What does AND 0xFF do

Question

In the following code   short     byte2  lt  lt  8     byte1  amp  0xFF     What is the purpose of  amp 0xFF  Because other somestimes I see it written as   short     byte2  lt  lt  8    byte1    And that seems to work fine too

User · Answer

amp  0xFF by itself only ensures that if bytes are longer than 8 bits  allowed by the language standard   the rest are ignored      And that seems to work fine too    If the result ends up greater than SHRT MAX  you get undefined behavior  In that respect both will work equally poorly

User · Answer

it clears the all the bits that are not in the first byte

User · Answer

Anding an integer with 0xFF leaves only the least significant byte  For example  to get the first byte in a short s  you can write s  amp  0xFF  This is typically referred to as  masking   If byte1 is either a single byte type  like uint8 t  or is already less than 256  and as a result is all zeroes except for the least significant byte  there is no need to mask out the higher bits  as they are already zero   See tristopiaPatrick Schl  ter s answer below when you may be working with signed types  When doing bitwise operations  I recommend working only with unsigned types

User · Answer

Assuming your byte1 is a byte 8bits   When you do a bitwise AND of a byte with 0xFF  you are getting the same byte   So byte1 is the same as byte1  amp  0xFF  Say byte1 is 01001101   then byte1  amp  0xFF   01001101  amp  11111111   01001101   byte1  If byte1 is of some other type say integer of 4 bytes  bitwise AND with 0xFF leaves you with least significant byte 8 bits  of the byte1

User · Answer

The danger of the second expression comes if the type of byte1 is char  In that case  some implementations can have it signed char  which will result in sign extension when evaluating  signed char byte1   0x80  signed char byte2   0x10   unsigned short value1     byte2  lt  lt  8     byte1  amp  0xFF    unsigned short value2     byte2  lt  lt  8    byte1    printf  quot value1  hu  hx n quot   value1  value1   printf  quot value2  hu  hx n quot   value2  value2    will print value1 4224 1080     right value2 65408 ff80    wrong    I tried it on gcc v3 4 6 on Solaris SPARC 64 bit and the result is the same with byte1 and byte2 declared as char  TL DR The masking is to avoid implicit sign extension  EDIT  I checked  it s the same behaviour in C    EDIT2  As requested explanation of sign extension  Sign extension is a consequence of the way C evaluates expressions  There is a rule in C called promotion rule  C will implicitly cast all small types to int before doing the evaluation  Let s see what happens to our expression  unsigned short value2     byte2  lt  lt  8    byte1    byte1 is a variable containing bit pattern 0xFF  If char is unsigned that value is interpreted as 255  if it is signed it is -128  When doing the calculation  C will extend the value to an int size  16 or 32 bits generally   This means that if the variable is unsigned and we will keep the value 255  the bit-pattern of that value as int will be 0x000000FF  If it is signed we want the value -128 which bit pattern is 0xFFFFFFFF  The sign was extended to the size of the tempory used to do the calculation  And thus oring the temporary will yield the wrong result  On x86 assembly it is done with the movsx instruction  movzx for the zero extend   Other CPU s had other instructions for that  6809 had SEX

User · Answer

if byte1 is an 8-bit integer type then it s pointless - if it is more than 8 bits it will essentially give you the last 8 bits of the value       0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1   amp   0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1     -------------------------------     0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1

User · Answer

The byte1  amp  0xff ensures that only the 8 least significant bits of byte1 can be non-zero   if byte1 is already an unsigned type that has only 8 bits  e g   char in some cases  or unsigned char in most  it won t make any difference is completely unnecessary   If byte1 is a type that s signed or has more than 8 bits  e g   short  int  long   and any of the bits except the 8 least significant is set  then there will be a difference  i e   it ll zero those upper bits before oring with the other variable  so this operand of the or affects only the 8 least significant bits of the result

[c] What does AND 0xFF do?

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