CODEWITHSUNDEEP
cruby
NickWEI
NickWEI

Reputation: 59

Why is `& 0xff` executed like this in Ruby source code?

in ruby.h, I find it difficult to understand this macro:

#define SYMBOL_P(x) (((VALUE)(x)&0xff)==SYMBOL_FLAG)

I have no idea why this & operation is executed with 0xff. Doesn't every number & 0xff equal itself?

Upvotes: 0

Views: 214

Answers (2)

David Grayson
David Grayson

Reputation: 87476

The VALUE type in the Ruby source code is usually 32-bit or 64-bit, so the & 0xFF sets all the bits except the lowest 8 to 0.

Upvotes: 0

milevyo
milevyo

Reputation: 2184

& is a bitwize operator (AND), (remember logic table?)

0 & 0 = 0
1 & 0 = 0
0 & 1 = 0
1 & 1 = 1

so what it does here?

0xff is the hexa of 255 in binary (DWORD): 00000000 00000000 00000000 11111111

so assuming a number x= any_value

the representation of x can be like this

???????? ???????? ???????? ????????

each ? can be either 1 or 0

so applying bitwize operator & (AND) with the mask 0xff gives

  ???????? ???????? ???????? ????????
&
  00000000 00000000 00000000 11111111
=
  00000000 00000000 00000000 ????????

for example

  00000000 00000000 00000011 00000011
&
  00000000 00000000 00000000 11111111
=
  00000000 00000000 00000000 00000011
  ^________________________^ ^______^
            zeroed             kept

Upvotes: 1

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