CompSci 100 20.1 From bits to bytes to ints  At some level everything is stored as either a zero or a one  A bit is a binary digit a byte is a binary.

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Presentation transcript:

CompSci From bits to bytes to ints  At some level everything is stored as either a zero or a one  A bit is a binary digit a byte is a binary term (8 bits)  We should be grateful we can deal with Strings rather than sequences of 0's and 1's.  We should be grateful we can deal with an int rather than the 32 bits that make an int  int values are stored as two's complement numbers with 32 bits, for 64 bits use the type long, a char is 16 bits  Standard in Java, different in C/C++  Facilitates addition/subtraction for int values  We don't need to worry about this, except to note: o Infinity + 1 = - Infinity o Math.abs(-Infinity) > Infinity

CompSci How are data stored?  To facilitate Huffman coding we need to read/write one bit  Why do we need to read one bit?  Why do we need to write one bit?  When do we read 8 bits at a time? Read 32 bits at a time?  We can't actually write one bit-at-a-time. We can't really write one char at a time either.  Output and input are buffered, minimize memory accesses and disk accesses  Why do we care about this when we talk about data structures and algorithms? o Where does data come from?

CompSci How do we buffer char input?  Done for us as part of InputStream and Reader classes  InputStreams are for reading bytes  Readers are for reading char values  Why do we have both and how do they interact? Reader r = new InputStreamReader(System.in);  Do we need to flush our buffers?  In the past Java IO has been notoriously slow  Do we care about I? About O?  This is changing, and the java.nio classes help o Map a file to a region in memory in one operation

CompSci Buffer bit output  To buffer bit output we need to store bits in a buffer  When the buffer is full, we write it.  The buffer might overflow, e.g., in process of writing 10 bits to 32-bit capacity buffer that has 29 bits in it  How do we access bits, add to buffer, etc.?  We need to use bit operations  Mask bits -- access individual bits  Shift bits – to the left or to the right  Bitwise and / or / negate bits

CompSci Bit Logical Operations  Work on integers types in binary (by bit)  long s, ints, char s, shorts, and byte s  Three binary operators  And: &  Or: |  Exclusive Or (xor): ^  What is result of  27 & 14?  27 | 14?  27 ^ 14? aba&ba|ba^b

CompSci Bit Logical Operations  Need to work bit position by bit position = 27 (many leading zeros not shown) = & = | = ^ =  Also have unary negation (not): ~ = 27 ~ = -26  Use “masks” with the various operators to  Set or clear bits  Test bits  Toggle bits  (Example later)  What is result of  27 & 14?  27 | 14?  27 ^ 14”

CompSci Bit Shift Operations  Work on same types as logical ops  One left shift and two right shifts  Left shift: << = << = 108 (shifting left is like? )  Logical right shift: >>> = >>> = 6 (shifting right is like? )  Arithmetic right shift: >> = >> = = >>> 16 (for contrast) = 65575

CompSci Representing pixels  A pixel typically stores RGB and alpha/transparency values  Each RGB is a value in the range 0 to 255  The alpha value is also in range 0 to 255 Pixel red = new Pixel(255,0,0,0); Pixel white = new Pixel(255,255,255,0);  Typically store these values as int values, a picture is simply an array of int values void process(int pixel){ int blue = pixel & 0xff; int green = (pixel >> 8) & 0xff; int red = (pixel >> 16) & 0xff; }

CompSci Representing pixels  A pixel typically stores RGB and alpha/transparency values  Each RGB is a value in the range 0 to 255  The alpha value is also in range 0 to 255 Pixel red = new Pixel(255,0,0,0); Pixel white = new Pixel(255,255,255,0);  Typically store these values as int values, a picture is simply an array of int values void process(int pixel){ int blue = pixel & 0xff; int green = (pixel >> 8) & 0xff; int red = (pixel>> 16) & 0xff; }

CompSci Bit masks and shifts void process(int pixel){ int blue = pixel & 0xff; int green = (pixel >> 8) & 0xff; int red = (pixel >> 16) & 0xff; }  Hexadecimal number: 0,1,2,3,4,5,6,7,8,9,a,b,c,d,e,f  Note that f is 15, in binary this is 1111, one less than  The hex number 0xff is an 8 bit number, all ones  The bitwise & operator creates an 8 bit value, 0— 255 (why)  1&1 == 1, otherwise we get 0, similar to logical and  Similarly we have |, bitwise or

CompSci Bit operations revisited  How do we write out all of the bits of a number / ** * writes the bit representation of a int * to standard out */ void bits(int val) {