Bit macro in linux

Test_bit macro in C/C++

I’m trying to read input of a linux device via ioctl() and I saw on many example of code with the «test_bit» macro but the only code I found about it is this : #define test_bit(bit, array) (array[bit / 8] & (1

void test()< uint8_t key_b[KEY_MAX/8 + 1]; /* the events (up to 64 at once) */ const char *keyboard = "/dev/input/keyboard0"; int keybrdToCapture; int yalv; keybrdToCapture = open(keyboard, O_RDONLY); memset(key_b, 0, sizeof(key_b)); ioctl(keybrdToCapture, EVIOCGKEY(sizeof(key_b)), key_b); for (yalv = 0; yalv < KEY_MAX; yalv++) < if (test_bit(yalv, key_b)) < switch ( yalv) < case 0x1c : dial->setMessage("Enter"); dial->show(); break; case 0x66 : dial->setMessage("Home"); dial->show(); break; case 0x3b : dial->setMessage("F1"); dial->show(); break; case 0x3c : dial->setMessage("F2"); dial->show(); break; default: dial->setMessage("Unknow for now"); dial->show(); > > > 

How does #define test_bit(bit, array) (array[bit / 8] & (1

Thank you for that answer, I’m quite new to c++ and linux and from what I’ve understood this code is supposed to check that a specific bit is set to 1 in a specific byte. (Maybe I’m totaly wrong I did not get many help on the explanation of this from now).But when I press a key in debug I can see one bit in my byte change from 0 to 1 but I never enter in my test_bit condition. Maybe it’s is simply a problem of misunderstanding.

array[bit / 8] divides bit number by 8 to address the proper byte of array . bit % 8 computes remainder of division by 8 (actually, separates last 3 bits storing 0 . 7). 1

array should be a variable (or expression) with type array of unsigned char . ( char or signed char might work as well but signed-ness can cause issues.)

From your previous comment I change with this hoping it’s better : #ifndef test_bit #define test_bit(bit, array) (array[bit / 8] & (1

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[Solved]-What’s the meaning of BIT() in this linux kernel macro?-C

looks like you can find the answer inside the first header file included, i.e. bitops.h!

i.e. BIT defines a bit mask for the specified bit number from 0 (least significant, or rightmost bit) to whatever fits into an unsigned long.
So BIT(10) should evaluate to the numeric value of 1024 (which is 1

The BIT macro shifts the value 1 left by the value given to it, so BIT(10) == (1

BIT is a macro defined in include/linux/bitops.h in the kernel tree:

So BIT(10) is basically an unsigned long with tenth bit set.

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