Does a function exist to convert a 128-bit hex number to a string?
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Re: 128-bit hex number ---> string buffer
ern said:
There is no such thing as a "hex number". Hexadecimal is a numericDoes a function exist to convert a 128-bit hex number to a string?
representation system, not a kind of number.
If you have a hex representation, you already *have* a string, and
converting a string to a string is simple enough.
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
email: rjh at above domain (but drop the www, obviously)
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Re: 128-bit hex number ---> string buffer
"ern" <erniedude@gmai l.comwrites:
What format is the number in to start out with? I ask becauseDoes a function exist to convert a 128-bit hex number to a string?
it's strange to specify that a number is in hex if it's simply in
an integer variable. Normally, one thinks of a numeric variable
as simply having a value, and the base is not important. Also: C
doesn't guarantee the existence of an 128-bit integer type.
--
"Am I missing something?"
--Dan Pop
Comment
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Re: 128-bit hex number ---> string buffer
Richard Heathfield wrote:Let me rephrase. I have a 128 bit value, which I would like to beern said:
>>Does a function exist to convert a 128-bit hex number to a string?
There is no such thing as a "hex number". Hexadecimal is a numeric
representation system, not a kind of number.
>
If you have a hex representation, you already *have* a string, and
converting a string to a string is simple enough.
>
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
email: rjh at above domain (but drop the www, obviously)
converted to a hexadecimally represented string.
For example,
If my 128 bit value (represented hexadecimally) was:
4D9479E256DD4E8 A923E32065141A9 11
I would want this function to populate a char buffer such that:
charBuffer = "4D9479E256DD4E 8A923E32065141A 911"
Last time I checked... http://www.asciitable.com/If you have a hex representation, you already *have* a string, and
converting a string to a string is simple enough.
the string representation "4" has the hex value 0x34, contradicting
your above statement. You *always* have a string, though (as in this
case) it might not be the one you need.
Comment
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Re: 128-bit hex number ---> string buffer
In article <1158784086.598 461.44300@i3g20 00cwc.googlegro ups.com>,
ern <erniedude@gmai l.comwrote:
There is no library function to do that directly, but you can>Does a function exist to convert a 128-bit hex number to a string?
write it easily enough if you are willing to assume a specific
character set (such as ASCII), and if you know the encoding
represented by the number.
For example, one that would work for one encoding combination is,
#include <stdio.h>
int main(void) {
const char hexnum[33] = "54686520616e73 776572206973203 432";
unsigned short tmpbuf[16];
int i;
sscanf( hexnum,
"%2hx%2hx%2hx%2 hx%2hx%2hx%2hx% 2hx%2hx%2hx%2hx %2hx%2hx%2hx%2h x%2hx",
&tmpbuf[0], &tmpbuf[1], &tmpbuf[2], &tmpbuf[3],
&tmpbuf[4], &tmpbuf[5], &tmpbuf[6], &tmpbuf[7],
&tmpbuf[8], &tmpbuf[9], &tmpbuf[10], &tmpbuf[11],
&tmpbuf[12], &tmpbuf[13], &tmpbuf[14], &tmpbuf[15] );
for (i=0; i<16; i++) printf("%c", (unsigned char) tmpbuf[i] );
printf("\n");
return 0;
}
--
If you lie to the compiler, it will get its revenge. -- Henry Spencer
Comment
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Re: 128-bit hex number ---> string buffer
ern said:
Okay. Since C doesn't guarantee the existence of 128-bit integer types, let>
Richard Heathfield wrote:Let me rephrase. I have a 128 bit value, which I would like to be>ern said:
>>>>Does a function exist to convert a 128-bit hex number to a string?
>There is no such thing as a "hex number". Hexadecimal is a numeric
>representati on system, not a kind of number.
>>
>If you have a hex representation, you already *have* a string, and
>converting a string to a string is simple enough.
>>
converted to a hexadecimally represented string.
us assume that you have your value in an array of unsigned char, which we
will treat as a bit array.
Here's some preprocessing stuff:
#include <stddef.h>
#include <limits.h>
#define BYTE(x) ((x) / CHAR_BIT)
#define BIT(x) ((x) % CHAR_BIT)
#define SET_BIT(a, b) \
(a)[BYTE(b)] |= (1 << (BIT(b)))
#define CLEAR_BIT(a, b) \
(a)[BYTE(b)] &= ~(1 << (BIT(b)))
#define TEST_BIT(a, b) \
(!!(((a)[BYTE(b)]) & (1 << (BIT(b)))))
Okay, here's our routine, which requires a pointer to the first of len bytes
of input and a pointer to the first of len * 2 + 1 bytes of output. I
should warn you that I have *not* tested it very much! Especially the code
that deals with weird byte sizes. So it's likely to have a hole or two. But
it should give you the general idea, and is not limited to 128-bit inputs
(but if that is what you want and CHAR_BIT is 8, then this routine expects
len to be 16):
void to_hex(char *out, unsigned char *in, size_t len)
{
char hexabet[] = "0123456789ABCD EF";
size_t bits = len * CHAR_BIT;
size_t bit = 0;
int ch = 0;
if(bits & 7)
{
bit = bits & 7; bits &= ~7;
while(bit != 0)
{
ch <<= 1;
ch |= TEST_BIT(in, bit);
--bit;
}
*out++ = hexabet[ch];
}
for(; bit < bits; bit += 8)
{
*out++ = hexabet[(TEST_BIT(in, bit + 7) << 3) |
(TEST_BIT(in, bit + 6) << 2) |
(TEST_BIT(in, bit + 5) << 1) |
TEST_BIT(in, bit + 4)];
*out++ = hexabet[(TEST_BIT(in, bit + 3) << 3) |
(TEST_BIT(in, bit + 2) << 2) |
(TEST_BIT(in, bit + 1) << 1) |
TEST_BIT(in, bit)];
}
*out = '\0';
}
And here's a driver:
#include <stdio.h>
int main(void)
{
unsigned char in[] =
{
0x4D, 0x94, 0x79, 0xE2,
0x56, 0xDD, 0x4E, 0x8A,
0x92, 0x3E, 0x32, 0x06,
0x51, 0x41, 0xA9, 0x11
};
char out[33] = {0};
to_hex(out, in, 16);
printf("%s\n", out);
return 0;
}
ASCII has nothing to do with C.>>If you have a hex representation, you already *have* a string, and
>converting a string to a string is simple enough.
Last time I checked... http://www.asciitable.com/
You seem to be confusing character sets, representations , and values. Thethe string representation "4" has the hex value 0x34,
string "4" comprises two bytes, { '4', '\0' }. The code point of '4' is
implementation-defined. On some systems it is indeed 0x34, but on others
it's 0xF4 or some other value.
If it contradicts what I said, then it's wrong, because what I said iscontradicting your above statement.
right.
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
email: rjh at above domain (but drop the www, obviously)
Comment
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Re: 128-bit hex number ---> string buffer
Walter Roberson said:
You don't actually need to assume this. See my parallel reply.In article <1158784086.598 461.44300@i3g20 00cwc.googlegro ups.com>,
ern <erniedude@gmai l.comwrote:
>>>>Does a function exist to convert a 128-bit hex number to a string?
There is no library function to do that directly, but you can
write it easily enough if you are willing to assume a specific
character set (such as ASCII),
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
email: rjh at above domain (but drop the www, obviously)
Comment
-
Re: 128-bit hex number ---> string buffer
Walter Roberson wrote:Thanks for the code, but that's the inverse of what I need.In article <1158784086.598 461.44300@i3g20 00cwc.googlegro ups.com>,
ern <erniedude@gmai l.comwrote:
>>Does a function exist to convert a 128-bit hex number to a string?
There is no library function to do that directly, but you can
write it easily enough if you are willing to assume a specific
character set (such as ASCII), and if you know the encoding
represented by the number.
>
For example, one that would work for one encoding combination is,
>
>
#include <stdio.h>
>
int main(void) {
const char hexnum[33] = "54686520616e73 776572206973203 432";
unsigned short tmpbuf[16];
int i;
>
sscanf( hexnum,
"%2hx%2hx%2hx%2 hx%2hx%2hx%2hx% 2hx%2hx%2hx%2hx %2hx%2hx%2hx%2h x%2hx",
&tmpbuf[0], &tmpbuf[1], &tmpbuf[2], &tmpbuf[3],
&tmpbuf[4], &tmpbuf[5], &tmpbuf[6], &tmpbuf[7],
&tmpbuf[8], &tmpbuf[9], &tmpbuf[10], &tmpbuf[11],
&tmpbuf[12], &tmpbuf[13], &tmpbuf[14], &tmpbuf[15] );
>
for (i=0; i<16; i++) printf("%c", (unsigned char) tmpbuf[i] );
printf("\n");
>
return 0;
}
>
--
If you lie to the compiler, it will get its revenge. -- Henry Spencer
Let me make the problem simpler. Say I have a 32-bit integer. I what
the hexadecimally represented string version.
So the integer is the INPUT, and the hexadecimally represented string
is the OUTPUT.
Comment
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Re: 128-bit hex number ---> string buffer
ern posted:
If you have an aversion to the Standard Library, then maybe something alongIf my 128 bit value (represented hexadecimally) was:
>
4D9479E256DD4E8 A923E32065141A9 11
>
I would want this function to populate a char buffer such that:
>
charBuffer = "4D9479E256DD4E 8A923E32065141A 911"
the lines of the following. (This sample is capped at 32-Bit, but this is
easily adjusted.)
#include <assert.h>
#define MAX_VAL 0xFFFFFFFFu
#define MAX_DIVISOR 0x10000000u
#define MAX_HEX_DIGITS 8
typedef long unsigned IntType;
void IntToHexStr(Int Type val,char *const buf)
{
int const assert_dummy = ( assert(val <= MAX_VAL),
assert(!!buf),
0);
static char const digits[16] = "0123456789ABCD EF";
/* No terminating null character */
IntType divisor = MAX_DIVISOR;
char *p = buf;
do
{
/* Maybe "ldiv" should be used... ? */
unsigned const digit_val = val / divisor;
val %= divisor;
divisor /= 0x10;
if(digit_val) *p++ = digits[digit_val];
} while(divisor);
if(buf==p) *p++ = '0';
*p = 0;
}
#include <stdlib.h>
#include <stdio.h>
int main(void)
{
char buf[MAX_HEX_DIGITS+ 1];
IntToHexStr(0x7 78FEE34,buf);
puts(buf);
return 0;
}
--
Frederick Gotham
Comment
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Re: 128-bit hex number ---> string buffer
Richard Heathfield wrote:Thanks for the help all. I will try and apply the help to solve theern said:
>>
Richard Heathfield wrote:Let me rephrase. I have a 128 bit value, which I would like to beern said:
>
Does a function exist to convert a 128-bit hex number to a string?
>
There is no such thing as a "hex number". Hexadecimal is a numeric
representation system, not a kind of number.
>
If you have a hex representation, you already *have* a string, and
converting a string to a string is simple enough.
>
converted to a hexadecimally represented string.
Okay. Since C doesn't guarantee the existence of 128-bit integer types, let
us assume that you have your value in an array of unsigned char, which we
will treat as a bit array.
>
Here's some preprocessing stuff:
>
#include <stddef.h>
#include <limits.h>
>
#define BYTE(x) ((x) / CHAR_BIT)
#define BIT(x) ((x) % CHAR_BIT)
#define SET_BIT(a, b) \
(a)[BYTE(b)] |= (1 << (BIT(b)))
#define CLEAR_BIT(a, b) \
(a)[BYTE(b)] &= ~(1 << (BIT(b)))
#define TEST_BIT(a, b) \
(!!(((a)[BYTE(b)]) & (1 << (BIT(b)))))
>
Okay, here's our routine, which requires a pointer to the first of len bytes
of input and a pointer to the first of len * 2 + 1 bytes of output. I
should warn you that I have *not* tested it very much! Especially the code
that deals with weird byte sizes. So it's likely to have a hole or two. But
it should give you the general idea, and is not limited to 128-bit inputs
(but if that is what you want and CHAR_BIT is 8, then this routine expects
len to be 16):
>
void to_hex(char *out, unsigned char *in, size_t len)
{
char hexabet[] = "0123456789ABCD EF";
size_t bits = len * CHAR_BIT;
size_t bit = 0;
int ch = 0;
>
if(bits & 7)
{
bit = bits & 7; bits &= ~7;
>
while(bit != 0)
{
ch <<= 1;
ch |= TEST_BIT(in, bit);
--bit;
}
*out++ = hexabet[ch];
}
>
for(; bit < bits; bit += 8)
{
*out++ = hexabet[(TEST_BIT(in, bit + 7) << 3) |
(TEST_BIT(in, bit + 6) << 2) |
(TEST_BIT(in, bit + 5) << 1) |
TEST_BIT(in, bit + 4)];
*out++ = hexabet[(TEST_BIT(in, bit + 3) << 3) |
(TEST_BIT(in, bit + 2) << 2) |
(TEST_BIT(in, bit + 1) << 1) |
TEST_BIT(in, bit)];
}
*out = '\0';
}
>
And here's a driver:
>
#include <stdio.h>
>
int main(void)
{
unsigned char in[] =
{
0x4D, 0x94, 0x79, 0xE2,
0x56, 0xDD, 0x4E, 0x8A,
0x92, 0x3E, 0x32, 0x06,
0x51, 0x41, 0xA9, 0x11
};
char out[33] = {0};
to_hex(out, in, 16);
printf("%s\n", out);
return 0;
}
>
>>Last time I checked... http://www.asciitable.com/If you have a hex representation, you already *have* a string, and
converting a string to a string is simple enough.
ASCII has nothing to do with C.
>>the string representation "4" has the hex value 0x34,
You seem to be confusing character sets, representations , and values. The
string "4" comprises two bytes, { '4', '\0' }. The code point of '4' is
implementation-defined. On some systems it is indeed 0x34, but on others
it's 0xF4 or some other value.
>>contradicting your above statement.
If it contradicts what I said, then it's wrong, because what I said is
right.
>
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
email: rjh at above domain (but drop the www, obviously)
problem.
Comment
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Re: 128-bit hex number ---> string buffer
ern wrote:Does a function exist to convert a 128-bit hex number to a string?
What's a 'hex number?' You mean like "Thirteen", or "666" ?
Or do you mean a "number" and its hexidecimal representation?
If so, then by "number", do you mean some sort of 128 bit Integer
format? There are int128_t and uint128_t on some platforms, which are
nonstandard. But there are also bignum libraries with 128 bit floating
point types.
If you are simply trying to take a 16 byte chunk of data, and convert
the value of each nybble to a hex digit in a string, isn't that just a
matter of using "%x" in some *printf() variety?
Comment
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Re: 128-bit hex number ---> string buffer
"ern" <erniedude@gmai l.comwrites:#v+Let me make the problem simpler. Say I have a 32-bit integer. I what
the hexadecimally represented string version.
>
So the integer is the INPUT, and the hexadecimally represented string
is the OUTPUT.
sprintf(buffer, "%08x", integer);
#v-
Anyways, you probably ask for:
#v+
void binary2hex(cons t char *data, size_t len, char *buffer) {
static const char digits[16] = '0123456789ABCD EF";
size_t i = 0;
for (; i<len; ++i, ++data) {
*buffer++ = digits[(data>>4) & 0xf];
*buffer++ = digits[ data & 0xf];
}
*buffer = 0;
}
#v-
Not tested but should work. Buffer has to have at least 2*len+1 chars.
--
Best regards, _ _
.o. | Liege of Serenly Enlightened Majesty of o' \,=./ `o
..o | Computer Science, Michal "mina86" Nazarewicz (o o)
ooo +--<mina86*tlen.pl >--<jid:mina86*jab ber.org>--ooO--(_)--Ooo--
Comment
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Re: 128-bit hex number ---> string buffer
Michal Nazarewicz <mina86@tlen.pl writes:"ern" <erniedude@gmai l.comwrites:>Let me make the problem simpler. Say I have a 32-bit integer. I what
>the hexadecimally represented string version.
>>
>So the integer is the INPUT, and the hexadecimally represented string
>is the OUTPUT.Duh... Even though it doesn't make big difference it shall be:Anyways, you probably ask for:
void binary2hex(cons t char *data, size_t len, char *buffer) {
#v+
void binary2hex(cons t unsigned char *data, size_t len, char *buffer) {
#v-
The code will of course fail if CHAR_BIT != 8.
--
Best regards, _ _
.o. | Liege of Serenly Enlightened Majesty of o' \,=./ `o
..o | Computer Science, Michal "mina86" Nazarewicz (o o)
ooo +--<mina86*tlen.pl >--<jid:mina86*jab ber.org>--ooO--(_)--Ooo--
Comment
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Re: 128-bit hex number ---> string buffer
ern wrote:
To do numeric types greater than 32-bits will require a nonstandard type.Thanks for the code, but that's the inverse of what I need.
>
Let me make the problem simpler. Say I have a 32-bit integer. I what
the hexadecimally represented string version.
>
So the integer is the INPUT, and the hexadecimally represented string
is the OUTPUT.
#include <stdio.h>
int main(int argc, char** argv){
long unsigned int num;
num = 0xCAFEBABEUL;
printf("0x%8lX\ n", num);
return 0;
}
Comment
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Re: 128-bit hex number ---> string buffer
"Ben Pfaff" <blp@cs.stanfor d.eduwrote in message
news:878xkejmud .fsf@benpfaff.o rg...We miss you, Dan. Elijah"ern" <erniedude@gmai l.comwrites:
>>>Does a function exist to convert a 128-bit hex number to a string?
What format is the number in to start out with? I ask because
it's strange to specify that a number is in hex if it's simply in
an integer variable. Normally, one thinks of a numeric variable
as simply having a value, and the base is not important. Also: C
doesn't guarantee the existence of an 128-bit integer type.
--
"Am I missing something?"
--Dan Pop
Comment
Comment