setting bits of a byte in C - c

Question

Im try to set the bits of a byte , based on this function signature
int setbits(int old_value, int position, int width, int set_value)
what am I doing wrong, because it keeps returning 0.
#include "stdafx.h"
int setbits(int old_value, int position, int width, int set_value);
int _tmain(int argc, _TCHAR* argv[])
{
//for example turn hex 0xA (10 base 2) into ( 8 base 2)
int old_value = 0xA;
int position = 2;
int width = 4;
int set_value = 0;
printf("setbits, old_value %x, position %i, width %i, set_value %x, output %x",old_value,7,width,set_value,0);
getchar();
}
//position - is the least significant bit of the old_value you wish to change
//old_value - is the value before you start changing it
//width - is the width in bits of the value you want to change (old_value width).
//set_value - is the value you want to use to modify the bits
int setbits(int old_value, int position, int width, int set_value)
{
int mask = 0x1;
int return_val = 0x0;
if (0 < position <= width)
{
//initialize mask
int c=0;
for (c; c<width ; c++)
{
mask = mask << 1;
}
printf("ini mask %x \n",mask);
//shift into position the set_value (aka state)
mask = mask >> position;
printf("shifted mask %x \n",mask);
//if state is 1
if (set_value)
{
return_val = mask | old_value;
}
else
{
return_val = (~mask) & old_value;
}
}
else
{
printf("setbits(), position is out of range, position : %x, width : %x", position, width);
}
return return_val;
}

Answer 1

Correcting your code
int setbits(int old_value, int position, int width, int set_value)
{
int mask = 0x1;
int return_val = 0x0;
if ((0 < position) && (position <= width) )
{
//initialize mask
mask = mask << width;
printf("ini mask 0x%02X \n",mask);
//shift into position the set_value (aka state)
mask = mask >> position;
printf("shifted mask 0x%02X \n",mask);
//if state is 1
if (set_value)
{
return_val = mask | old_value;
}
else
{
return_val = (~mask) & old_value;
}
printf("returned value: 0x%02X\n", return_val);
}
else
{
printf("setbits(), position is out of range, position : %x, width : %x", position, width);
}
return return_val;
}
With your example you'll get the output:
ini mask 0x10
shifted mask 0x04
returned value: 0x0A
That it is exactly what expected.
value --> 0x0A --> 0b00001010
mask --> 0x04 --> 0b00000100
~mask --> 0xFB--> 0b11111011
value & ~mask = 0x0A

Answer 2

Another way of approaching this problem that you might (or might not:) want to consider is with a few simple macros, e.g.,
/* ---
* bitfield macros (byte_bits=76543210, with lsb=bit#0 and 128=bit#7set)
* --------------------------------------------------------------------- */
#define getbit(x,bit) ( ((x)>>(bit)) & 1 ) /* get bit-th bit of x */
#define setbit(x,bit) ( (x) |= (1<<(bit)) ) /* set bit-th bit of x */
#define clearbit(x,bit) ( (x) &= ~(1<<(bit)) ) /* clear bit-th bit of x */
#define putbit(x,bit,val) \
if(((int)(val))==0) clearbit((x),(bit)); else setbit((x),(bit))
#define bitmask(nbits) ((1<<(nbits))-1) /* a mask of nbits 1's */
#define getbitfield(x,bit1,nbits) (((x)>>(bit1)) & (bitmask(nbits)))
#define putbitfield(x,bit1,nbits,val) /* x:bit1...bit1+nbits-1 = val */ \
if ( (nbits)>0 && (bit1)>=0 ) { /* check input */ \
(x) &= (~((bitmask((nbits))) << (bit1))); /*set field=0's*/ \
(x) |= (((val)&(bitmask((nbits)))) << (bit1)); /*set field=val*/ \
} else /* let user supply final ; */