Need help - pointers

[Nezhate]

Hi There,
I'm writing a program that takes a decimal number and returns it's
binary equivalent.
the problem is that I don't know how to return (from function) the
binary value as an array of cocatenated characters. For example : I
want to convert number 3 using 4 bits, ie 3 <----"0011".
After executing, I'm getting this warning:
warning: return from incompatible pointer type
line:49: warning: function returns address of local variable

Here it is my code:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int power (int base, int m);
char* dec2bin(int input, int m);

int main ()
{
int input = 0;
int m = 0;

printf("Number to convert\n");
scanf("%d", &input);
printf ("Number of bits\n");
scanf ("%d",&m);
printf("input %d is converted to %s\n",input,dec 2bin(input, m));
exit (0);
}

char* dec2bin(int input, int m)
{
static int i = 0;
char* array[m+1];
int limit = 0;
limit = power(2, m);
const int MASK = limit/2; // the binary equivalent is: 1 + m 0 bits

array[m] = '\0';
for (i= 0; i < m ;i++)
{
if (input & MASK){
strcpy(array[i],"1");}
else{
strcpy(array[i],"0");}
input = input << 1;
}
return (array);
}

int power (int base, int m)
{
static int i;
int p=1;
for (i = 1; i <= m; ++i)
{
p =p * base;
}
return p;
}

[Richard Heathfield]

Martien Verbruggen said:

On Mon, 11 Aug 2008 12:02:39 +0000,
Richard Heathfield <rj*@see.sig.in validwrote:

>>
char *dec2bin(int n)

Following to your own explanation, above, shouldn't this be called
int2bin(), or maybe num2bin()?

int2bin would be favourite, I suppose. But of course I was lazy enough to
retain the OP's function name rather than take the trouble to think about
it.

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[Richard Heathfield]

Ben Bacarisse said:

Richard Heathfield <rj*@see.sig.in validwrites:

>Nezhate said:

>>I'm writing a program that takes a decimal number

<snip>

>>and returns it's binary equivalent.

Here's one way:

#include <limits.h>
#include <stddef.h>

char *dec2bin(int n)
{
static char bin[sizeof n * CHAR_BIT + 1];
size_t b = sizeof n * CHAR_BIT;
size_t i = 0;

while(b-- 0)
{
bin[i++] = '0' + !!(n & (1 << b));

Warning: this is UB in C99 (a signed shift into the sign bit
position[1]).

In 6.5.7(4), we have: "The result of E1 << E2 is E1 left-shifted E2 bit
positions; vacated bits are filled with zeros. If E1 has an unsigned type,
the value of the result is E1 × 2E2, reduced modulo one more than the
maximum value representable in the result type. If E1 has a signed type
and nonnegative value, and E1 × 2E2 is representable in the result type,
then that is the resulting value; otherwise, the behavior is undefined."

So yes, you're right, as a result of which I had another stab at this, but
fell foul of right-shifting an int (where the sign bit propagation is a
problem), and I thought about a couple of other approaches, but then I
decided it would be a lot easier just to drink my coffee and wake up
slowly. :-)

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[Iman S. H. Suyoto]

Richard Heathfield wrote:

Iman S. H. Suyoto said:

>Richard Heathfield wrote:

>>Here's one way:

#include <limits.h>
#include <stddef.h>

char *dec2bin(int n)
{
static char bin[sizeof n * CHAR_BIT + 1];

CMIIW. If the + 1 is for the '\0' terminator, don't we need that
terminator there?

Yes, and it's there. From 3.5.7 (Initialization ):

"If an object that has static storage duration is not initialized
explicitly, it is initialized implicitly as if every member that has
arithmetic type were assigned 0 and every member that has pointer type
were assigned a null pointer constant. If an object that has
automatic storage duration is not initialized explicitly, its value is
indeterminate."

Right. Sorry, I forgot about that.

Thanks, Richard.