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ASN.1 / DER Encoding of Integers
ASN Basic Encoding Rule of an integerWhat is base 64 encoding used for?Encode URL in JavaScript?How should I interpret the ASN.1 BER standard for REALs?How to do Base64 encoding in node.js?Two's complement in ASN.1 Unaligned PERASN Basic Encoding Rule of an integerEncoding and decoding ASN.1 REAL with BERReading encode asn file manuallyASN.1 BER Handling Tag Number 31 or HigherInterpreting ASN.1 indefinite-lenght encoding with multiple encapsulated octet-strings
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
I'm currently starting to work with DER (Distinguished Encoding Rules) encoding and have problems understanding the encoding of integers.
In the reference document https://www.itu.int/ITU-T/studygroups/com17/languages/X.690-0207.pdf this encoding is defined as follows:
8.3.1 The encoding of an integer value shall be primitive. The contents octets shall consist of one or more octets.
8.3.2 If the contents octets of an integer value encoding consist of more than one octet, then the bits of the first octet and bit 8 of the second octet:
shall not all be ones; and
shall not all be zero.
NOTE – These rules ensure that an integer value is always encoded in the smallest possible number of octets.
8.3.3 The contents octets shall be a two's complement binary number equal to the integer value, and consisting of bits 8 to 1 of the first octet, followed by bits 8 to 1 of the second octet, followed by bits 8 to 1 of each octet in turn up to and including the last octet of the contents octets.
On another site, https://docs.microsoft.com/en-us/windows/desktop/seccertenroll/about-integer, it is explained that for positive numbers whose binary representation starts with a 1, a zero byte is added at the front. This is also mentioned in the answers to a former question on stackoverflow: ASN Basic Encoding Rule of an integer.
Unfortunately, from these answers I cannot see how this latter instruction can be deduced from the rules of the reference document.
For example, if I want to encode the number 128, why can't I do this as
[tag byte] [length byte] 10000000?
I know that the correct encoding would be [tag byte] [length byte] 00000000 10000000, but which condition is injured by the variant above? Probably it has something do to with the two's complement, but isn't the two's complement of 128 again 10000000?
I hope you can help me understand why the description on the Microsoft site is equivalent to the original definition. Thank you.
encoding asn.1 der
asked Mar 26 at 13:05
BilboBilbo
235 bronze badges
add a comment |
I'm currently starting to work with DER (Distinguished Encoding Rules) encoding and have problems understanding the encoding of integers.
In the reference document https://www.itu.int/ITU-T/studygroups/com17/languages/X.690-0207.pdf this encoding is defined as follows:
8.3.1 The encoding of an integer value shall be primitive. The contents octets shall consist of one or more octets.
8.3.2 If the contents octets of an integer value encoding consist of more than one octet, then the bits of the first octet and bit 8 of the second octet:
shall not all be ones; and
shall not all be zero.
NOTE – These rules ensure that an integer value is always encoded in the smallest possible number of octets.
8.3.3 The contents octets shall be a two's complement binary number equal to the integer value, and consisting of bits 8 to 1 of the first octet, followed by bits 8 to 1 of the second octet, followed by bits 8 to 1 of each octet in turn up to and including the last octet of the contents octets.
On another site, https://docs.microsoft.com/en-us/windows/desktop/seccertenroll/about-integer, it is explained that for positive numbers whose binary representation starts with a 1, a zero byte is added at the front. This is also mentioned in the answers to a former question on stackoverflow: ASN Basic Encoding Rule of an integer.
Unfortunately, from these answers I cannot see how this latter instruction can be deduced from the rules of the reference document.
For example, if I want to encode the number 128, why can't I do this as
[tag byte] [length byte] 10000000?
I know that the correct encoding would be [tag byte] [length byte] 00000000 10000000, but which condition is injured by the variant above? Probably it has something do to with the two's complement, but isn't the two's complement of 128 again 10000000?
I hope you can help me understand why the description on the Microsoft site is equivalent to the original definition. Thank you.
encoding asn.1 der
asked Mar 26 at 13:05
BilboBilbo
235 bronze badges
Maybe this will help
– pepo
Mar 26 at 14:49
Unfortunately it does not so far. I had already considered this document, but it also does not seem to mention that for negative integers there is an additional zero byte added. Instead, the document says "Encoding of negative integers has its own rules" and then describes the Two's complement in its own words. But in the reference document for ASN.1 the word "negative" only appears three times, and none of these refers to the encoding of integers. So I still wonder where these "own rules" come from.
– Bilbo
Mar 26 at 16:13
add a comment |
I'm currently starting to work with DER (Distinguished Encoding Rules) encoding and have problems understanding the encoding of integers.
In the reference document https://www.itu.int/ITU-T/studygroups/com17/languages/X.690-0207.pdf this encoding is defined as follows:
8.3.1 The encoding of an integer value shall be primitive. The contents octets shall consist of one or more octets.
8.3.2 If the contents octets of an integer value encoding consist of more than one octet, then the bits of the first octet and bit 8 of the second octet:
shall not all be ones; and
shall not all be zero.
NOTE – These rules ensure that an integer value is always encoded in the smallest possible number of octets.
8.3.3 The contents octets shall be a two's complement binary number equal to the integer value, and consisting of bits 8 to 1 of the first octet, followed by bits 8 to 1 of the second octet, followed by bits 8 to 1 of each octet in turn up to and including the last octet of the contents octets.
On another site, https://docs.microsoft.com/en-us/windows/desktop/seccertenroll/about-integer, it is explained that for positive numbers whose binary representation starts with a 1, a zero byte is added at the front. This is also mentioned in the answers to a former question on stackoverflow: ASN Basic Encoding Rule of an integer.
Unfortunately, from these answers I cannot see how this latter instruction can be deduced from the rules of the reference document.
For example, if I want to encode the number 128, why can't I do this as
[tag byte] [length byte] 10000000?
I know that the correct encoding would be [tag byte] [length byte] 00000000 10000000, but which condition is injured by the variant above? Probably it has something do to with the two's complement, but isn't the two's complement of 128 again 10000000?
I hope you can help me understand why the description on the Microsoft site is equivalent to the original definition. Thank you.
encoding asn.1 der
asked Mar 26 at 13:05
BilboBilbo
235 bronze badges
I'm currently starting to work with DER (Distinguished Encoding Rules) encoding and have problems understanding the encoding of integers.
In the reference document https://www.itu.int/ITU-T/studygroups/com17/languages/X.690-0207.pdf this encoding is defined as follows:
8.3.1 The encoding of an integer value shall be primitive. The contents octets shall consist of one or more octets.
8.3.2 If the contents octets of an integer value encoding consist of more than one octet, then the bits of the first octet and bit 8 of the second octet:
shall not all be ones; and
shall not all be zero.
NOTE – These rules ensure that an integer value is always encoded in the smallest possible number of octets.
8.3.3 The contents octets shall be a two's complement binary number equal to the integer value, and consisting of bits 8 to 1 of the first octet, followed by bits 8 to 1 of the second octet, followed by bits 8 to 1 of each octet in turn up to and including the last octet of the contents octets.
On another site, https://docs.microsoft.com/en-us/windows/desktop/seccertenroll/about-integer, it is explained that for positive numbers whose binary representation starts with a 1, a zero byte is added at the front. This is also mentioned in the answers to a former question on stackoverflow: ASN Basic Encoding Rule of an integer.
Unfortunately, from these answers I cannot see how this latter instruction can be deduced from the rules of the reference document.
For example, if I want to encode the number 128, why can't I do this as
[tag byte] [length byte] 10000000?
I know that the correct encoding would be [tag byte] [length byte] 00000000 10000000, but which condition is injured by the variant above? Probably it has something do to with the two's complement, but isn't the two's complement of 128 again 10000000?
I hope you can help me understand why the description on the Microsoft site is equivalent to the original definition. Thank you.
encoding asn.1 der
encoding asn.1 der
asked Mar 26 at 13:05
BilboBilbo
235 bronze badges
asked Mar 26 at 13:05
BilboBilbo
235 bronze badges
edited Mar 26 at 13:17
Bilbo
asked Mar 26 at 13:05
BilboBilbo
235 bronze badges
asked Mar 26 at 13:05
BilboBilbo
235 bronze badges
asked Mar 26 at 13:05
BilboBilbo
235 bronze badges
235 bronze badges
Maybe this will help
– pepo
Mar 26 at 14:49
Unfortunately it does not so far. I had already considered this document, but it also does not seem to mention that for negative integers there is an additional zero byte added. Instead, the document says "Encoding of negative integers has its own rules" and then describes the Two's complement in its own words. But in the reference document for ASN.1 the word "negative" only appears three times, and none of these refers to the encoding of integers. So I still wonder where these "own rules" come from.
– Bilbo
Mar 26 at 16:13
add a comment |
Maybe this will help
– pepo
Mar 26 at 14:49
Unfortunately it does not so far. I had already considered this document, but it also does not seem to mention that for negative integers there is an additional zero byte added. Instead, the document says "Encoding of negative integers has its own rules" and then describes the Two's complement in its own words. But in the reference document for ASN.1 the word "negative" only appears three times, and none of these refers to the encoding of integers. So I still wonder where these "own rules" come from.
– Bilbo
Mar 26 at 16:13
Maybe this will help
– pepo
Mar 26 at 14:49
Maybe this will help
– pepo
Mar 26 at 14:49
Unfortunately it does not so far. I had already considered this document, but it also does not seem to mention that for negative integers there is an additional zero byte added. Instead, the document says "Encoding of negative integers has its own rules" and then describes the Two's complement in its own words. But in the reference document for ASN.1 the word "negative" only appears three times, and none of these refers to the encoding of integers. So I still wonder where these "own rules" come from.
– Bilbo
Mar 26 at 16:13
Unfortunately it does not so far. I had already considered this document, but it also does not seem to mention that for negative integers there is an additional zero byte added. Instead, the document says "Encoding of negative integers has its own rules" and then describes the Two's complement in its own words. But in the reference document for ASN.1 the word "negative" only appears three times, and none of these refers to the encoding of integers. So I still wonder where these "own rules" come from.
– Bilbo
Mar 26 at 16:13
add a comment |
1 Answer
1
active
oldest
votes
The Two's complement rule (8.3.3) says that if the high bit of the first (lowest index) content byte is set that the number is negative.
02 01 80 has contents 0b1000_0000. Since the high bit is set, the number is negative.
Flip all the bits (0b0111_1111), then add one: 0b1000_0000; meaning that it represents negative 128.
For a less degenerate example, 0b1000_0001 => 0b0111_1110 => 0b0111_1111, showing that 0x81 is negative 127.
For the number (positive) 127, since the high bit isn't set, the number is interepreted as positive, so the contents are just 0b0111_1111 aka 0x7F, resulting in 02 01 7F
answered Mar 26 at 15:22
bartonjsbartonjs
15.6k1 gold badge30 silver badges64 bronze badges
Sorry, but I don't see how this answers my question. Maybe I failed to make my point clear. I basically understand how the representation of numbers in Two's complement works. My question refers to the sentence from the Microsoft website "If the integer is positive but the high order bit is set to 1, a leading 0x00 is added to the content to indicate that the number is not negative." I still don't see where this comes from?
– Bilbo
Mar 26 at 16:03
The extra 0 byte means the high bit of the first byte is not set, so the value is the positive 15 but number 0b000_0000_1000_0000 (128). Without it the high bit is set, so the number is negative: the two’s complement of the 7 bit value 0b000_0000, which is (in 8-bit land)-128.
– bartonjs
Mar 26 at 16:21
Ok, I think now I see where I was wrong. I thought as 1000 0000 being the binary representation of 128; but this is NOT the representation as a Two's complement number. If we consider signed bytes, we NEED a 0 at the start, so we would like to put a zero bit at the start. And since we must use full bytes in our encoding, we then must put a complete zero byte at the start.
– Bilbo
Mar 26 at 18:19
@Bilbo Yep, that’s pretty much it.
– bartonjs
Mar 26 at 18:20
Just to add another piece that contributed to my confusion: The Microsoft article says "The Value field of the TLV triplet contains the encoded integer if it is positive, or its two's complement if it is negative." Here the OPERATION of taking the two's complement (by switching bits and adding 1) must be meant; apparently in contrast, the reference document says that always(!) "the contents octets shall be a two's complement binary number equal to the integer value". Here not the operation, but the REPRESENTATION in two's complement is meant, which switches the bits only for negative numbers.
– Bilbo
Mar 26 at 21:53
|
show 1 more comment
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The Two's complement rule (8.3.3) says that if the high bit of the first (lowest index) content byte is set that the number is negative.
02 01 80 has contents 0b1000_0000. Since the high bit is set, the number is negative.
Flip all the bits (0b0111_1111), then add one: 0b1000_0000; meaning that it represents negative 128.
For a less degenerate example, 0b1000_0001 => 0b0111_1110 => 0b0111_1111, showing that 0x81 is negative 127.
For the number (positive) 127, since the high bit isn't set, the number is interepreted as positive, so the contents are just 0b0111_1111 aka 0x7F, resulting in 02 01 7F
answered Mar 26 at 15:22
bartonjsbartonjs
15.6k1 gold badge30 silver badges64 bronze badges
Sorry, but I don't see how this answers my question. Maybe I failed to make my point clear. I basically understand how the representation of numbers in Two's complement works. My question refers to the sentence from the Microsoft website "If the integer is positive but the high order bit is set to 1, a leading 0x00 is added to the content to indicate that the number is not negative." I still don't see where this comes from?
– Bilbo
Mar 26 at 16:03
The extra 0 byte means the high bit of the first byte is not set, so the value is the positive 15 but number 0b000_0000_1000_0000 (128). Without it the high bit is set, so the number is negative: the two’s complement of the 7 bit value 0b000_0000, which is (in 8-bit land)-128.
– bartonjs
Mar 26 at 16:21
Ok, I think now I see where I was wrong. I thought as 1000 0000 being the binary representation of 128; but this is NOT the representation as a Two's complement number. If we consider signed bytes, we NEED a 0 at the start, so we would like to put a zero bit at the start. And since we must use full bytes in our encoding, we then must put a complete zero byte at the start.
– Bilbo
Mar 26 at 18:19
@Bilbo Yep, that’s pretty much it.
– bartonjs
Mar 26 at 18:20
Just to add another piece that contributed to my confusion: The Microsoft article says "The Value field of the TLV triplet contains the encoded integer if it is positive, or its two's complement if it is negative." Here the OPERATION of taking the two's complement (by switching bits and adding 1) must be meant; apparently in contrast, the reference document says that always(!) "the contents octets shall be a two's complement binary number equal to the integer value". Here not the operation, but the REPRESENTATION in two's complement is meant, which switches the bits only for negative numbers.
– Bilbo
Mar 26 at 21:53
|
show 1 more comment
The Two's complement rule (8.3.3) says that if the high bit of the first (lowest index) content byte is set that the number is negative.
02 01 80 has contents 0b1000_0000. Since the high bit is set, the number is negative.
Flip all the bits (0b0111_1111), then add one: 0b1000_0000; meaning that it represents negative 128.
For a less degenerate example, 0b1000_0001 => 0b0111_1110 => 0b0111_1111, showing that 0x81 is negative 127.
For the number (positive) 127, since the high bit isn't set, the number is interepreted as positive, so the contents are just 0b0111_1111 aka 0x7F, resulting in 02 01 7F
answered Mar 26 at 15:22
bartonjsbartonjs
15.6k1 gold badge30 silver badges64 bronze badges
Sorry, but I don't see how this answers my question. Maybe I failed to make my point clear. I basically understand how the representation of numbers in Two's complement works. My question refers to the sentence from the Microsoft website "If the integer is positive but the high order bit is set to 1, a leading 0x00 is added to the content to indicate that the number is not negative." I still don't see where this comes from?
– Bilbo
Mar 26 at 16:03
The extra 0 byte means the high bit of the first byte is not set, so the value is the positive 15 but number 0b000_0000_1000_0000 (128). Without it the high bit is set, so the number is negative: the two’s complement of the 7 bit value 0b000_0000, which is (in 8-bit land)-128.
– bartonjs
Mar 26 at 16:21
Ok, I think now I see where I was wrong. I thought as 1000 0000 being the binary representation of 128; but this is NOT the representation as a Two's complement number. If we consider signed bytes, we NEED a 0 at the start, so we would like to put a zero bit at the start. And since we must use full bytes in our encoding, we then must put a complete zero byte at the start.
– Bilbo
Mar 26 at 18:19
@Bilbo Yep, that’s pretty much it.
– bartonjs
Mar 26 at 18:20
Just to add another piece that contributed to my confusion: The Microsoft article says "The Value field of the TLV triplet contains the encoded integer if it is positive, or its two's complement if it is negative." Here the OPERATION of taking the two's complement (by switching bits and adding 1) must be meant; apparently in contrast, the reference document says that always(!) "the contents octets shall be a two's complement binary number equal to the integer value". Here not the operation, but the REPRESENTATION in two's complement is meant, which switches the bits only for negative numbers.
– Bilbo
Mar 26 at 21:53
|
show 1 more comment
The Two's complement rule (8.3.3) says that if the high bit of the first (lowest index) content byte is set that the number is negative.
02 01 80 has contents 0b1000_0000. Since the high bit is set, the number is negative.
Flip all the bits (0b0111_1111), then add one: 0b1000_0000; meaning that it represents negative 128.
For a less degenerate example, 0b1000_0001 => 0b0111_1110 => 0b0111_1111, showing that 0x81 is negative 127.
For the number (positive) 127, since the high bit isn't set, the number is interepreted as positive, so the contents are just 0b0111_1111 aka 0x7F, resulting in 02 01 7F
answered Mar 26 at 15:22
bartonjsbartonjs
15.6k1 gold badge30 silver badges64 bronze badges
The Two's complement rule (8.3.3) says that if the high bit of the first (lowest index) content byte is set that the number is negative.
02 01 80 has contents 0b1000_0000. Since the high bit is set, the number is negative.
Flip all the bits (0b0111_1111), then add one: 0b1000_0000; meaning that it represents negative 128.
For a less degenerate example, 0b1000_0001 => 0b0111_1110 => 0b0111_1111, showing that 0x81 is negative 127.
For the number (positive) 127, since the high bit isn't set, the number is interepreted as positive, so the contents are just 0b0111_1111 aka 0x7F, resulting in 02 01 7F
answered Mar 26 at 15:22
bartonjsbartonjs
15.6k1 gold badge30 silver badges64 bronze badges
answered Mar 26 at 15:22
bartonjsbartonjs
15.6k1 gold badge30 silver badges64 bronze badges
answered Mar 26 at 15:22
bartonjsbartonjs
15.6k1 gold badge30 silver badges64 bronze badges
answered Mar 26 at 15:22
bartonjsbartonjs
15.6k1 gold badge30 silver badges64 bronze badges
15.6k1 gold badge30 silver badges64 bronze badges
Sorry, but I don't see how this answers my question. Maybe I failed to make my point clear. I basically understand how the representation of numbers in Two's complement works. My question refers to the sentence from the Microsoft website "If the integer is positive but the high order bit is set to 1, a leading 0x00 is added to the content to indicate that the number is not negative." I still don't see where this comes from?
– Bilbo
Mar 26 at 16:03
The extra 0 byte means the high bit of the first byte is not set, so the value is the positive 15 but number 0b000_0000_1000_0000 (128). Without it the high bit is set, so the number is negative: the two’s complement of the 7 bit value 0b000_0000, which is (in 8-bit land)-128.
– bartonjs
Mar 26 at 16:21
Ok, I think now I see where I was wrong. I thought as 1000 0000 being the binary representation of 128; but this is NOT the representation as a Two's complement number. If we consider signed bytes, we NEED a 0 at the start, so we would like to put a zero bit at the start. And since we must use full bytes in our encoding, we then must put a complete zero byte at the start.
– Bilbo
Mar 26 at 18:19
@Bilbo Yep, that’s pretty much it.
– bartonjs
Mar 26 at 18:20
Just to add another piece that contributed to my confusion: The Microsoft article says "The Value field of the TLV triplet contains the encoded integer if it is positive, or its two's complement if it is negative." Here the OPERATION of taking the two's complement (by switching bits and adding 1) must be meant; apparently in contrast, the reference document says that always(!) "the contents octets shall be a two's complement binary number equal to the integer value". Here not the operation, but the REPRESENTATION in two's complement is meant, which switches the bits only for negative numbers.
– Bilbo
Mar 26 at 21:53
|
show 1 more comment
Sorry, but I don't see how this answers my question. Maybe I failed to make my point clear. I basically understand how the representation of numbers in Two's complement works. My question refers to the sentence from the Microsoft website "If the integer is positive but the high order bit is set to 1, a leading 0x00 is added to the content to indicate that the number is not negative." I still don't see where this comes from?
– Bilbo
Mar 26 at 16:03
The extra 0 byte means the high bit of the first byte is not set, so the value is the positive 15 but number 0b000_0000_1000_0000 (128). Without it the high bit is set, so the number is negative: the two’s complement of the 7 bit value 0b000_0000, which is (in 8-bit land)-128.
– bartonjs
Mar 26 at 16:21
Ok, I think now I see where I was wrong. I thought as 1000 0000 being the binary representation of 128; but this is NOT the representation as a Two's complement number. If we consider signed bytes, we NEED a 0 at the start, so we would like to put a zero bit at the start. And since we must use full bytes in our encoding, we then must put a complete zero byte at the start.
– Bilbo
Mar 26 at 18:19
@Bilbo Yep, that’s pretty much it.
– bartonjs
Mar 26 at 18:20
Just to add another piece that contributed to my confusion: The Microsoft article says "The Value field of the TLV triplet contains the encoded integer if it is positive, or its two's complement if it is negative." Here the OPERATION of taking the two's complement (by switching bits and adding 1) must be meant; apparently in contrast, the reference document says that always(!) "the contents octets shall be a two's complement binary number equal to the integer value". Here not the operation, but the REPRESENTATION in two's complement is meant, which switches the bits only for negative numbers.
– Bilbo
Mar 26 at 21:53
Sorry, but I don't see how this answers my question. Maybe I failed to make my point clear. I basically understand how the representation of numbers in Two's complement works. My question refers to the sentence from the Microsoft website "If the integer is positive but the high order bit is set to 1, a leading 0x00 is added to the content to indicate that the number is not negative." I still don't see where this comes from?
– Bilbo
Mar 26 at 16:03
Sorry, but I don't see how this answers my question. Maybe I failed to make my point clear. I basically understand how the representation of numbers in Two's complement works. My question refers to the sentence from the Microsoft website "If the integer is positive but the high order bit is set to 1, a leading 0x00 is added to the content to indicate that the number is not negative." I still don't see where this comes from?
– Bilbo
Mar 26 at 16:03
The extra 0 byte means the high bit of the first byte is not set, so the value is the positive 15 but number 0b000_0000_1000_0000 (128). Without it the high bit is set, so the number is negative: the two’s complement of the 7 bit value 0b000_0000, which is (in 8-bit land)
-128.– bartonjs
Mar 26 at 16:21
The extra 0 byte means the high bit of the first byte is not set, so the value is the positive 15 but number 0b000_0000_1000_0000 (128). Without it the high bit is set, so the number is negative: the two’s complement of the 7 bit value 0b000_0000, which is (in 8-bit land)
-128.– bartonjs
Mar 26 at 16:21
Ok, I think now I see where I was wrong. I thought as 1000 0000 being the binary representation of 128; but this is NOT the representation as a Two's complement number. If we consider signed bytes, we NEED a 0 at the start, so we would like to put a zero bit at the start. And since we must use full bytes in our encoding, we then must put a complete zero byte at the start.
– Bilbo
Mar 26 at 18:19
Ok, I think now I see where I was wrong. I thought as 1000 0000 being the binary representation of 128; but this is NOT the representation as a Two's complement number. If we consider signed bytes, we NEED a 0 at the start, so we would like to put a zero bit at the start. And since we must use full bytes in our encoding, we then must put a complete zero byte at the start.
– Bilbo
Mar 26 at 18:19
@Bilbo Yep, that’s pretty much it.
– bartonjs
Mar 26 at 18:20
@Bilbo Yep, that’s pretty much it.
– bartonjs
Mar 26 at 18:20
Just to add another piece that contributed to my confusion: The Microsoft article says "The Value field of the TLV triplet contains the encoded integer if it is positive, or its two's complement if it is negative." Here the OPERATION of taking the two's complement (by switching bits and adding 1) must be meant; apparently in contrast, the reference document says that always(!) "the contents octets shall be a two's complement binary number equal to the integer value". Here not the operation, but the REPRESENTATION in two's complement is meant, which switches the bits only for negative numbers.
– Bilbo
Mar 26 at 21:53
Just to add another piece that contributed to my confusion: The Microsoft article says "The Value field of the TLV triplet contains the encoded integer if it is positive, or its two's complement if it is negative." Here the OPERATION of taking the two's complement (by switching bits and adding 1) must be meant; apparently in contrast, the reference document says that always(!) "the contents octets shall be a two's complement binary number equal to the integer value". Here not the operation, but the REPRESENTATION in two's complement is meant, which switches the bits only for negative numbers.
– Bilbo
Mar 26 at 21:53
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Maybe this will help
– pepo
Mar 26 at 14:49
Unfortunately it does not so far. I had already considered this document, but it also does not seem to mention that for negative integers there is an additional zero byte added. Instead, the document says "Encoding of negative integers has its own rules" and then describes the Two's complement in its own words. But in the reference document for ASN.1 the word "negative" only appears three times, and none of these refers to the encoding of integers. So I still wonder where these "own rules" come from.
– Bilbo
Mar 26 at 16:13