Get n msb and lsb bits

[lh-]

hi

i'm writing a program in c++ where i have registers. they are 8 bit each. however, i may have values bigger than 255, in which case i need to place the value in say 2 registers if it has 16 bits, 3 registers for 24 bits etc.

so i need to extract first 8 bits (which for me are msb) and second 8 bits (lsb).

getting lsb didn't require me much of thinking:

register = value & 0x00FF

for msb, this is what i come up with:

register = (value & 0xFF00) >> 8;

it works but i'm here to ask if there's a better solution.

thanks

 

[Easyrider83]

Re: get n msb and lsb bits

Convert to 16-bit pointer and read as array. Also, you can use union. But compilied assembly will be equal I beileve.

 

[lh-]

Re: get n msb and lsb bits

hmm. are those 2 really better solutions? i mean, bitwise operations are much faster than any other operation, let alone a for loop executing 16 times.

and i still don't get what is the purpose of this array or union. my register storage is itself an array of bytes.

 

[andre_luis]

Re: get n msb and lsb bits

and i still don't get what is the purpose of this array or union. my register storage is itself an array of bytes.

As Easyrider83 mentioned, by handling the variable of 2 bytes length as an array, you can read one byte without the need of any shift opperation, as for example if you declare this:

uint8_t Variable[2];
uint16_t * ptr = Variable;
*(ptr) = 0x1234

In theory, you should be able to read its 8 MSB by something like that (not tested):

uint8_t Value_msb = *(ptr); // 0x12

And to read the 8 LSB, by something like that:

uint8_t Value_lsb = *(ptr+1); // 0x34

 

[Easyrider83]

Re: get n msb and lsb bits

No need to declare new pointer. Just pointer-type conversion which takes no code