setting bits of a byte in 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; }

Possible duplicate of How do you set, clear and toggle a single bit in C/C++? — Box Box Box Box
– Box Box Box Box, Commented Feb 8, 2016 at 7:13
Also, if (0 < position <= width) is wrong. it should be if ((0 < position) && (position <= width)). — Balaraman L
– Balaraman L, Commented Feb 8, 2016 at 7:16

LPs · Accepted Answer · 2016-02-08 07:46:22Z

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

John Forkosh · Accepted Answer · 2016-02-08 12:14:31Z

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 ; */

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