Ok, I try to explain with a simplified example
#include <stdio.h>
/* Just for print in binary mode */
char *chartobin(unsigned char c)
{
static char a[9];
int i;
for (i = 0; i < 8; i++)
a[7 - i] = (c & (1 << i)) == (1 << i) ? '1' : '0';
a[8] = '\0';
return a;
}
int main(void)
{
unsigned char u = 0xf;
printf("%s\n", chartobin(u));
u >>= 2; // Shift bits 2 positions (to the right)
printf("%s\n", chartobin(u));
printf("%s\n", chartobin(u & 0x1)); // Check if the last bit is on
return 0;
}
Output:
00001111
00000011
00000001
Do I replace the above line with #define AR5K_EEPROM_FF_DIS(_v) 1
?
Nooooo!!
If you initialize u
with 0xb
instead of 0xf
you get:
00001011
00000010
00000000
As you can see (((_v) >> 2) & 0x1 != 1