Next, we include the several conversion from type to type.

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1 Number Conversions: Binary Decimal; Floating Points In order to communicate with a computer, we need, at some point, to speak the same language. The words of our language are made up of combinations of letters in an alphabet, just like in our spoken languages. But the alphabet we use is much simpler than one used for most spoken languages. It has only two letters, or characters: 0 and 1. And the words in our new language usually have an even stranger constraint: the number of letters often (but not always) is some positive integer power of 2! Imagine, for a moment, how different English would be if each word was restricted to a length of 1, 2, 4, 8, 16, 32 letters. Or 64. Or 128. We might encounter sentences such as Welcomme to Occident Colg. We hope u chuz a majr in Computer Sins. And then, to make matters more complicated, somebody decides that some words should qualities that contextualize it s meaning. So our patters of 0 s and 1 s, found usually in lengths of 2 k, might be decomposed into various fields, such as mantissas (to ensure precision), exponents (to ensure a wide range), and signs (to denote signs)!. Or perhaps we encounter an operation on a word to give it an opposite meaning, such as attaching the prefix un to an English word, or finding the two s complement of a binary number. Languages run into trouble when two entities who speak different languages try to communicate with each other. So in class, we ve identified several types of words living in a computer, and then processes with which to translate these words to values which we humans might speak. The biggest difficulty is that humans speak about numbers with base 10. Computers only understand base 2, or binary, numbers. And in some sense, base 16 numbers, (hexadecimal numbers) are a compromise. They re really groups of four binary digits (bits), but they use all the characters in our base 10 system (and the six letters A, B, C, D, E, F). Since we ve established an understanding of these numbers in class, here we present a series of examples to highlight aspects of our translation processes. We will translate numbers back and forth from bases 2, 10, and 16. We will also convert numerical values back and forth between these bases and the integer and the floating point data types we used in the introductory programming course. As a reminder, the two tables below show the translation of the number 0 through in bases 2, 10, and 16. decimal binary hexadecimal decimal binary hexadecimal 8 9 A B C D E F Next, we include the several conversion from type to type. 1

2 1. Convert 5B2C 16 to binary and then decimal: b k b b 14 b 13 b 12 b 11 b 10 b 9 b 8 b 7 b 6 b 5 b 4 b 3 b 2 b 1 b 0 bit k B2C 16 = = = Convert F ADE 16 to binary and then decimal: b k b b 14 b 13 b 12 b 11 b 10 b 9 b 8 b 7 b 6 b 5 b 4 b 3 b 2 b 1 b 0 bit k F ADE 16 = = = Convert to binary and hexadecimal: n = = = AD Remainders

3 4. Convert to binary and hexadecimal: n = = = 7E Convert to binary and hexadecimal Remainders : : r = = = 0.5A9FB

4 6. Convert ( 4 ) 10 to binary and hexadecimal : : r = ( 4 ) 10 = = = = Convert the float given in hexadecimal as r = A as a base 10 decimal number. r s 1 e m i) Since s = 1, the number is negative. ii) The value e is computed as the unsigned bits : e = 12. iii) The value m is computed as the 23 unsigned bits divided by That is: = 7804 divided by Therefore: m = = Using the float value computation formula, we have: r = ( 1) s 1.m 2 (e 127) =

5 8. Express the number r = in 32 bit float format. Revisiting our float value computation formula, we have: r = ( 1) s 1.m 2 (e 127) = First, since r 0, the sign bit s = 0. r Next, the value e 127 is value for which lies in the interval [1, 2). 2 (e 127) Since r lies between 2 million and 4 million, we might guess a value of e 127 = 21, since 2 21 = 20972, or simply, 2-meg. This proves to be the correct values, since we see that: r = = which is definitely within the interval [1, 2). We conclude that e = = 148 = Finally, to determine m, we subtract 1 and proceed as before for a decimal number: 5

6 We have found However, the float format only has room for 23 mantissa bits: We combine s, e, m into the float format and see: r Notice that we rounded up in our mantissa! Finally, let s ask how close our approximation is. The exact value of the float we computed is: ( ) 2 ( ) = = = Compared to r = , we see that the closest float is actually off by

7 9. What is the difference between the two largest floats? The largest float is the one with the largest exponent and the largest mantissa. The closest float to this, (the 2nd largest) will have the same exponent, and almost the same mantissa... in fact, the mantissa only differs in the least significant bit. Therefore, our largest float is: r 1 = = 7FFFFFFF 16 with s 1 = 0, e 1 = 255 and m 1 = = So, in decimal form, r 1 = ( ) 2128 r 1 = Our second largest float is: r 2 = = 7FFFFFFE 16 with s 2 = 0, e 2 = 255 and m 2 = = So, in decimal form, r 2 = ( ) 2128 r 2 = The difference between the two numbers can be found by subtracting r 1 r 2. But it s also interesting to note that we don t even need these values to compute the difference. Since the only difference in the two numbers comes from the least significant bit in the mantissa, we can proceed as: s 1 = s 2 = 0 e 1 = e 2 = 255 m 1 m 2 = Therefore: r 1 r 2 =((1 + m 1 ) (1 + m 2 )) 2 e = = = 40, 564, 819, 207, 303, 340, 847, 894, 502, 572, 032. This number is ENORMOUS! It is approximately 40 novillion! Take a moment to let this sink in. That over the entire range real numbers which can be expressed exactly in 32 bit float format, the largest two such numbers have a gap between them of over 40 novillion! This is more than a million times the well known constant: Avagadro s Number. 7

8 10. What is the difference between the two smallest (in magnitude) floats? We approach this almost identically to the previous question, the only difference being that this time, our smallest magnitude will have exponent and mantissa will consisting entirely of 0 s instead of 1 s. The smallest magnitude for a float is the one with the smallest exponent and the smallest mantissa. The closest float to this, (the 2nd smallest) will have the same exponent, and almost the same mantissa... in fact, the mantissa only differs in the least significant bit. Therefore, our smallest magnitude is: r 3 = = with s 3 = 0, e 3 = 0 and m 3 = 0 So, in decimal form: r 1 = (1 + 0) = Our second largest float is: r 4 = = with s 4 = 0, e 4 = 0 and m 4 = = So, in decimal form,: r 2 = ( ) = The difference between the two numbers can be found by subtracting r 4 r 3. But it s also interesting to note that we don t even need these values to compute the difference. Since the only difference in the two numbers comes from the least significant bit in the mantissa, we can proceed as: s 3 = s 4 = 0 e 3 = e 4 = 0 m 3 m 4 = Therefore: r 4 r 3 =((1 + m 4 ) (1 + m 3 )) 2 e = = 2 0 = This number is TINY! Now the distance between two consecutive floats is on the order of We seldom need to be that precise. As a comparison, the attractive gravitational pull between the star Betelgeuse, in the constellation Orion, and an average Oxy student, is 200 novillion times this number! 8

9 11. Not to beat a dead horse, but what are the largest and smallest (in magnitude) values representable by a 64 bit double, and what are the distances between each and it s nearest next double? Clearly, this asks the last two questions above, but for a double. To answer this, we would carry out the same steps as we did for floats. Since we will only focus on positive values, we will acknowledge that the sign bits will always be equal to s = 0, and not mention sign again. This leaves us with: For the largest two doubles, d 1, d 2 : e 1 = e 2 = = 7FF 16 = m 1 = = FFFFFFFFFFFFF 16 = m 2 = = FFFFFFFFFFFFE 16 = So our largest double is: d 1 = 1.m 1 2 (e ) = ( ) ( ) = = (2 53 1) The difference between this and the next largest double is given by: d 1 d 2 =((1 + m ) ((1 + m )) 2 (e ) = = For the smallest two doubles, d 3, d 4 : e 3 = e 4 = = 0 16 = 0 m 3 = = = 0 m 4 = = = 1 9

10 So our smallest double is: d 3 =1.m 3 2 (e ) = = The difference between this and the next largest double is given by: d 4 d 3 = ((1 + m ) ((1 + m )) 2 (e ) = = Approximately how many digits accuracy are there in a float? The precision of a float data type involves the mantissa. This is see this, consider the relative difference between two consecutive floats. For simplicity we consider only positive values, and presume that for both floats, the exponents are the same: f 2 f 1 f 1 = 1.m 2 2 (e 127) 1.m 1 2 (e 127) 1.m 1 2 (e 127) = 1.m 2 1.m 1 1.m 1 = m We note that the value of 1.m can lie anywhere in the interval [1, 2), which won t change our results by more than a factor of 2. We conclude that we can rely on (6) digits precision for a float. We will often realize seven (7) digits precision. 13. Approximately how many digits accuracy are there in a double? The precision of a double data type involves the mantissa. This is see this, consider the relative difference between two consecutive double. For simplicity we consider only positive values, and presume that for both double, the exponents are the same: d 2 d 1 d 1 = 1.m 2 2 (e 1023) 1.m 1 2 (e 1023) 1.m 1 2 (e 1023) = 1.m 2 1.m 1 1.m 1 = m We conclude that we can rely on () digits precision for a float. We will often realize seven (16) digits precision. 10

Numbering Systems. Contents: Binary & Decimal. Converting From: B D, D B. Arithmetic operation on Binary.

Numbering Systems. Contents: Binary & Decimal. Converting From: B D, D B. Arithmetic operation on Binary. Numbering Systems Contents: Binary & Decimal. Converting From: B D, D B. Arithmetic operation on Binary. Addition & Subtraction using Octal & Hexadecimal 2 s Complement, Subtraction Using 2 s Complement.