ALL ABOUT INTEGER REPRESENTATION. --------------------------------- Computers operate on binary values (as a result of being built from transistors). there are different binary representations for integers possible qualifications: 1. positive numbers only 2. positive and negative numbers 3. ease of human readability 4. speed of computer operations there are 4 commonly known (1 not common) integer reprentations. All have been used at various times for various reasons. 1. unsigned 2. sign magnitude 3. one's complement 4. two's complement 5. biased (not commonly known) virtually all modern computers operate based on 2's complement representation. why? 1. hardware is faster 2. hardware is simpler (which makes it faster) UNSIGNED -------- the standard binary encoding already given only positive values range: 0 to 2**n - 1, for n bits example: 4 bits, values 0 to 15 n=4, 2**4 -1 is 15 binary decimal hex binary decimal hex 0000 0 0 1000 8 8 0001 1 1 1001 9 9 0010 2 2 1010 10 a 0011 3 3 1011 11 b 0100 4 4 1100 12 c 0101 5 5 1101 13 d 0110 6 6 1110 14 e 0111 7 7 1111 15 f SIGN MAGNITUDE -------------- a human readable way of getting both positive and negative integers. The hw that does arithmetic on sign magnitude integers is not fast, and it is more complex than the hw that does arithmetic on 1's comp. and 2's comp. integers. use 1 bit of integer to represent the sign of the integer let sign bit of 0 be positive, 1 be negative. the rest of the integer is a magnitude, using same encoding as unsigned integers example: 4 bits 0101 is 5 1101 is -5 to get the additive inverse of a number, just flip (not, invert, complement, negate) the sign bit. range: -(2**(n-1)) + 1 to 2**(n-1) -1 where n is the number of bits 4 bits, -7 to +7 n=4, - 2**3 + 1 to 2**3 - 1 -8 + 1 to 8 - 1 because of the sign bit, there are 2 representations for 0. This is a problem for hardware. . . 0000 is 0, 1000 is 0 the computer must do all calculations such that they come out correctly and the same whichever representation is used. ONE's COMPLEMENT ---------------- historically important, and we use this representation to get 2's complement integers. Now, nobody builds machines that are based on 1's comp. integers. In the past, early computers built by Semour Cray (while at CDC) were based on 1's comp. integers. positive integers use the same representation as unsigned. 00000 is 0 00111 is 7, etc. negation (finding an additive inverse) is done by taking a bitwise complement of the positive representation. COMPLEMENT. INVERT. NOT. FLIP. NEGATE. a logical operation done on a single bit the complement of 1 is 0. the complement of 0 is 1. -1 --> take +1, 00001 complement each bit 11110 that is -1. don't add or take away any bits. EXAMPLES: 11100 this must be a negative number. to find out which, find the additive inverse! 00011 is +3 by sight, so 11100 must be -3 things to notice: 1. any negative number will have a 1 in the MSB. 2. there are 2 representations for 0, 00000 and 11111. TWO's COMPLEMENT ---------------- a variation on 1's complement that does NOT have 2 representations for 0. This makes the hardware that does arithmetic faster than for the other representations. a 3 bit example: bit pattern: 100 101 110 111 000 001 010 011 1's comp: -3 -2 -1 0 0 1 2 3 2's comp.: -4 -3 -2 -1 0 1 2 3 the negative values are all "slid" down by one, eliminating the extra zero representation. how to get an integer in 2's comp. representation: positive values are the same as for sign/mag and 1's comp. they will have a 0 in the MSB (but it is NOT a sign bit!) positive: just write down the value as before negative: take the positive value 00101 (+5) take the 1's comp. 11010 (-5 in 1's comp) add 1 + 1 ------ 11011 (-5 in 2's comp) to get the additive inverse of a 2's comp integer, 1. take the 1's comp. 2. add 1 to add 1 without really knowing how to add: start at LSB, for each bit (working right to left) while the bit is a 1, change it to a 0. when a 0 is encountered, change it to a 1 and stop. the rest of the bits remain the same as they were. EXAMPLE: what decimal value does the two's complement 110011 represent? It must be a negative number, since the most significant bit is a 1. Therefore, find the additive inverse: 110011 (2's comp. ?) 001100 (after taking the 1's complement) + 1 ------ 001101 (2's comp. +13) Therefore, it's additive inverse (110011) must be -13. A LITTLE BIT ON ADDING ---------------------- we'll see how to really do this in the next chapter, but here's a brief overview. carry in a b sum carry out 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1 a 0011 +b +0001 -- ----- sum 0100 its really just like we do for decimal! 0 + 0 = 0 1 + 0 = 1 1 + 1 = 2 which is 10 in binary, sum is 0 and carry the 1. 1 + 1 + 1 = 3 sum is 1, and carry a 1. BIASED REPRESENTATION --------------------- an integer representation that skews the bit patterns so as to look just like unsigned but actually represent negative numbers. examples: given 4 bits, we BIAS values by 2**3 (8) TRUE VALUE to be represented 3 add in the bias +8 ---- unsigned value 11 so the bit pattern of 3 in 4-bit biased-8 representation will be 1011 going the other way, suppose we were given a biased-8 representation as 0110 unsigned 0110 represents 6 subtract out the bias - 8 ---- TRUE VALUE represented -2 this representation allows operations on the biased numbers to be the same as for unsigned integers, but actually represents both positive and negative values. choosing a bias: the bias chosen is most often based on the number of bits available for representing an integer. To get an approx. equal distribution of true values above and below 0, the bias should be 2 ** (n-1) or (2**(n-1)) - 1 SIGN EXTENSION -------------- how to change an integer with a smaller number of bits into the same integer (same representation) with a larger number of bits. this must be done a lot by arithmetic units, so it is best to go over it. by representation: unsigned: xxxxx --> 000xxxxx copy the original integer into the LSBs, and put 0's elsewhere sign magnitude: sxxxx --> s000xxxx copy the original integer's magnitude into the LSBs, put the original sign into the MSB, and put 0's elsewhere 1's and 2's complement: called SIGN EXTENSION. copy the original integer into the LSBs, take the MSB of original integer and copy it elsewhere. example: 0010101 000 0010101 11110000 11111111 11110000 OVERFLOW -------- sometimes a value cannot be represented in the limited number of bits allowed. Examples: unsigned, 3 bits: 8 would require at least 4 bits (1000) sign mag., 4 bits: 8 would require at least 5 bits (01000) when a value cannot be represented in the number of bits allowed, we say that overflow has occurred. Overflow occurs when doing arithmetic operations. example: 3 bit unsigned representation 011 (3) + 110 (6) --------- ? (9) it would require 4 bits (1001) to represent the value 9 in unsigned rep. CHARACTER REPRESENTATION ------------------------ everything represented by a computer is represented by binary sequences. A common non-integer needed to be represented is characters. We use standard encodings (binary sequences) to repreesent characters. REMEMBER: bit patterns do NOT imply a representation Many I/O devices work with 8 bit quantities. A standard code ASCII (American Standard for Computer Information Interchange) defines what character is represented by each sequence. examples: 0100 0001 is 41 (hex) or 65 (decimal). It represents `A' 0100 0010 is 42 (hex) or 66 (decimal). It represents `B' Different bit patterns are used for each different character that needs to be represented. The code has some nice properties. If the bit patterns are compared, (pretending they represent integers), then `A' < `B' This is good, because it helps with sorting things into alphabetical order. Notes: `a' (61 hex) is different than `A' (41 hex) `8' (38 hex) is different than the integer 8 the digits: `0' is 48 (decimal) or 30 (hex) `9' is 57 (decimal) or 39 (hex) Because of this, code must deal with both characters and integers correctly. Consider the following example: in1 db ? result db ? get_ch in1 moveb result, in1 iadd result, in1 put_ch result suppose the user types `3' result <- 51 + 51 = 102 (decimal) put_ch prints out `f', since the ASCII code for 102(decimal) is `f' What if we wanted the value 6 to be printed? What we need is an algorithm for translating character strings to the integers they represent, and visa versa. Then, do the arithmetic on integers, and print out the integer result. ALGORITHM: character string --> integer the steps: for `3' `5' `4' read `3' translate `3' to 3 read `5' translate `5' to 5 integer = 3 * 10 + 5 = 35 read `4' translate `4' to 4 integer = 35 * 10 + 4 = 354 the algorithm: integer = 0 while there are more characters get character digit <- character - 48 integer <- integer * 10 + digit In the real world, we also have the equivalent problem, only dealing with printing out integers. ALGORITHM: integer --> character string the steps: for 354 (an integer), figure out how many characters there are (3) 354 div 100 gives 3 translate 3 to `3' and print it out 354 mod 100 gives 54 54 div 10 gives 5 translate 5 to `5' and print it out 54 mod 10 gives 4 4 div 1 gives 4 translate 4 to `4' and print it out 4 mod 1 gives 0, so you're done NOTE: These algorithms work the same in bases other than 10. Just change the 10 to the desired base. (There is some extra work in bases > 10, since have to translate the digits > 9 into the characters. For example, 12 (decimal) is c (hex).) REPRESENTATION OF FLOATING POINT NUMBERS ---------------------------------------- Computers represent real values in a form similar to that of scientific notation. There are standards which define what the representation means so that across computers there will be consistancy. Note that this is not the only way to represent floating point numbers, it is just the IEEE standard way of doing it. Here's what we do: the representation ------------------- | S | E | F | ------------------- S is one bit representing the sign of the number E is an 8 bit biased integer representing the exponent F is an unsigned integer the value represented is: S e (-1) x f x 2 where e = E - bias n f = F/2 + 1 for single precision numbers (the emphasis in this class) n = 23 bias = 127 Now, what does all this mean? (Everything here is for IEEE single precision floating point representation.) --> S, E, F all represent fields within a representation. Each is just a bunch of bits. --> S is just a sign bit. 0 for positive, 1 for negative. --> E is an exponent field. The E field is a biased-127 representation. So, the true exponent represented is (E - bias). The radix for the number is ALWAYS 2. Note: Computers that did not use this representation, like those built before the standard, did not always use a radix of 2. Example: some IBM machines had radix of 16. --> F is the mantissa. It is in a somewhat modified form. There are 23 bits available for the mantissa. It turns out that if fl. pt. numbers are always stored in a normalized form, then the leading bit (the one on the left, or MSB) is always a 1. So, why store it at all? It gets put back into the number (giving 24 bits of precision for the mantissa) for any calculation, but we only have to store 23 bits. This MSB is called the HIDDEN BIT. An example: put the decimal number 64.2 into the IEEE single precision representation. first step: get a binary representation for 64.2 to do this, get binary reps. for the stuff to the left and right of the decimal point separately. 64 is 1000000 .2 can be gotten using the algorithm: .2 x 2 = 0.4 0 (msb) .4 x 2 = 0.8 0 .8 x 2 = 1.6 1 .6 x 2 = 1.2 1 .2 x 2 = 0.4 0 now this whole pattern (0011) repeats. .4 x 2 = 0.8 0 .8 x 2 = 1.6 1 .6 x 2 = 1.2 1 so a binary representation for .2 is .001100110011. . . Putting the halves back together again: 64.2 is 1000000.0011001100110011. . . second step: normalize the binary representation. (make it look like scientific notation) 6 1.000000 00110011. . . x 2 third step: 6 is the true exponent. For the standard form, it needs to be in biased-127 form. 6 + 127 ----- 133 (value of biased exponent) 133 in 8 bit, unsigned representation is 1000 0101 this is bit pattern used for E in the standard form. fourth step: the mantissa stored (F) is the stuff to the right of the radix point in the normal form. We need 23 bits of it. 000000 00110011001100110 put it all together (and include the correct sign bit): S E F 0 10000101 00000000110011001100110 the values are often given in hex, so here it is 0100 0010 1000 0000 0110 0110 0110 0110 0x 4 2 8 0 6 6 6 6 or 42806666h Some extra details: --> Since floating point numbers are always stored in normal form, how do we represent 0? (What does the fl. pt. number 0x3f80 0000 represent?) (What does the fl. pt. number 0x0080 0000 represent?) We take the bit patterns 0x0000 0000 and 0x8000 0000 to represent the value 0. (What fl. pt. numbers cannot be represented because of this?) --> Other special values: +infinity 0 11111111 00000... (0x7f80 0000) -infinity 1 11111111 00000... (0xff80 0000) NaN (Not a Number) ? 11111111 ?????... (S is either 0 or 1, E=0xff, and F is anything but all zeros)