sprintf syntax for string to hex

[huey_jiang]

Hi All,

I am trying to figure out a right syntax to convert an integer array
into hex array. sprintf worked for me on doing single integer:

int i, Iarray[8], n=15;
char buf[9];
sprintf(buf, "0x%02x", n);

The above code worked. Howeve, what I am trying to do is to convert an
array. I tried the following syntax, no good:

for (i=0; i<9; i++) sprintf(buff, "0x%02x", Iarray);
for (i=0; i<9; i++) sprintf(buff + i, "0x%02x", Iarray);

Anybody can help? Thanks a lot!

Huey

 

[Jogi Kuenstner]

Hi All,

I am trying to figure out a right syntax to convert an integer array
into hex array. sprintf worked for me on doing single integer:

int i, Iarray[8], n=15;
char buf[9];
sprintf(buf, "0x%02x", n);

The above code worked. Howeve, what I am trying to do is to convert an
array. I tried the following syntax, no good:

for (i=0; i<9; i++) sprintf(buff, "0x%02x", Iarray);
for (i=0; i<9; i++) sprintf(buff + i, "0x%02x", Iarray);

Anybody can help? Thanks a lot!

Huey

You are incrementing the destination-buffer ( buff ) via i.
But you are writing on each run in the for-loop 4 Characters to it:
Assume the first three bytes in your buffer being ab, cd and ef.
You want to write : 0xab0xcd0xef (according to your format string without a
blank)
But in fact you write:
0xab
0xcd
0xef
=
000xef

Regards
Jogi

P.S.:
1.) you should look to the size of your destination-buffer, if your program
with the current fromat-string should work, it must be at least four times
bigger than the source !
2.) you should try to use the return-value of sprintf, which tells you how
much was written, so you could end up in something like:
p = buff;
for(...)
p += sprintf( p, "0x%02x", Iarray);

 

[Michael Mair]

Hi All,

I am trying to figure out a right syntax to convert an integer array
into hex array. sprintf worked for me on doing single integer:

Nope, you are converting an array of int into a string containing
a hex representation. You probably want to convert the array of
int into an array of strings which each contain the hex number.

int i, Iarray[8], n=15;
char buf[9];
sprintf(buf, "0x%02x", n);

The above code worked. Howeve, what I am trying to do is to convert an
array. I tried the following syntax, no good:

for (i=0; i<9; i++) sprintf(buff, "0x%02x", Iarray);

&buff if any

for (i=0; i<9; i++) sprintf(buff + i, "0x%02x", Iarray);

This effectively gives you a buffer overrun as you write
at least four characters plus string terminator to buff[0], buff[1],
...., buff[8], i.e. you occupy buff[0]..buff[12] even though you
only are allowed to access buff[0]..buff[8].
In addition, you access Iarray[8] which also is not available
by you. Even if buf is large enough and Iarray[8] exists, you
effectively could as well write
sprintf(buff, "000000000x%02x", Iarray[8]);
instead of your loops.

Apart from that, %x denotes an argument of type unsigned int,
so we need (unsigned int)Iarray.

The right size for buf to contain only n for arbitrary
unsigned int values is
(sizeof(unsigned int)*CHAR_BIT+3)/4
for the hex digits plus 2 for the leading '0' and 'x' plus the
string terminator '\0'. On typical implementations with 32-Bit
unsigned int, this is 11. If you use C99, you should
use snprintf() instead of sprintf() as sprintf() is inherently
unsafe. Then the correct minimum buffer size does not play
as much of a role.

Do not use "magic numbers" explicitly in your code; instead
#define them in one place and only use them.

#define LARGE_ENOUGH 11
#define ARRAY_SIZE 9

size_t i;
int Iarray[ARRAY_SIZE], ret;
char buffers[ARRAY_SIZE][LARGE_ENOUGH];

.....

for (i=0; i<ARRAY_SIZE; i++) {
ret = sprintf(buf, "0x%02x", (unsigned) Iarray);
if (ret >= LARGE_ENOUGH) {
/* We already have overwritten the array bounds and are
** deep in undefined behaviour; with a bit of luck, we
** will get this out: */
fprintf(stderr, "LARGE_ENOUGH is not large enough by %d\n",
ret+1-LARGE_ENOUGH);
/* Probably, data was corrupted; die. */
exit(EXIT_FAILURE);
} else if (ret < 0) {
fprintf(stderr, "sprintf failed for i=%lu\n",
(unsigned long) i);
}
}

In C99, we can do the loop somewhat cleaner:

for (i=0; i<ARRAY_SIZE; i++) {
ret = snprintf(buf, LARGE_ENOUGH, "0x%02x", (unsigned) Iarray);
if (ret >= LARGE_ENOUGH) {
/* We have not overwritten anything but: */
fprintf(stderr, "LARGE_ENOUGH is not large enough by %d\n",
ret+1-LARGE_ENOUGH);
} else if (ret < 0) {
fprintf(stderr, "sprintf failed for i=%zu\n", i);
}
}


Cheers
Michael

 

[CBFalconer]

I am trying to figure out a right syntax to convert an integer array
into hex array. sprintf worked for me on doing single integer:

int i, Iarray[8], n=15;
char buf[9];
sprintf(buf, "0x%02x", n);

The above code worked. Howeve, what I am trying to do is to convert an
array. I tried the following syntax, no good:

for (i=0; i<9; i++) sprintf(buff, "0x%02x", Iarray);
for (i=0; i<9; i++) sprintf(buff + i, "0x%02x", Iarray);

Anybody can help? Thanks a lot!

#define NINE 9
#define FIFTEEN 15 /* or suitable names for the purpost */
int i, Iarray[NINE], n = FIFTEEN;
char buf[NINE * FIFTEEN]; /* holds it all */
char *bptr;

bptr = buf;
for (i = 0; i < NINE; i++) {
bptr += sprintf(bptr, ......);
}

(UNTESTED) but the point is advancing bptr by the amount written,
and having a large enouth buf for the whole schmeer.