Reputation: 8092
I want find the first non-zero bit in the binary representation of a u32
. leading_zeros
/trailing_zeros
looks like what I want:
let x: u32 = 0b01000;
println!("{}", x.trailing_zeros());
This prints 3 as expected and described in the docs. What will happen on big-endian machines, will it be 3 or some other number?
The documentation says
Returns the number of trailing zeros in the binary representation
is it related to machine binary representation (so the result of trailing_zeros
depends on architecture) or base-2 numeral system (so result will be always 3)?
Upvotes: 4
Views: 3406
Reputation: 601779
The type u32
respresents binary numbers with 32 bits as an abstract concept. You can imagine them as abstract, mathematical numbers in the range from 0 to 232-1. The binary representation of these numbers is written in the usual convention of starting with the most significant bit (MSB) and ending with the least significant bit (LSB), and the trailing_zeros()
method returns the number of trailing zeros in that representation.
Endianness only comes into play when serializing such an integer to bytes, e.g. for writing it to a bytes buffer, a file or the network. You are not doing any of this in your code, so it doesn't matter here.
As mentioned above, writing a number starting with the MSB is also just a convention, but this convention is pretty much universal today for numbers written in positional notation. For programming, this convention is only relevant when formatting a number for display, parsing a number from a string, and maybe for naming methods like trailing_zeros()
. When storing an u32
in a register, the bits don't have any defined order.
Upvotes: 10