probably there is a smart way to do that , but anyway i get error on this :
-(int*)decimalBinary:(int)decimal
{
int i=0;
int *bin;
while (decimal!=0)
{
bin[i]=decimal%2;
decimal=decimal/2;
i++;
}
return bin;
}
on the modulo line . why ? And whats the better way to get it to array ?
Declaring
int *bin;
sets aside space for a pointer but doesn't make it point to an object. It is crucial to initialize bin before using it.
To solve your problem you can declare an array bin[4] in caller function (int main) and then pass *bin to your calling function.
The following code is adapted from This answer on how to print an integer in binary format. Storing "binary digits" into an int array is added into the code below:
#include <stdio.h> /* printf */
#include <stdlib.h> /* strtol */
const char *byte_to_binary(long x);
int main(void)
{
long lVal;
int i, len, array[18];
char buf[18];
{ /* binary string to int */
char *tmp;
char *b = "11010111001010110";
lVal=strtol(b, &tmp, 2); //convert string in "base 2" format to long int
printf("%d\n", lVal);
}
{
printf("%s", byte_to_binary(lVal));
/* byte to binary string */
sprintf(buf,"%s", byte_to_binary(lVal));
}
len = strlen(buf);
for(i=0;i<len;i++)
{ //store binary digits into an array.
array[i] = (buf[i]-'0');
}
getchar();
return 0;
}
const char *byte_to_binary(long x)
{
static char b[17]; //16 bits plus '\0'
b[0] = '\0';
char *p = b;
int z;
for (z = 65536; z > 0; z >>= 1) //2^16
{
*p++ = (x & z) ? '1' : '0';
}
return b;
}
int *bintoint bin[sizeof(int) + 1. Set a point to end ofbufand fill backwards, decrementing the pointer each step.