Im relatievly new to the whole bit shifting, and c++.
Let's say i have an uint8_t 00100100 (36) and i want to check if the 3th bit is set. Here is the code how im doing it now, for only one bit.
uint8_t x = 36; if(x&1<<3) printf("is set"); how can i check if the 3th OR the 6th bit is set? I want to check several combinations of bits like 5th or 7th or 8th.
what is the most elegant way to do it?
Checking bits by numeric position is one of the correct ways to do so but it makes the code depend on magic numbers, which makes it harder to read and maintain.
Usually, when checking for bitmasks, there is a goal of checking for some specific flag, let's say a hardware register for example.
Imagine for example that each bit of your integer represents a specific light in your house, and you want to check whether the bathroom and the kitchen are on:
enum LightMask { ENTRANCE = 0x01, LIVING_ROOM = 0x02, RESTROOM = 0x04, KITCHEN = 0x08, BATHROOM = 0x10, BEDROOM1 = 0x20, BEDROOM2 = 0x40, ATTIC = 0x80 }; uint8_t lightsOn = GetLightsOn(); if (lightsOn & (BATHROOM | KITCHEN)) { // ... } This is elegant, easy to understand and can be modified pretty easily.
The enum could also be expressed in terms of bit shifting at no cost for the compiler if you want to explicitely use bit positions instead of constants:
enum LightMask { ENTRANCE = 1 << 0, LIVING_ROOM = 1 << 1, RESTROOM = 1 << 2, KITCHEN = 1 << 3, BATHROOM = 1 << 4, BEDROOM1 = 1 << 5, BEDROOM2 = 1 << 6, ATTIC = 1 << 7 }; 
If you don't know which bit position you want to check until runtime, one way is to make a function so you can call it for any nth bit you want to check:
bool IsBitSet(uint8_t num, int bit) { return 1 == ( (num >> bit) & 1); } uint8_t x = 37; //00100101 for (int i = 0; i < 8; ++i) { if ( IsBitSet(x, i) ) printf("%dth bit is set\n", i); else printf("%dth bit not set\n", i); } The output would be:
0th bit is set 1th bit is not set 2th bit is set 3th bit is not set 4th bit is not set 5th bit is set 6th bit is not set 7th bit is not set If I wanted to check whether bit 3 or bit 6 is set:
uint8_t x = 36; //00100100 if ( IsBitSet(x, 2) || IsBitSet(x, 5) ) printf("bit 3 and/or bit 6 is set\n"); You could also make this function inline to possibly increase efficiency.
uint8_t x = 36; if (x & ((1 << 2) | (1 << 5))) printf("is set"); or if you know hex:
uint8_t x = 36; if (x & 0x24) printf("is set"); 
(1 << 2) | (1 << 5) is the equivalent of (4) + (32), which is 36.
<< 2to shift the third bit to the 1's place.(a & b) || (a & c)is equivalent to(a & b) | (a & c)which is algebraically equivalent toa & (b | c). Iow, you can combine your masks by oring them and apply them all at once at the end