Convert 4 bytes to int

Question

I m reading a binary file like this   InputStream in   new FileInputStream  file    byte   buffer   new byte 1024   while    in read buffer    gt  -1         int a               What I want to do it to read up to 4 bytes and create a int value from those but  I don t know how to do it    I kind of feel like I have to grab 4 bytes at a time  and perform one  byte  operation   like     lt  lt      amp  FF and stuff like that   to create the new int   What s the idiom for this    EDIT  Ooops this turn out to be a bit more complex   to explain      What I m trying to do is  read a  file   may be ascii  binary  it doesn t matter    and extract the integers it may have    For instance suppose the binary  content   in base 2      00000000 00000000 00000000 00000001 00000000 00000000 00000000 00000010   The integer representation should be 1   2 right   -    1 for the first 32 bits  and 2 for the remaining 32 bits    11111111 11111111 11111111 11111111   Would be -1   and   01111111 11111111 11111111 11111111   Would be  Integer MAX VALUE   2147483647

User · Answer

For reading unsigned 4 bytes as integer we should use a long variable, because the sign bit is considered as part of the unsigned number.

long result = (((bytes[0] << 8 & bytes[1]) << 8 & bytes[2]) << 8) & bytes[3]; 
result = result & 0xFFFFFFFF;

This is tested well worked function

User · Answer

If you have them already in a byte   array  you can use      int result   ByteBuffer wrap bytes  getInt      source  here

User · Answer

You should put it into a function like this   public static int toInt byte   bytes  int offset      int ret   0    for  int i 0  i lt 4  amp  amp  i offset lt bytes length  i          ret  lt  lt   8      ret     int bytes i   amp  0xFF        return ret      Example   byte   bytes   new byte   -2  -4  -8  -16   System out println Integer toBinaryString toInt bytes  0       Output   11111110111111001111100011110000   This takes care of running out of bytes and correctly handling negative byte values   I m unaware of a standard function for doing this   Issues to consider    Endianness  different CPU architectures put the bytes that make up an int in different orders  Depending on how you come up with the byte array to begin with you may have to worry about this  and Buffering  if you grab 1024 bytes at a time and start a sequence at element 1022 you will hit the end of the buffer before you get 4 bytes  It s probably better to use some form of buffered input stream that does the buffered automatically so you can just use readByte   repeatedly and not worry about it otherwise  Trailing Buffer  the end of the input may be an uneven number of bytes  not a multiple of 4 specifically  depending on the source  But if you create the input to begin with and being a multiple of 4 is  guaranteed   or at least a precondition  you may not need to concern yourself with it    to further elaborate on the point of buffering  consider the BufferedInputStream   InputStream in   new BufferedInputStream new FileInputStream file   1024     Now you have an InputStream that automatically buffers 1024 bytes at a time  which is a lot less awkward to deal with  This way you can happily read 4 bytes at a time and not worry about too much I O   Secondly you can also use DataInputStream   InputStream in   new DataInputStream new BufferedInputStream                       new FileInputStream file   1024    byte b   in readByte      or even   int i   in readInt      and not worry about constructing ints at all

User · Answer

The easiest way is   RandomAccessFile in   new RandomAccessFile  filename    r     int i   in readInt      -- or --  DataInputStream in   new DataInputStream new BufferedInputStream      new FileInputStream  filename       int i   in readInt

User · Answer

You can also use BigInteger for variable length bytes  You can convert it to Long  Integer or Short  whichever suits your needs   new BigInteger bytes  intValue      or to denote polarity   new BigInteger 1  bytes  intValue

User · Answer

ByteBuffer has this capability  and is able to work with both little and big endian integers   Consider this example         read the file into a byte array File file   new File  file bin    FileInputStream fis   new FileInputStream file   byte    arr   new byte  int file length     fis read arr        create a byte buffer and wrap the array ByteBuffer bb   ByteBuffer wrap arr        if the file uses little endian as apposed to network      big endian  Java s native  format      then set the byte order of the ByteBuffer if use little endian      bb order ByteOrder LITTLE ENDIAN        read your integers using ByteBuffer s getInt        four bytes converted into an integer  System out println bb getInt        Hope this helps

User · Answer

try something like this   a   buffer 3   a   a 256   buffer 2   a   a 256   buffer 1   a   a 256   buffer 0     this is assuming that the lowest byte comes first  if the highest byte comes first you might have to swap the indices  go from 0 to 3    basically for each byte you want to add  you first multiply a by 256  which equals a shift to the left by 8 bits  and then add the new byte

User · Answer

The following code reads 4 bytes from array  a byte    at position index and returns a int   I tried out most of the code from the other answers on Java 10 and some other variants I dreamed up   This code used the least amount of CPU time but allocates a ByteBuffer until Java 10 s JIT gets rid of the allocation   int result   result   ByteBuffer     wrap array      getInt index     This code is the best performing code that does not allocate anything   Unfortunately  it consumes 56  more CPU time compared to the above code   int result  short data0  data1  data2  data3   data0     short   array index     amp  0x00FF   data1     short   array index     amp  0x00FF   data2     short   array index     amp  0x00FF   data3     short   array index     amp  0x00FF   result    data0  lt  lt  24     data1  lt  lt  16     data2  lt  lt  8    data3

User · Answer

just see how DataInputStream readInt   is implemented       int ch1   in read        int ch2   in read        int ch3   in read        int ch4   in read        if   ch1   ch2   ch3   ch4   lt  0          throw new EOFException        return   ch1  lt  lt  24     ch2  lt  lt  16     ch3  lt  lt  8     ch4  lt  lt  0

User · Answer

Converting a 4-byte array into integer     Explictly declaring anInt -4  byte-by-byte byte   anInt     byte 0xff  byte 0xff  byte 0xff  byte 0xfc      Equals -4   And now you have a 4-byte array with an integer equaling -4      Converting back to integer from 4-bytes    result    int    anInt 0  lt  lt 24      anInt 1  lt  lt 24  gt  gt  gt 8        anInt 2  lt  lt 24  gt  gt  gt 16       anInt 3  lt  lt 24  gt  gt  gt 24

User · Answer

for  int i   0  i  lt  buffer length  i         a    a  lt  lt  8    buffer i      if  i   3    0               a is ready       a   0

[java] Convert 4 bytes to int

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