Strange output problem

[Materialised]

Hi Everyone,

I am having a few issues while attempting to use Kevin Easton's base64
encryption algorythm. (See
http://groups.google.com/[email protected])

here is a sample program I am using to test the algorythm.

----- start main.c -----

#include <stdio.h>
#include "base64.h"

int main(void)
{
char string[5] = "hello";
int len;
char string2[10];
len = strlen(string);
printf("%s\n", string);
b64_encode(string2, string, len);
printf("%s\n", string2);

return 0;
}


----- end main.c -----

I compile the program using:

gcc base64.c main.c -o base

And when I run it I get the following output.
mick@codegurus $ ./base
hello?(????

aGVsbG/wBCjG7L+/AQ==??????(

Does anyone know where its going wrong?

 

[Ed Morton]

Hi Everyone,

I am having a few issues while attempting to use Kevin Easton's base64
encryption algorythm. (See

char string[5] = "hello";

Needs to be size "6" to hold the null character.

Ed.

 

[Pedro Graca]

I'll have a go! NEWBIE ANSWER -- probably full of errors too

#include <stdio.h>
#include "base64.h"

int main(void)
{
char string[5] = "hello";

string[0] = 'h'; /* this */
string[1] = 'e'; /* makes */
string[2] = 'l'; /* string */
string[3] = 'l'; /* 5 chars */
string[4] = 'o'; /* long! */
/* "hello" needs at least 6 chars: 'h', 'e', 'l', 'l', 'o', '\0' */

no more chars have been allocated

--> UB (undefined behaviour) ???

you are trying to copy '\0' to a place in memory you do not own.


You may want to let the compiler decide on the array size:
char string[] = "hello";

or specify enough size for your constants
char string[32] = "hello";

int len;
char string2[10];
len = strlen(string);

I think strlen() will go on after the final 'o' checking for the null
terminator which can be _anywhere_

printf("%s\n", string);

same thing here

b64_encode(string2, string, len);

and here

printf("%s\n", string2);

return 0;
}


----- end main.c -----

I compile the program using:

gcc base64.c main.c -o base

And when I run it I get the following output.
mick@codegurus $ ./base
hello?(????

Lucky you got a '\0' not very far away

 

[Michael Fyles]

#include <stdio.h>
#include "base64.h"

int main(void)
{
char string[5] = "hello";

string[0] = 'h'; /* this */
string[1] = 'e'; /* makes */
string[2] = 'l'; /* string */
string[3] = 'l'; /* 5 chars */
string[4] = 'o'; /* long! */
/* "hello" needs at least 6 chars: 'h', 'e', 'l', 'l', 'o', '\0' */

no more chars have been allocated

--> UB (undefined behaviour) ???

Not undefined behaviour.

you are trying to copy '\0' to a place in memory you do not own.

It's actually a special case of character array
initialisation -- if there is not enough space for the
terminating null, it will not be assigned.

 

[Mac]

#include <stdio.h>
#include "base64.h"

int main(void)
{
char string[5] = "hello";

string[0] = 'h'; /* this */
string[1] = 'e'; /* makes */
string[2] = 'l'; /* string */
string[3] = 'l'; /* 5 chars */
string[4] = 'o'; /* long! */
/* "hello" needs at least 6 chars: 'h', 'e', 'l', 'l', 'o', '\0' */

no more chars have been allocated

--> UB (undefined behaviour) ???

Not undefined behaviour.

Until you call a function which expects a C string. Without the
terminating NUL, it is not a string.

It's actually a special case of character array
initialisation -- if there is not enough space for the
terminating null, it will not be assigned.

--Mac

 

[Christopher Benson-Manica]

Not undefined behaviour.

But OP's attempt to print it was, correct?

 

[Bliss]

I am having a few issues while attempting to use Kevin Easton's base64
encryption algorythm.

here is a sample program I am using to test the algorythm.

----- start main.c -----

#include <stdio.h>
#include "base64.h"

int main(void)
{
char string[5] = "hello";

Bzzzt-----------------^^
Here is your problem. "hello" takes six chars (when you include the
terminating 0)
make that:
char string[] = "hello";
and let the compiler count the number of chars needed.

int len;
char string2[10];
len = strlen(string);
printf("%s\n", string);
b64_encode(string2, string, len);
printf("%s\n", string2);

return 0;
}


----- end main.c -----

I compile the program using:

gcc base64.c main.c -o base

And when I run it I get the following output.
mick@codegurus $ ./base
hello?(????

aGVsbG/wBCjG7L+/AQ==??????(

Does anyone know where its going wrong?

Yep you thought that "hello" was five chars long

Roger

 

[Thomas Stegen]

here is a sample program I am using to test the algorythm.

^^^^^^^^^

I like this word Yeah, I got good algorythm.

 

[Eric Sosman]

^^^^^^^^^

I like this word Yeah, I got good algorythm.

Algorhythm,
Algomusic,
Algorhythm,
Who could ask for anything more?

 

[Lew Pitcher]

-----BEGIN PGP SIGNED MESSAGE-----
Hash: SHA1

Eric Sosman wrote:

| Thomas Stegen wrote:
|
|>Materialised wrote:
|>
|>>here is a sample program I am using to test the algorythm.
|>
|> ^^^^^^^^^
|>
|>I like this word Yeah, I got good algorythm.
|
|
| Algorhythm,
| Algomusic,
| Algorhythm,
| Who could ask for anything more?

"Algorythm" sounds like a method of birth control for programmers ;-)


- --

Lew Pitcher, IT Consultant, Enterprise Application Architecture
Enterprise Technology Solutions, TD Bank Financial Group

(Opinions expressed here are my own, not my employer's)
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[August Derleth]

But OP's attempt to print it was, correct?

Indeed. printf() (and all other string-expecting C functions, whether or
not the Standard mentions them) expects a null-terminated string.

 

[Dan Pop]

Indeed. printf() (and all other string-expecting C functions, whether or
not the Standard mentions them) expects a null-terminated string.

Wrong!

fangorn:~/tmp 38> cat test.c
#include <stdio.h>

int main(void)
{
char word[5] = "hello";

printf("%.5s\n", word);
return 0;
}
fangorn:~/tmp 39> gcc test.c
fangorn:~/tmp 40> ./a.out
hello

And, to prove that it worked by design and not by accident:

s If no l length modifier is present, the argument shall
be a pointer to the initial element of an array of
character type. Characters from the array are
written up to (but not including) the terminating null
character. If the precision is specified, no more than
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
that many bytes are written. If the precision is not
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
specified or is greater than the size of the array,
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
the array shall contain a null character.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

The standard is usually very precise WRT when a null-terminated string
is actually expected and when any ordinary character array will do.
No need for bogus rules of thumb.

Dan

 

[Eric Sosman]

Indeed. printf() (and all other string-expecting C functions, whether or
not the Standard mentions them) expects a null-terminated string.

Nit-pick: If the "%s" conversion is used with a precision
(e.g. "%.10s" or "%.*s") the corresponding argument need only
point to the start of a character array, not necessarily zero-
terminated.

char hello[5] = "hello"; /* no terminator */
printf ("%s\n", hello); /* undefined behavior */
printf ("%.4s\n", hello); /* well-defined; prints "hell\n" */

 

[Christopher Benson-Manica]

char word[5] = "hello";
printf("%.5s\n", word);

Or, better yet,

printf( "%.*s\n", sizeof word, word );

Right?

 

[Eric Sosman]

char word[5] = "hello";
printf("%.5s\n", word);

Or, better yet,

printf( "%.*s\n", sizeof word, word );

Right?

printf ("%.*s\n", (int)(sizeof word), word);

 

[Dan Pop]

char word[5] = "hello";
printf("%.5s\n", word);

Or, better yet,

printf( "%.*s\n", sizeof word, word );

Right?

Wrong. Undefined behaviour is never better than well defined behaviour.

Even if correctly done, your variant is ludicrous for my trivial example
and unlikely to be of much use in real life code, where the actual value
is seldom dictated by the size of the array (whose definition may not
even be in scope).

Dan