add functions to encode and decode integers in variable length binary

formats
master
leitner 13 years ago
parent ad6c2d9ce7
commit d17c2d1012

@ -0,0 +1,18 @@
.TH fmt_asn1derlength 3
.SH NAME
fmt_asn1derlength \- encode unsigned integer like ASN.1 DER length
.SH SYNTAX
.B #include <fmt.h>
size_t \fBfmt_asn1derlength\fP(char *\fIdest\fR,unsigned long long \fIsource\fR);
.SH DESCRIPTION
fmt_asn1derlength encodes an unsigned integer using the UTF-8 rules. This
can take from 1 byte (0-0x7f) up to sizeof(source)+1 bytes.
If \fIdest\fR equals FMT_LEN (i.e. is NULL), fmt_asn1derlength returns the
number of bytes it would have written.
For convenience, fmt.h defines the integer FMT_ASN1LENGTH to be big
enough to contain every possible fmt_asn1derlength output.
.SH "SEE ALSO"
scan_asn1derlength(3)

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#include "fmt.h"
/* write int in least amount of bytes, return number of bytes */
/* as used in ASN.1 length */
size_t fmt_asn1derlength(char* dest,unsigned long long l) {
/* encoding is either l%128 or (0x80+number of bytes,bytes) */
size_t needed=(sizeof l),i;
if (l<128) {
if (dest) *dest=l&0x7f;
return 1;
}
for (i=1; i<needed; ++i)
if (!(l>>(i*8)))
break;
if (dest) {
int j=i;
*dest=0x80+i; ++dest;
while (j) {
--j;
*dest=((l>>(j*8))&0xff);
++dest;
}
}
return i+1;
}

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.TH fmt_utf8 3
.SH NAME
fmt_utf8 \- encode 31-bit unsigned integer using UTF-8 rules
.SH SYNTAX
.B #include <fmt.h>
size_t \fBfmt_utf8\fP(char *\fIdest\fR,uint32_t \fIsource\fR);
.SH DESCRIPTION
fmt_utf8 encodes a 31-bit unsigned integer using the UTF-8 rules. This
can take from 1 byte (0-0x7f) up to 5 bytes (0x4000000-0x7fffffff).
Values larger than 0x7fffffff cannot be represented in this encoding.
If \fIdest\fR equals FMT_LEN (i.e. is NULL), fmt_utf8 returns the
number of bytes it would have written.
For convenience, fmt.h defines the integer FMT_UTF8 to be big enough to
contain every possible fmt_utf8 output.
.SH "SEE ALSO"
scan_utf8(3)

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#include "fmt.h"
size_t fmt_utf8(char *dest,uint32_t n) {
size_t i,j;
if (n<=0x7f) {
if (dest) *dest=n;
return 1;
}
for (i=0x3f,j=1; i<0x7fffffff; i=(i<<5)|0x1f, ++j) {
if (i>=n) {
--j;
if (dest) {
size_t k=j*6;
*dest++=((char)0xc0 >> (j-1)) | (n >> k);
while (k) {
*dest++=0x80 | ((n >> (k-6)) & 0x3f);
k-=6;
}
}
return j+1;
}
}
/* we were asked to encode a value that cannot be encoded */
return 0;
}

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.TH scan_asn1length 3
.SH NAME
scan_asn1length \- decode an unsigned integer from ASN.1 DER length encoding
.SH SYNTAX
.B #include <scan.h>
size_t \fBscan_asn1length\fP(const char *\fIsrc\fR,size_t \fIlen\fR,unsigned long long *\fIdest\fR);
.SH DESCRIPTION
scan_asn1length decodes an unsigned integer in ASN.1 DER length encoding
from a memory area holding binary data. It writes the decode value in
\fIdest\fR and returns the number of bytes it read from \fIsrc\fR.
scan_asn1length never reads more than \fIlen\fR bytes from \fIsrc\fR. If the
sequence is longer than that, or the memory area contains an invalid
sequence, scan_asn1length returns 0 and does not touch \fIdest\fR.
The length of the longest ASN.1 DER length sequence is 128 bytes. In
practice the largest sequence is sizeof(*dest)+1.
.SH "SEE ALSO"
fmt_asn1length(3)

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#include "scan.h"
size_t scan_asn1derlength(const char* src,size_t len,unsigned long long* length) {
const char* orig=src;
const char* max=orig+len;
if (src>=max) return 0;
/* If the highest bit of the first byte is clear, the byte is the length.
* Otherwise the next n bytes are the length (n being the lower 7 bits) */
if (*src&0x80) {
int chars=*src&0x7f;
unsigned long long l=0;
while (chars>0) {
if (++src>=max) return 0;
if (l>(((unsigned long long)-1)>>8)) return 0; /* catch integer overflow */
l=l*256+(unsigned char)*src;
--chars;
}
*length=l;
} else
*length=*src&0x7f;
src++;
if (src+*length>max) return 0; /* catch integer overflow */
if ((uintptr_t)src+*length<(uintptr_t)src) return 0; /* gcc 4.1 removes this check without the cast to uintptr_t */
return src-orig;
}

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.TH scan_utf8 3
.SH NAME
scan_utf8 \- decode an unsigned integer from UTF-8 encoding
.SH SYNTAX
.B #include <scan.h>
size_t \fBscan_utf8\fP(const char *\fIsrc\fR,size_t \fIlen\fR,uint32_t *\fIdest\fR);
.SH DESCRIPTION
scan_utf8 decodes an unsigned integer in UTF-8 encoding from a memory
area holding binary data. It writes the decode value in \fIdest\fR and
returns the number of bytes it read from \fIsrc\fR.
scan_utf8 never reads more than \fIlen\fR bytes from \fIsrc\fR. If the
sequence is longer than that, or the memory area contains an invalid
sequence, scan_utf8 returns 0 and does not touch \fIdest\fR.
The length of the longest UTF-8 sequence is 5. If the buffer is longer
than that, and scan_utf8 fails, then the data was not a valid UTF-8
encoded sequence.
.SH "SEE ALSO"
fmt_utf8(3)

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#include "fmt.h"
size_t scan_utf8(const char* in,size_t len,uint32_t* num) {
uint32_t i,k,m;
const char* orig=in;
if (len==0) return 0;
i=(*(unsigned char*)in++); /* grab first byte */
if (i>=0xfe || /* 0xfe and 0xff are invalid encodings in utf-8 for the first byte */
(i&0xc0)==0x80) return 0; /* first bits being 10 marks continuation chars, invalid sequence for first byte */
for (k=0; i&0x80; i<<=1, ++k); /* count leading 1 bits */
if (!k) {
if (num) *num=i;
return 1;
}
if (k>len) return 0;
i=(i&0xff)>>k; /* mask the leading 1 bits */
/* The next part is a little tricky.
* UTF-8 says that the encoder has to choose the most efficient
* encoding, and the decoder has to reject other encodings. The
* background is that attackers encoded '/' not as 0x2f but as 0xc0
* 0xaf, and that evaded bad security checks just scan for the '/'
* byte in pathnames.
* At this point k contains the number of bytes, so k-1 is the number
* of continuation bytes. For each additional continuation byte, we
* gain 6 bits of storage space, but we lose one in the signalling in
* the initial byte. So we have 6 + (k-1) * 5 bits total storage
* space for this encoding. The minimum value for k bytes is the
* maximum number for k-1 bytes plus 1. If the previous encoding has
* 11 bits, its maximum value is 11 1-bits or 0x7ff, and the minimum
* value we are looking for is 0x800 or 1<<11. For 2 bytes, UTF-8 can
* encode 11 bits, after that each additional byte gains 5 more bits.
* So for k>2, we want
* 1 << (11+(k-3)*5)
* or optimized to get rid of the -3
* 1 << (k*5-4)
* but for k==2 the delta is 4 bits (not 5), so we want
* 1 << 7
* abusing the fact that a boolean expression evaluates to 0 or 1, the
* expression can be written as
* 1 << (k*5-4+(k==2))
*/
m=(1<<(k*5-4+(k==2)));
while (k>1) {
i=(i<<6) | ((*in++)&0x3f);
--k;
}
if (i<m) return 0; /* if the encoded value was less than m, reject */
if (num) *num=i;
return in-orig;
}
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