Storing a Binary Value into an unsigned int array - c

Question

I've been having some trouble storing a binary value into an unsigned int array. I was trying to pass through the binary representation as a char array however it's not working as I had hoped. What I was doing was going through the char array using a while loop and assigning each number to an element to the unsigned int array but that is completely wrong. I just pretty much want to know how to store a binary value into an unsigned int. How do I show zeros in front? I've tried putting in the array just the binary value without 0s at the front but that wasn't working. Could I convert the binary to int value and then convert it back at print time?
Here is the basic code for the function
void setstring(unsigned int array[10], char *bitString) {
len=strlen(bitString);
for (int i=1; i<=10; i++) {
for (int p=1; d%32!=0; d++) {
array[10-i]=bitString[len-];
}
}
}
The printing is just meant to print the bitString or the array but at the moment nothing is printing at all. The print is simply a for loop which iterates through the unsigned int array.

Answer 1

You might be looking for a function to set particular bits of an unsigned char variable.
Try using this to set bits
void setBit(unsigned char *target, int pos)
{
//pos must be < sizeof(unsigned char)
unsigned char mask=1<<pos;
*target = *target | mask;
}
And this to unset bits
void unsetBit(unsigned char *target, int pos)
{
unsigned char mask=~(1<<pos);
*target = *target & mask;
}
Note that pos begins from 0.
You could use these functions for displaying the bits:
int getBit(unsigned char target, int pos)
{
target = target>>pos;
return target & 1;
}
void printBits(unsigned char target)
{
int i;
for(i=sizeof(target)*8-1; i>=0; --i)
{
printf("%d", getBit(target, i));
}
}
In these functions the target variable is passed by reference.
Example:
unsigned char a=0;
setBit(&a, 0);
setBit(&a, 1);
setBit(&a, 6);
printf("\nBit pattern is: ");
printBits(a);
printf(". Value is %d.", a);
would print
Bit pattern is: 01000011. Value is 67.
Further
unsetBit(&a, 1);
printf("\nBit pattern is: ");
printBits(a);
printf(". Value is %d.", a);
would give
Bit pattern is: 01000001. Value is 65.
EDIT: This is a good place to learn bit manipulation.

Answer 2

Here is your (set) of functions :
int binary_string_to_integer(const char *str, unsigned int *vl)
{
unsigned int value = 0;
int result = 0;
if (str == NULL || vl == NULL || !strlen(str)) result = -1;
if (!result)
{
while (!result && *str)
{
if (*str != '1' && *str != '0')
{
result = -1;
}
value <<= 1;
value += (*str++ == '1');
}
}
if (!result) *vl = value;
return result;
}
int binary_string_to_integer_array_of_0_or_1(const char *str, int *vl, int skipzeroes)
{
int result = 0;
if (str == NULL || vl == NULL || !strlen(str)) result = -1;
if (!result)
{
while (*str)
{
if (*str != '1' && *str != '0')
{
result = -1;
break;
}
if (skipzeroes && !result && *str == '0')
{
str++;
continue;
}
*(vl + result++) = (*str++ == '1');
}
}
return result;
}
char *reverse(char *str)
{
char tmp;
size_t len;
if (str != NULL)
{
len = strlen(str);
for (size_t i = 0; i < len / 2; i++)
{
tmp = *(str + i);
*(str + i) = *(str + len - i - 1);
*(str + len - i - 1) = tmp;
}
}
return str;
}
char *uint_to_binary_string(unsigned int vl, char *buff)
{
char *ptr = buff;
if (buff != NULL)
{
while (vl)
{
*ptr++ = '0' + (vl & 1);
vl >>= 1;
}
}
*ptr = 0;
return reverse(buff);
}