Problem printing Hexadecimal

[abhishekkarnik]

Hi,
I am trying to read an exe file and print it out character by
character in hexadecimal format. The file goes something like this in
hexadecimal
0x4d 0x5a 0x90 0x00 0x03 .... so on
When I read it into my array " two_bytes" which has 16 characters in
total, I get each of these values read correctly, but when I attempt
printing them out, it prints something like this:
0x4d 0x5a 0xffffff90 0x00 0x03 .... so on. I have no clue why. How can
I correct this

FILE *fp = NULL;
int count = 0;
char nibble = NULL;
char two_bytes[16];
int i;
fp = fopen("xyz.exe" ,"r");
if(fp == NULL)
{
printf("Unable to Open file\n");
exit(0);
}
while(!feof(fp) )
{
fscanf(fp,"%c", &two_bytes[count++]);
if(count % 16 == 0)
{
for(i=0; i<count ; i++)
printf("%#x ",two_bytes[i]);
printf(" || ");
for(i=0; i<count ; i++)
printf("%4c ",two_bytes[i]);
count = 0;
printf("\n");
}
}
fclose(fp);
return 0;

[Hallvard B Furuseth]

ab************@ gmail.com writes:

char two_bytes[16];
(...)
printf("%#x ",two_bytes[i]);

'char' can be signed. A negative char is promoted to int, and
%#x prints it as unsigned int, thus char 0x90 becomes 0xffffff90.

Use unsigned char two_bytes[16];
or printf("%#x ", (unsigned char) two_bytes[i]);

--
Regards,
Hallvard

[Walter Roberson]

In article <11************ *********@n59g2 000hsh.googlegr oups.com>,
<ab************ @gmail.comwrote :

I am trying to read an exe file and print it out character by
character in hexadecimal format. The file goes something like this in
hexadecimal
0x4d 0x5a 0x90 0x00 0x03 .... so on
When I read it into my array " two_bytes" which has 16 characters in
total, I get each of these values read correctly, but when I attempt
printing them out, it prints something like this:
0x4d 0x5a 0xffffff90 0x00 0x03 .... so on. I have no clue why. How can
I correct this

>FILE *fp = NULL;
int count = 0;
char nibble = NULL;
char two_bytes[16];

make two_bytes unsigned char

> int i;
fp = fopen("xyz.exe" ,"r");
if(fp == NULL)
{
printf("Unable to Open file\n");
exit(0);
}
while(!feof(fp) )
{
fscanf(fp,"%c", &two_bytes[count++]);

You've made a classic error there. feof() is not set until
you try and fail to read a character -- it does NOT "peek ahead"
to see whether the next read would fail or not. So you are going
to have to try to read the character and then determine whether
the read failed.

There is no good reason to fscanf with a single %c format:
see fgetc().

> if(count % 16 == 0)
{
for(i=0; i<count ; i++)
printf("%#x ",two_bytes[i]);
printf(" || ");
for(i=0; i<count ; i++)
printf("%4c ",two_bytes[i]);

Oh, you don't want to do that. What if it's a newline or a bell
or something like that?

You should first test that it's a printable character, and if
it is then you can putchar() it; if it isn't, then print
a placeholder instead ('.' is typical.)

Or instead of a placeholder, if it isn't important that the
text equivilent always be exactly the same length, you could
print a representation -- for example, the character pair
'\' 't' for tab, '\' 'b' for backspace. ^ followed by a letter
is a common representation for ASCII control characters.

That just leaves you with a choice of representations when
the character is above the ASCII printable range, such as if
it is 0xac ...

> count = 0;
printf("\n");
}
}
fclose(fp);
return 0;

--
Programming is what happens while you're busy making other plans.

[abhishekkarnik]

Thank you !!! That fixed it. I realized I'd forgotten to make it a
"unsigned char"

On Mar 19, 5:13 pm, Hallvard B Furuseth <h.b.furus...@u sit.uio.no>
wrote:

abhishekkar...@ gmail.com writes:

char two_bytes[16];
(...)
printf("%#x ",two_bytes[i]);

'char' can be signed. A negative char is promoted to int, and
%#x prints it as unsigned int, thus char 0x90 becomes 0xffffff90.

Use unsigned char two_bytes[16];
or printf("%#x ", (unsigned char) two_bytes[i]);

--
Regards,
Hallvard

[Default User]

ab************@ gmail.com wrote:

Thank you !!! That fixed it. I realized I'd forgotten to make it a
"unsigned char"

Please don't top-post. Your replies belong following or interspersed
with properly trimmed quotes. See the majority of other posts in the
newsgroup, or:
<http://www.caliburn.nl/topposting.html >

[Tobias Rischer]

ab************@ gmail.com wrote:

Hi,
I am trying to read an exe file and print it out character by
character in hexadecimal format. The file goes something like this in
hexadecimal
0x4d 0x5a 0x90 0x00 0x03 .... so on
When I read it into my array " two_bytes" which has 16 characters in
total, I get each of these values read correctly, but when I attempt
printing them out, it prints something like this:
0x4d 0x5a 0xffffff90 0x00 0x03 .... so on. I have no clue why.

Because the char is a small integer which can be positive or negative.
Before being used by printf, it is implicitly converted to a normal
integer. Every byte from 0x80 to 0xFF will be interpreted as a negative
number, and the print result is based on a four-byte-representation of
that number.

How can
I correct this

FILE *fp = NULL;
int count = 0;
char nibble = NULL;
char two_bytes[16];

Here: declare your bytes as unsigned char.

Good luck!

Tobias

[Kenny McCormack]

In article <56************ *@mid.individua l.net>,
Default User <de***********@ yahoo.comwrote:

>ab************ @gmail.com wrote:

>Thank you !!! That fixed it. I realized I'd forgotten to make it a
"unsigned char"

Please don't top-post. Your replies belong following or interspersed
with properly trimmed quotes. See the majority of other posts in the
newsgroup, or:
<http://www.caliburn.nl/topposting.html >

Get a life!

[Old Wolf]

On Mar 20, 9:19 am, [e-mail] c-cnrc.gc.ca (Walter Roberson)
wrote:

<abhishekkar... @gmail.comwrote :

char two_bytes[16];

make two_bytes unsigned char

fscanf(fp,"%c", &two_bytes[count++]);

There is no good reason to fscanf with a single %c format:
see fgetc().

Well, fgetc() returns a value in the range for unsigned char.
If he were still reading into a char, then using fgetc() would
cause implementation-defined behaviour due to out-of-range
assignment. However, fscanf with "%c" does not suffer that same
problem.

[pete]

Old Wolf wrote:

>
On Mar 20, 9:19 am, [e-mail] c-cnrc.gc.ca (Walter Roberson)
wrote:

<abhishekkar... @gmail.comwrote :

char two_bytes[16];

make two_bytes unsigned char

fscanf(fp,"%c", &two_bytes[count++]);

There is no good reason to fscanf with a single %c format:
see fgetc().

Well, fgetc() returns a value in the range for unsigned char.
If he were still reading into a char, then using fgetc() would
cause implementation-defined behaviour due to out-of-range
assignment.
However, fscanf with "%c" does not suffer that same problem.

fgetc returns type int
and is capable of returning a negative value equal to EOF.

--
pete

[Walter Roberson]

In article <11************ **********@o5g2 000hsb.googlegr oups.com>,
Old Wolf <ol*****@inspir e.net.nzwrote:

>On Mar 20, 9:19 am, [e-mail] c-cnrc.gc.ca (Walter Roberson)
wrote:

> <abhishekkar... @gmail.comwrote :

char two_bytes[16];

>make two_bytes unsigned char

fscanf(fp,"%c", &two_bytes[count++]);

>There is no good reason to fscanf with a single %c format:
see fgetc().

>Well, fgetc() returns a value in the range for unsigned char.
If he were still reading into a char, then using fgetc() would
cause implementation-defined behaviour due to out-of-range
assignment.

You cannot "read into a char" when using fgetc(): you have
to assign its value to something. You would take the usual
precautions to assign to an int, check the value for EOF, and
only then assign from the int into the unsigned char.

>However, fscanf with "%c" does not suffer that same problem.

You would have to check the return value of fscanf(), which
is going to return EOF when the input ends. If you are prone
to make the mistake of assigning the return value to an unsigned
value for fgetc(), you are equally likely to be prone to assigning
the return value from fscanf() to an unsigned int.
--
Okay, buzzwords only. Two syllables, tops. -- Laurie Anderson