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libowfat/scan/scan_utf8.c

52 lines
2.1 KiB

#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=((uint32_t)1<<(k*5-4+(k==2)));
while (k>1) {
if ((*in&0xc0)!=0x80) return 0;
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 (size_t)(in-orig);
}