Printing a hex dump of a buffer

[A. Farber]

Hello,

I'm trying to print out a const char* buffer
by first printing 16 bytes as hex codes,
then printing them again as characters
or dots (if they aren't printable) and so on:

20 A5 96 74 00 00 00 00 05 AD 4A 7C D3 FF 70 00 ..t......J|..p.
31 B0 66 7F 9F 3D AC 00 9D FF C2 02 AB 10 28 76 1.f..=........(v
49 9A 5E 7C F7 3D E4 00 35 F9 3A 02 83 00 E0 76 I.^|.=..5.:....v

Right now it's too late here in Germany
and I have come up with this fugliness only:

void
dump_buffer(unsigned n, const unsigned char* buf)
{
const unsigned char *p, *end;
unsigned i, j;

end = buf + n;

for (i = 0; ; i += 16) {
p = buf + i;
for (j = 0; j < 16; j++) {
fprintf(stderr, "%02X ", p[j]);
if (p + j >= end)
goto BREAKOUT;
}
fprintf(stderr, " ");
p = buf + i;
for (j = 0; j < 16; j++) {
fprintf(stderr, "%c", isprint(p[j]) ? p[j] :
'.');
if (p + j >= end)
goto BREAKOUT;
}
fprintf(stderr, "\n");
}
BREAKOUT:
return;
}

I know that it's horrible. Maybe I'll do better tomorrow.

Does maybe any kind soul has a nice solution
for this probably often encountered problem?

I've looked at the Vim's xxd source code, but
it is too much for me to understand right now...

Thank you
Alex

 

[Malcolm McLean]

Hello,

I'm trying to print out a const char* buffer
by first printing 16 bytes as hex codes,
then printing them again as characters
or dots (if they aren't printable) and so on:

20 A5 96 74 00 00 00 00 05 AD 4A 7C D3 FF 70 00 ..t......J|..p.
31 B0 66 7F 9F 3D AC 00 9D FF C2 02 AB 10 28 76 1.f..=........(v
49 9A 5E 7C F7 3D E4 00 35 F9 3A 02 83 00 E0 76 I.^|.=..5.:....v

Right now it's too late here in Germany
and I have come up with this fugliness only:

void
dump_buffer(unsigned n, const unsigned char* buf)
{
const unsigned char *p, *end;
unsigned i, j;

end = buf + n;

for (i = 0; ; i += 16) {
p = buf + i;
for (j = 0; j < 16; j++) {
fprintf(stderr, "%02X ", p[j]);
if (p + j >= end)
goto BREAKOUT;
}
fprintf(stderr, " ");
p = buf + i;
for (j = 0; j < 16; j++) {
fprintf(stderr, "%c", isprint(p[j]) ? p[j] :
'.');
if (p + j >= end)
goto BREAKOUT;
}
fprintf(stderr, "\n");
}
BREAKOUT:
return;
}

I know that it's horrible. Maybe I'll do better tomorrow.

Does maybe any kind soul has a nice solution
for this probably often encountered problem?

I've looked at the Vim's xxd source code, but
it is too much for me to understand right now...

Thank you
Alex

Untested but

void dumpbuff(char *buff, int N)
{
int i;

for(i=0;i<N;i+=16
if(i + 16 <= N)
printline(data+i, 16);
else
printline(data+1, N - i);
}

void printline(char *data, int N)
{
int i;

for(i=0;i<N;i++)
printf("%02x ", data):
printf(" ");
for(i=0;i<N;i++)
if(isgraph((unsigned) data))
printf("%c", data);
else
printf(".");
printf("\n");
}

As long as you break things up the right way the function can look nice and
clean.

 

[Ben Bacarisse]

I'm trying to print out a const char* buffer
by first printing 16 bytes as hex codes,
then printing them again as characters
or dots (if they aren't printable) and so on:

20 A5 96 74 00 00 00 00 05 AD 4A 7C D3 FF 70 00 ..t......J|..p.
31 B0 66 7F 9F 3D AC 00 9D FF C2 02 AB 10 28 76 1.f..=........(v
49 9A 5E 7C F7 3D E4 00 35 F9 3A 02 83 00 E0 76 I.^|.=..5.:....v

Does maybe any kind soul has a nice solution
for this probably often encountered problem?

I'd do this:

void dump_buffer(unsigned n, const unsigned char* buf)
{
int on_this_line = 0;
while (n-- > 0) {
fprintf(stderr, "%02X ", *buf++);
on_this_line += 1;
if (on_this_line == 16 || n == 0) {
int i;
fputs(" ", stderr);
for (i = on_this_line; i < 16; i++)
fputs(" ", stderr);
for (i = on_this_line; i > 0; i--)
fputc(isprint(buf[-i]) ? buf[-i] : '.', stderr);
fputs("\n", stderr);
on_this_line = 0;
}
}
}

 

[Peter Nilsson]

Hello,

I'm trying to print out a const char* buffer
by first printing 16 bytes as hex codes,
then printing them again as characters
or dots (if they aren't printable) and so on:

20 A5 96 74 00 00 00 00 05 AD 4A 7C D3 FF 70 00   ..t......J|..p.
31 B0 66 7F 9F 3D AC 00 9D FF C2 02 AB 10 28 76  1.f..=........(v
49 9A 5E 7C F7 3D E4 00 35 F9 3A 02 83 00 E0 76  I.^|.=..5.:....v

Right now it's too late here in Germany
and I have come up with this fugliness only:

void
dump_buffer(unsigned n, const unsigned char* buf)

Consider... [untested]

#include <stdio.h>
#include <limits.h>

#define BYTES_PER_LINE 16
#define DOT '.'

#define NIBBLES_PER_BYTE ((CHAR_BIT + 3) / 4)

void dump_buffer(FILE *fp, const void *vp, size_t n)
{
const unsigned char *b = vp;

while (n)
{
const unsigned char *p = b;
size_t i, w = n < BYTES_PER_LINE ? n : BYTES_PER_LINE;

for (i = 0; i < w; i++)
{
if (i != 0) putc(' ', fp);
printf("%0*X", NIBBLES_PER_BYTE, (unsigned) p);
}

for (; i < BYTES_PER_LINE; i++)
{
printf("%*c", NIBBLES_PER_BYTE + 1, ' ');
}

fputs(" ", fp);

for (i = 0; i < w; i++)
{
if ( p == ' '
|| (isprint(p) && !isspace(p)) )
putc(p, fp);
else
putc(DOT, fp);
}

putc('\n', fp);
n -= w;
b += BYTES_PER_LINE;
}
}

fprintf(stderr, "%02X ", p[j]);

You're assuming 8 bit bytes (if you want good formatting).

fprintf(stderr, "%c", isprint(p[j]) ? p[j] : '.');

Note that isprint() for \n, \t, \b, \r, etc... are likely
to be true.

 

[A. Farber]

Thank you all!

 

[Ben Bacarisse]

fprintf(stderr, "%c", isprint(p[j]) ? p[j] : '.');

Note that isprint() for \n, \t, \b, \r, etc... are likely
to be true.

That is at odds with my reading of the standard (specifically
paragraph 3 of 7.4). Am I misunderstanding what you mean?

 

[santosh]

fprintf(stderr, "%c", isprint(p[j]) ? p[j] : '.');

Note that isprint() for \n, \t, \b, \r, etc... are likely
to be true.

That is at odds with my reading of the standard (specifically
paragraph 3 of 7.4). Am I misunderstanding what you mean?

Certainly '\b', '\r' and possibly '\n' should be classified as control
characters as their function is to change the active position of the
device connected to the stream. There is nothing to be printed.

 

[Peter Nilsson]

fprintf(stderr, "%c", isprint(p[j]) ? p[j] : '.');

Note that isprint() for \n, \t, \b, \r, etc... are likely
to be true.

That is at odds with my reading of the standard (specifically
paragraph 3 of 7.4).  Am I misunderstanding what you mean?

No, I was wrong. Thanks for the correction.