
The way real instruction sets usually handle this is by having multiple flags that can be used to detect different types of overflow.
– plugwash
Mar 2 '18 at 13:23
1
1
Since this is an instruction set emulator I would expect that integer overflow/underflow would behave in the usual way, i.e. wrap around modulo 2^N, so addition and subtraction would be the same for both signed and unsigned.
– Paul R
Jan 31 '14 at 8:34
Since this is an instruction set emulator I would expect that integer overflow/underflow would behave in the usual way, i.e. wrap around modulo 2^N, so addition and subtraction would be the same for both signed and unsigned.
– Paul R
Jan 31 '14 at 8:34
The way real instruction sets usually handle this is by having multiple flags that can be used to detect different types of overflow.
– plugwash
Mar 2 '18 at 13:23
The way real instruction sets usually handle this is by having multiple flags that can be used to detect different types of overflow.
– plugwash
Mar 2 '18 at 13:23
add a comment |
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Somewhat related: Which 2's complement integer operations can be used without zeroing high bits in the inputs, if only the low part of the result is wanted?
– Peter Cordes
Jul 9 '18 at 23:02