I am really confused, but cannot to do simple task as it seems:
I simply need to set number of bits in byte.
For example:
I need 5 bits set. So I need 0xb00011111.
Is it possible to do this without loop?
Also I'd not like to write lot of #defines too.
For any integer n less than the number of bits in the word, the mask you need is:
const unsigned int mask = (1u << n) - 1;
No loop required.
A simple function using this:
unsigned int set_lsbs(unsigned int n)
{
return (1u << n) - 1;
}
The ten first results are:
0: 0x0
1: 0x1
2: 0x3
3: 0x7
4: 0xf
5: 0x1f
6: 0x3f
7: 0x7f
8: 0xff
9: 0x1ff
Note: the syntax 0xb00011111 is not not a binary literal, that 'b' is simply interpreted as a hex digit.
0xb00011111 just for simplicity. Thank you very much!
0xb00011111 is valid C, but it's not like what the OP expected
0xb00011111 is valid C but a very big hex literal :) The OP intended 0b00011111 which is not valid C, since C does not have binary literals.Generation of a mask to set the least-significant 5 bits for any integer type can be done thus:
mask = ~((~0u) << 5) ;
This will create an integer value with the least significant 5 bits set to 1 and all higher-order bits regardless of integer type set to 0.
Specifically for the 8 bit type in question:
uint8_t mask = ~((~0u) << 5) ;
To explain how this works (ignoring intermediate integer types which are larger than 8 bits):
~0u (one's compliment of zero) = 0x111111110x111000000x00011111 Is it possible to do this without loop?
yes, it's possible to set n bits in given number without loop but not random bits, you can set n consecutive bits.
To set a single bit at given position
num = num | 1 << pos ;
For setting 5 consecutive bits in given no you can use below logic
num = num | 31 << pos;
Here
31 ==> 1x2^0 + 1x2^1 + .. => 31 is the sum of 5 ones(1 1111)
case? No, only two bitwise operations make same code, fast and easy.double pow(float, int); returns double. Did you read answers before yours? This is very bad example and very bad advice. Better delete it.