Rappresentazione dell'informazione

Rappresentazione dell'informazione Corrispondenza tra informazione I e sua rappresentazione P(I), composta da cifre ai di un alfabeto di simboli A I P(i) = {ai}, ai A Esempi di alfabeto: {a, b,..., z} {0, 1,..., 9} {0, 1} I simboli possono essere simboli su carta, suoni, livelli di tensione, segnali di fumo... Alfabeti diversi possono rappresentare la stessa informazione: 20 10100 venti XX

Capacità di rappresentazione Dato un alfabeto composto da n simboli, quante informazioni diverse posso rappresentare con k cifre? n k Quanti oggetti si possono rappresentare con 3 cifre decimali? 10 3 Quanti oggetti si possono rappresentare con n bit? 2 n Date C informazioni, quante cifre di un alfabeto con n simboli sono necessarie per rappresentarle tutte? logn(c) Quanti bit sono necessari per rappresentare 30 oggetti? log2(16) = 4 log2(32) = 5

Funzioni floor e ceil Dato un numero reale x, la funzione floor di x, indicata con x, ritorna il più grande numero intero minore o uguale a x. Esempi: 0.5 = 0, 23 = 23, -0.3 = -1, -2 = -2 Dato un numero reale x, la funzione ceil di x, indicata con x, ritorna il più piccolo numero intero maggiore o uguale a x Esempi: 0.5 = 1, 23 = 23, -0.3 = 0, -2 = -2

Numerazione Se l'informazione da rappresentare è una quantità, allora la sua rappresentazione è detta numerazione L'alfabeto B è detto base Numerazione DECIMALE (base 10) B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} Numerazione ESADECIMALE (base 16) B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, a, b, c, d, e, f} Numerazione BINARIA (base 2) B = {0, 1} Numerazione OTTALE (base 8) B = {0, 1, 2, 3, 4, 5, 6, 7} Numerazione ROMANA semplificata B = {I, V, X, L, C, D, M}

Sistema di numerazione Un sistema di numerazione è composto da 1. una base B di n elementi (quantità elementari) 2. una funzione biunivoca (codifica) da informazione a rappresentazione Sistema di numerazione a conteggio: ogni cifra rappresenta sempre la stessa quantità Sistema di numerazione posizionale: ogni cifra rappresenta una quantità diversa a seconda della sua posizione

Codifica posizionale Base B con n simboli B = {b0, b1, b2,..., bn-1} rappresenta i valori { 0, 1, 2,..., N-1} Ciascun numero E (quantità) può essere rappresentato come combinazione lineare di pesi legati alla base: kx v k N i E: [ck, ck-1,..., c0] E =. i=0 Base 2: pesi = {..., 16, 8, 4, 2, 1, 0.5, 0.25, 0.125,...} Base 10: pesi = {..., 1000, 100, 10, 1, 0.1, 0.01,...} Esempi 1210 = 1 x 10 1 + 2 x 10 0 1002 = 1 x 2 2 + 0 x 2 1 + 0 x 2 0

Rappresentazioni Numeriche Decimale 3510 = 3 x 10 1 + 5 x 10 0 Binario 001000112 = 1 x 2 5 + 1 x 2 1 + 1 x 2 0 Esadecimale 0x23 or 23hex = 2 x 16 1 + 3 x 16 0 0-15 (base 10) 0-9, a-f (base 16) Dec Bin Hex 0 0000 0 1 0001 1 2 0010 2 3 0011 3 Dec Bin Hex 4 0100 4 5 0101 5 6 0110 6 7 0111 7 Dec Bin Hex 8 1000 8 9 1001 9 10 1010 a 11 1011 b Dec Bin Hex 12 1100 c 13 1101 d 14 1110 e 15 1111 f

Base 2 2 0 1 2 1 2 2 2 4 2 3 8 2 4 16 2 5 32 2 6 64 2 7 128 2 8 256 2 9 512 2 10 1024 2 11 2048 2 12 4096 2 13 8192 2 16 65536 2 10 1024 Ki 2 20 Mi 2 30 Gi 2 40 Ti 2 50 Pi

Rappresentazione Binaria Il numero binario bit più significativo (MSB) bit meno significativo (LSB) 01011000 00010101 00101110 11100111 rappresenta la quantità 0 x 2 31 + 1 x 2 30 + 0 x 2 29 + + 1 x 2 0 Con 32 bit è possibile rappresentare 2 32 numeri tra 0 e 2 32 1 Questa rappresentazione è chiamata unsigned (senza segno) poiché si assume che i numeri siano sempre positivi NOTA: 2 n in base due è un 1 seguito da n zeri: 2 5 = 100000 NOTA: 2 n 1 in base due è rappresentato da n uni: 2 5 1= 11111

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sha1_base64="acccimbsb2aibvn+sbhznphdqdy=">aaab8nicbvc7tsnaedzzdoevokq5esfrrtzcaroigspewirsykxnyyaccj5bd3ugymox0ejfhwj5iqr+bdu4gisprjo72tkjeykmuu6ns7s8srq2xtmobm5t7+zw9vbvtgw1b5/hmtbdkbmqqogpaiv0ew0sciv0wsl17nceqbsrq1ucjhbebkzeshcgmdr+hntqbsmtqbejv5i6kdea1l76w5jbcbryyyzpew6cqco0ci5hvu1bawnjezagxkyvi8aearf0ro+tyrjtbdqvkhyi/n5iwwtmnaqzyyjhvzn3cve/r2dxdbgkqiuwqfh8eaojxshdtcgaadoughbznhyoujqzzrbbc8o4z0sbvvln+vdmv18k/mnjsug1z+rnq7kycjkkr+seeoscnmknargfcalkitytf8c6r86b8/4zuusuowfkd5ypb+8ekvq=</latexit> Intero Senza Segno Data una sequenza di bit di lunghezza, a quale intero senza segno corrisponde? x = B2U w (X) = w 1 X i=0 X i 2 i Tutti i valori di interi senza segno sono considerati nonnegativi

Binario < > Esadecimale

Numeri negativi (con segno) Su 32 bit si possono rappresentare unicamente 2 32 numeri. Se volessimo rappresentare anche i negativi, potremmo rappresentare 2 31 numeri positivi e 2 31 numeri negativi 0000 0000 0000 0000 0000 0000 0000 0000due = 0 0000 0000 0000 0000 0000 0000 0000 0001due = 1 0111 1111 1111 1111 1111 1111 1111 1111due = 2 31-1 1000 0000 0000 0000 0000 0000 0000 0000due = -2 31 1000 0000 0000 0000 0000 0000 0000 0001due = -(2 31 1) 1000 0000 0000 0000 0000 0000 0000 0010due = -(2 31 2) 1111 1111 1111 1111 1111 1111 1111 1110due = -2 1111 1111 1111 1111 1111 1111 1111 1111due = -1

Complemento a 2 0000 0000 0000 0000 0000 0000 0000 0000due = 0 0000 0000 0000 0000 0000 0000 0000 0001due = 1 0111 1111 1111 1111 1111 1111 1111 1111due = 2 31-1 1000 0000 0000 0000 0000 0000 0000 0000due = -2 31 1000 0000 0000 0000 0000 0000 0000 0001due = -(2 31 1) 1000 0000 0000 0000 0000 0000 0000 0010due = -(2 31 2) Perché questa rappresentazione è utilizzata? Si consideri la somma di 1 e 2 si ottiene 1 Si consideri la somma di 2 e 1 si ottiene 1 Questo formato permette di sommare senza nessuna trasformazione aggiuntiva! Ogni numero rappresenta la quantità x31( 2 31 ) + x302 30 + x292 29 + + x12 1 + x02 0 Su 32 bit si rappresentano tutti i numeri da 2 31 1 a 2 31 1111 1111 1111 1111 1111 1111 1111 1110due = -2 1111 1111 1111 1111 1111 1111 1111 1111due = -1

Complemento a 2 0000 0000 0000 0000 0000 0000 0000 0000due = 0 0000 0000 0000 0000 0000 0000 0000 0001due = 1 0111 1111 1111 1111 1111 1111 1111 1111due = 2 31-1 1000 0000 0000 0000 0000 0000 0000 0000due = -2 31 1000 0000 0000 0000 0000 0000 0000 0001due = -(2 31 1) 1000 0000 0000 0000 0000 0000 0000 0010due = -(2 31 2) 1111 1111 1111 1111 1111 1111 1111 1110due = -2 1111 1111 1111 1111 1111 1111 1111 1111due = -1 La somma di un numero x e del suo complemento a 1 x è sempre uguale a una stringa di 32 uni (corrispondente al numero 1) x + x = 1 x = x + 1 Analogamente, la somma di x e x è uguale a una stringa di 32 zeri, con un riporto di 1. In realtà x + ( x) = 2 n, da cui il nome complemento a 2

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sha1_base64="acccimbsb2aibvn+sbhznphdqdy=">aaab8nicbvc7tsnaedzzdoevokq5esfrrtzcaroigspewirsykxnyyaccj5bd3ugymox0ejfhwj5iqr+bdu4gisprjo72tkjeykmuu6ns7s8srq2xtmobm5t7+zw9vbvtgw1b5/hmtbdkbmqqogpaiv0ew0sciv0wsl17nceqbsrq1ucjhbebkzeshcgmdr+hntqbsmtqbejv5i6kdea1l76w5jbcbryyyzpew6cqco0ci5hvu1bawnjezagxkyvi8aearf0ro+tyrjtbdqvkhyi/n5iwwtmnaqzyyjhvzn3cve/r2dxdbgkqiuwqfh8eaojxshdtcgaadoughbznhyoujqzzrbbc8o4z0sbvvln+vdmv18k/mnjsug1z+rnq7kycjkkr+seeoscnmknargfcalkitytf8c6r86b8/4zuusuowfkd5ypb+8ekvq=</latexit> Intero in Complemento a 2 Data una sequenza di bit di lunghezza, a quale intero in complemento a 2 corrisponde? x = B2C w (X) = X w 1 2 w 1 + w 2 X i=0 X i 2 i La trasformazione è simile a quella degli interi senza segno, eccetto per il bit più significativo (detto bit del segno), che ha peso negativo. 0 per non-negativo 1 per negativo Decimal Esadecimale Binario 15213 3B 6D 00111011 01101101-15213 C4 93 11000100 10010011

Esercizio Calcolare le rappresentazioni in complemento a 2 su 32 bit dei seguenti numeri decimali: 5, 5, 6 5: 0000 0000 0000 0000 0000 0000 0000 0101 5: 1111 1111 1111 1111 1111 1111 1111 1011 6: 1111 1111 1111 1111 1111 1111 1111 1010 Dato 5 verificare che facendone il complemento a 1 e sommando 1 si ottiene 5 Calcolare le stesse rappresentazioni usando solo 4 bit Calcolare le stesse rappresentazioni usando solo 3 bit

Estensione di segno Occasionalmente sarà necessario estendere numeri con segno rappresentati su 16 bit su 32 bit, Vuol dire trasformare la rappresentazione di un numero con segno su 16 bit nel numero intero rappresentato, e calcolarne la rappresentazione su 32 bit 210: 0000 0000 0000 0010 0000 0000 0000 0000 0000 0000 0000 0010 210: 1111 1111 1111 1110 1111 1111 1111 1111 1111 1111 1111 1110 La conversione è banale: si prende il bit più significativo e lo si usa per riempire i bit aggiuntivi sulla sinistra

Rappresentazioni Alternative Le seguenti due rappresentazioni intuitive furono scartate poiché richiedevano dei passi di conversione addizionali prima di poter effettuare operazioni aritmetiche Modulo e segno: il bit più significativo rappresenta +/ e i bit rimanenti rappresentano il valore assoluto del numero Complemento a 1: x è rappresentato invertendo tutti i bit di x Entrambe le rappresentazioni descritte soffrono del problema della doppia rappresentazione dello zero

Overflow Si verifica overflow (tracimazione) quando il risultato di un'operazione non può essere rappresentato con l'hardware a disposizione, per es. 32 bit. Quando si può verificare overflow con una somma? Quando si può verificare overflow con una sottrazione?

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sha1_base64="b4nsxb6ygo4ntlma0uu8+7jioxo=">aaacfhicbza9swnbeib34lemx1flm9wgwb27iqhd0myygjgbxah7m0lcsvfb7pwqjqv9cf4kw62sxnbewv/ijv6h0alennegmxn9wculjh04pbn5hcwl8njlzxvtfao6uxvjo8riampir6btcwtahdbehrrasqer+bpa/uhi4rfuwfgvhdc4jqebiggobkokzfgvuntauw+rcszc12wz51ukwjjpcc961rpz2btox8elusnfnxrvt68fyssaekuw1ny4i7gbconkasgqxmihfnikhtdjzsgcsn10+kpg9xmrmkixgko0nul4ozgkwnpx4oedgcbbo+tn4h9ej8hbatdvyzwghhkycjwg6sirjcozatpxbhdf5hkgkqrsgieirlehzq6tplrkngef/f6vab65zy6/oq7vz4tgymsh7jfdwskjqznl0ibnisk9esrp5nl5cf6cv+ftu7xkfdpb5fc5719ff5rs</latexit> <latexit sha1_base64="b4nsxb6ygo4ntlma0uu8+7jioxo=">aaacfhicbza9swnbeib34lemx1flm9wgwb27iqhd0myygjgbxah7m0lcsvfb7pwqjqv9cf4kw62sxnbewv/ijv6h0alennegmxn9wculjh04pbn5hcwl8njlzxvtfao6uxvjo8riampir6btcwtahdbehrrasqer+bpa/uhi4rfuwfgvhdc4jqebiggobkokzfgvuntauw+rcszc12wz51ukwjjpcc961rpz2btox8elusnfnxrvt68fyssaekuw1ny4i7gbconkasgqxmihfnikhtdjzsgcsn10+kpg9xmrmkixgko0nul4ozgkwnpx4oedgcbbo+tn4h9ej8hbatdvyzwghhkycjwg6sirjcozatpxbhdf5hkgkqrsgieirlehzq6tplrkngef/f6vab65zy6/oq7vz4tgymsh7jfdwskjqznl0ibnisk9esrp5nl5cf6cv+ftu7xkfdpb5fc5719ff5rs</latexit> <latexit sha1_base64="b4nsxb6ygo4ntlma0uu8+7jioxo=">aaacfhicbza9swnbeib34lemx1flm9wgwb27iqhd0myygjgbxah7m0lcsvfb7pwqjqv9cf4kw62sxnbewv/ijv6h0alennegmxn9wculjh04pbn5hcwl8njlzxvtfao6uxvjo8riampir6btcwtahdbehrrasqer+bpa/uhi4rfuwfgvhdc4jqebiggobkokzfgvuntauw+rcszc12wz51ukwjjpcc961rpz2btox8elusnfnxrvt68fyssaekuw1ny4i7gbconkasgqxmihfnikhtdjzsgcsn10+kpg9xmrmkixgko0nul4ozgkwnpx4oedgcbbo+tn4h9ej8hbatdvyzwghhkycjwg6sirjcozatpxbhdf5hkgkqrsgieirlehzq6tplrkngef/f6vab65zy6/oq7vz4tgymsh7jfdwskjqznl0ibnisk9esrp5nl5cf6cv+ftu7xkfdpb5fc5719ff5rs</latexit> <latexit sha1_base64="b4nsxb6ygo4ntlma0uu8+7jioxo=">aaacfhicbza9swnbeib34lemx1flm9wgwb27iqhd0myygjgbxah7m0lcsvfb7pwqjqv9cf4kw62sxnbewv/ijv6h0alennegmxn9wculjh04pbn5hcwl8njlzxvtfao6uxvjo8riampir6btcwtahdbehrrasqer+bpa/uhi4rfuwfgvhdc4jqebiggobkokzfgvuntauw+rcszc12wz51ukwjjpcc961rpz2btox8elusnfnxrvt68fyssaekuw1ny4i7gbconkasgqxmihfnikhtdjzsgcsn10+kpg9xmrmkixgko0nul4ozgkwnpx4oedgcbbo+tn4h9ej8hbatdvyzwghhkycjwg6sirjcozatpxbhdf5hkgkqrsgieirlehzq6tplrkngef/f6vab65zy6/oq7vz4tgymsh7jfdwskjqznl0ibnisk9esrp5nl5cf6cv+ftu7xkfdpb5fc5719ff5rs</latexit> Intervalli Numerici Valori senza segno: U min =0 U max =2 w 1 <latexit sha1_base64="hvekpbwuokpc/mkpmjy6ldhlfbc=">aaacjhicbzc7tsnaeexx4r1eauqafrgihsiokiackykgeiqmshgixsskrfivrd0xefn5dj6br6cfigprumc3yicuvka6ondgm3pdrellrvvmlezgx8ynjqfk0zozc/ovhcutg6dgoc9ifzuzecwqqdensqrpeomqhqppw6v9wj+9rmnlri+pl2argq6whsmactsuegs8c2yh+/12fkrs9/kud3kqlh9wuc14/fygb3jtstwtuypif4u3ffu2rmn25t24ieuaosahwnqm5ybuyscqfar75sc1mic4gi42c6khqtvkbq/1+wpqgwkeoofs8qhe7xmzrnb2ojdvjiau7w+vgp95zzq6261m6iql1kjyrflhyjevruazib+qbomguby51fyaasi0komqouzzemt5ht7v7/8kv17bqxlhm9xg3jcysbbmvtg689gwa7addsh8jtgde2cp7mm5d56df+f1q7xkdgew2i9ypj4b8zwipw==</latexit> <latexit sha1_base64="hvekpbwuokpc/mkpmjy6ldhlfbc=">aaacjhicbzc7tsnaeexx4r1eauqafrgihsiokiackykgeiqmshgixsskrfivrd0xefn5dj6br6cfigprumc3yicuvka6ondgm3pdrellrvvmlezgx8ynjqfk0zozc/ovhcutg6dgoc9ifzuzecwqqdensqrpeomqhqppw6v9wj+9rmnlri+pl2argq6whsmactsuegs8c2yh+/12fkrs9/kud3kqlh9wuc14/fygb3jtstwtuypif4u3ffu2rmn25t24ieuaosahwnqm5ybuyscqfar75sc1mic4gi42c6khqtvkbq/1+wpqgwkeoofs8qhe7xmzrnb2ojdvjiau7w+vgp95zzq6261m6iql1kjyrflhyjevruazib+qbomguby51fyaasi0komqouzzemt5ht7v7/8kv17bqxlhm9xg3jcysbbmvtg689gwa7addsh8jtgde2cp7mm5d56df+f1q7xkdgew2i9ypj4b8zwipw==</latexit> <latexit sha1_base64="hvekpbwuokpc/mkpmjy6ldhlfbc=">aaacjhicbzc7tsnaeexx4r1eauqafrgihsiokiackykgeiqmshgixsskrfivrd0xefn5dj6br6cfigprumc3yicuvka6ondgm3pdrellrvvmlezgx8ynjqfk0zozc/ovhcutg6dgoc9ifzuzecwqqdensqrpeomqhqppw6v9wj+9rmnlri+pl2argq6whsmactsuegs8c2yh+/12fkrs9/kud3kqlh9wuc14/fygb3jtstwtuypif4u3ffu2rmn25t24ieuaosahwnqm5ybuyscqfar75sc1mic4gi42c6khqtvkbq/1+wpqgwkeoofs8qhe7xmzrnb2ojdvjiau7w+vgp95zzq6261m6iql1kjyrflhyjevruazib+qbomguby51fyaasi0komqouzzemt5ht7v7/8kv17bqxlhm9xg3jcysbbmvtg689gwa7addsh8jtgde2cp7mm5d56df+f1q7xkdgew2i9ypj4b8zwipw==</latexit> <latexit sha1_base64="hvekpbwuokpc/mkpmjy6ldhlfbc=">aaacjhicbzc7tsnaeexx4r1eauqafrgihsiokiackykgeiqmshgixsskrfivrd0xefn5dj6br6cfigprumc3yicuvka6ondgm3pdrellrvvmlezgx8ynjqfk0zozc/ovhcutg6dgoc9ifzuzecwqqdensqrpeomqhqppw6v9wj+9rmnlri+pl2argq6whsmactsuegs8c2yh+/12fkrs9/kud3kqlh9wuc14/fygb3jtstwtuypif4u3ffu2rmn25t24ieuaosahwnqm5ybuyscqfar75sc1mic4gi42c6khqtvkbq/1+wpqgwkeoofs8qhe7xmzrnb2ojdvjiau7w+vgp95zzq6261m6iql1kjyrflhyjevruazib+qbomguby51fyaasi0komqouzzemt5ht7v7/8kv17bqxlhm9xg3jcysbbmvtg689gwa7addsh8jtgde2cp7mm5d56df+f1q7xkdgew2i9ypj4b8zwipw==</latexit> 000...0 111...1 Valori complemento a 2: C min = 2 w 1 Esempio per U max C max C min 1 <latexit sha1_base64="+qqewasdo0ouxo0dzifzu1rjkw4=">aaacl3icbzdlssnqeizpvftvvzdudhbftusiggokxw5cklgvmlomx6kepdkj50y8epiypojp4vzxuldc+hymtyk3wf383wwz8wexkpzc99kzgbwahhkdgy9nte5nz5rn5w5tlbibtrgpybwhyffjju2sppa4nghhopaouggu/ogsjzwrpqcbgnshngnzlqiotzrlrwwe+rblg1kn9uopm77nq6sn6vxvy7jvl35guc7wf616nxlfrbm94n+f1xcv1q+9tvnfp41eeqimocdalufg1e7bkbqks5kfwixbxmaztnkpiutbtntfznwpsuarj9fwqxjpxo8tkytw3orb3hkcndvfrdd/y62euhvtvoo4idsiwersyw+rfubm8se/lqajolgcudrcgaeinjkdelmz5hmw8jy839//fc3v2mbn21+r1hf6wyyxbbbivpjh1lmd7bi91msc3bj79saentvnyxl13j5bb5z+zdz7uc77b9hvpjk=</latexit> C max =2 w 1 1 w = 16 Decimale Esadecimale Binario 65535 FF FF 11111111 11111111 32767 7F FF 01111111 11111111-32768 80 00 10000000 00000000-1 FF FF 11111111 11111111 0 00 00 00000000 00000000 100...0 011...1

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sha1_base64="ocjwwmx8wld+t4qqlpgu+xmesca=">aaab8nicbvc7tsnaedyhvwivacxniqijkriretbf0fameiareis6xzbhlppzuttdiqx8as1uviiwh6lgx7cnc0iyajszq52dmjhcoot+opwv1bx1jepmbwt7z3evvn9wb2krofg8lrhuhcyafap8fcihl2hgusihg05vcr/7cnqiwn3hliegyhmlxoizzksohdybbtmtqjejv5igkdee1r8go5jbcbryyyzpe26cqco0ci5hxhtyawnjuzabfkyvi8aearf0tk+syrjtbdqvkhyi/n5iwwtmlaqzyyjhg1n0cve/r29xfbmkqiuwqfh8eaojxshdtcgaadosghbznhyoujqzzrbbc8o4z0sbvvll+vawv18m/lnzqul1zhut67kykjkix+sueosctmgtarofcalkitytf8c6r86b8/4zwnhknupyb87hn+vokvi=</latexit> <latexit sha1_base64="ocjwwmx8wld+t4qqlpgu+xmesca=">aaab8nicbvc7tsnaedyhvwivacxniqijkriretbf0fameiareis6xzbhlppzuttdiqx8as1uviiwh6lgx7cnc0iyajszq52dmjhcoot+opwv1bx1jepmbwt7z3evvn9wb2krofg8lrhuhcyafap8fcihl2hgusihg05vcr/7cnqiwn3hliegyhmlxoizzksohdybbtmtqjejv5igkdee1r8go5jbcbryyyzpe26cqco0ci5hxhtyawnjuzabfkyvi8aearf0tk+syrjtbdqvkhyi/n5iwwtmlaqzyyjhg1n0cve/r29xfbmkqiuwqfh8eaojxshdtcgaadosghbznhyoujqzzrbbc8o4z0sbvvll+vawv18m/lnzqul1zhut67kykjkix+sueosctmgtarofcalkitytf8c6r86b8/4zwnhknupyb87hn+vokvi=</latexit> v<latexit sha1_base64="rpnqdvxuku79q5givxsvybo/ahy=">aaab8nicbvc7tsnaedzzdoevokq5esfrrtzcaroigspewirsykxnyyaccj5bd3uriitfqasvfallhyj4f2zjahkmgs3samcntkqw6lqfzsrq2vrgzmwrur2zu7dfozh8mlhvhhwey1h3q2zacgu+cptqttswkjtqcse3ud+zgjyivvc4syci2fijkeamm6k9hdtqbsmtqjejv5i6kdea1l76w5jbcbryyyzpew6cqco0ci5hxu1bawnjezagxkyvi8aearf0tk+tyrjtbdqvkhyi/n5iwwtmlaqzyyjho1n0cve/r2dxdbwkqiuwqfh8eaojxshdtcgaadoughbznhyoujqzzrbbc8o4z0sbvvln+vawv18m/nnjuug1l+rnm7kycjkmj+smeossnmkdargfcalkitytf8c6r86b8/4zuukuo0fkd5ypb+12kvm=</latexit> <latexit sha1_base64="rpnqdvxuku79q5givxsvybo/ahy=">aaab8nicbvc7tsnaedzzdoevokq5esfrrtzcaroigspewirsykxnyyaccj5bd3uriitfqasvfallhyj4f2zjahkmgs3samcntkqw6lqfzsrq2vrgzmwrur2zu7dfozh8mlhvhhwey1h3q2zacgu+cptqttswkjtqcse3ud+zgjyivvc4syci2fijkeamm6k9hdtqbsmtqjejv5i6kdea1l76w5jbcbryyyzpew6cqco0ci5hxu1bawnjezagxkyvi8aearf0tk+tyrjtbdqvkhyi/n5iwwtmlaqzyyjho1n0cve/r2dxdbwkqiuwqfh8eaojxshdtcgaadoughbznhyoujqzzrbbc8o4z0sbvvln+vawv18m/nnjuug1l+rnm7kycjkmj+smeossnmkdargfcalkitytf8c6r86b8/4zuukuo0fkd5ypb+12kvm=</latexit> <latexit sha1_base64="rpnqdvxuku79q5givxsvybo/ahy=">aaab8nicbvc7tsnaedzzdoevokq5esfrrtzcaroigspewirsykxnyyaccj5bd3uriitfqasvfallhyj4f2zjahkmgs3samcntkqw6lqfzsrq2vrgzmwrur2zu7dfozh8mlhvhhwey1h3q2zacgu+cptqttswkjtqcse3ud+zgjyivvc4syci2fijkeamm6k9hdtqbsmtqjejv5i6kdea1l76w5jbcbryyyzpew6cqco0ci5hxu1bawnjezagxkyvi8aearf0tk+tyrjtbdqvkhyi/n5iwwtmlaqzyyjho1n0cve/r2dxdbwkqiuwqfh8eaojxshdtcgaadoughbznhyoujqzzrbbc8o4z0sbvvln+vawv18m/nnjuug1l+rnm7kycjkmj+smeossnmkdargfcalkitytf8c6r86b8/4zuukuo0fkd5ypb+12kvm=</latexit> <latexit sha1_base64="rpnqdvxuku79q5givxsvybo/ahy=">aaab8nicbvc7tsnaedzzdoevokq5esfrrtzcaroigspewirsykxnyyaccj5bd3uriitfqasvfallhyj4f2zjahkmgs3samcntkqw6lqfzsrq2vrgzmwrur2zu7dfozh8mlhvhhwey1h3q2zacgu+cptqttswkjtqcse3ud+zgjyivvc4syci2fijkeamm6k9hdtqbsmtqjejv5i6kdea1l76w5jbcbryyyzpew6cqco0ci5hxu1bawnjezagxkyvi8aearf0tk+tyrjtbdqvkhyi/n5iwwtmlaqzyyjho1n0cve/r2dxdbwkqiuwqfh8eaojxshdtcgaadoughbznhyoujqzzrbbc8o4z0sbvvln+vawv18m/nnjuug1l+rnm7kycjkmj+smeossnmkdargfcalkitytf8c6r86b8/4zuukuo0fkd5ypb+12kvm=</latexit> <latexit sha1_base64="v0em0urh/km2emyuumrubdvgzhw=">aaab9hicbvdlsgnbejynrxhfuy9ebomgcgfxbpuw9oixomscsqi9k04cmvtgplcjsz7bq548ivf/x4p/4u66b02su1hvtvexfylpyly/rdlc4tlysnm1sra+sblv3d65m2gsbboivkfue2bqyqbdkqswhwke31py8saxmd96qg1kgnzsjmked6nadquasqwbxyonx63zdtshnydoqwqsqlnf/eooqhh7gjbqyezhsspqjabjcoxtsjc2gieywwg7kq3ar9nl8qhtfhabojbhqllupbfx90ycvjet30snfab7m+tl4n9ej6bhws+rqrqtbii7rfjhfsgildmoka+krilikioxareggqi15cbekszpkzw0d2f2+3nihtfp6871sa1xurrtzntsnx0yh52ybrtiteyywubsit2zf+vrerxerpef0zjv7oyyp7a+vghjephe</latexit> <latexit sha1_base64="v0em0urh/km2emyuumrubdvgzhw=">aaab9hicbvdlsgnbejynrxhfuy9ebomgcgfxbpuw9oixomscsqi9k04cmvtgplcjsz7bq548ivf/x4p/4u66b02su1hvtvexfylpyly/rdlc4tlysnm1sra+sblv3d65m2gsbboivkfue2bqyqbdkqswhwke31py8saxmd96qg1kgnzsjmked6nadquasqwbxyonx63zdtshnydoqwqsqlnf/eooqhh7gjbqyezhsspqjabjcoxtsjc2gieywwg7kq3ar9nl8qhtfhabojbhqllupbfx90ycvjet30snfab7m+tl4n9ej6bhws+rqrqtbii7rfjhfsgildmoka+krilikioxareggqi15cbekszpkzw0d2f2+3nihtfp6871sa1xurrtzntsnx0yh52ybrtiteyywubsit2zf+vrerxerpef0zjv7oyyp7a+vghjephe</latexit> <latexit sha1_base64="v0em0urh/km2emyuumrubdvgzhw=">aaab9hicbvdlsgnbejynrxhfuy9ebomgcgfxbpuw9oixomscsqi9k04cmvtgplcjsz7bq548ivf/x4p/4u66b02su1hvtvexfylpyly/rdlc4tlysnm1sra+sblv3d65m2gsbboivkfue2bqyqbdkqswhwke31py8saxmd96qg1kgnzsjmked6nadquasqwbxyonx63zdtshnydoqwqsqlnf/eooqhh7gjbqyezhsspqjabjcoxtsjc2gieywwg7kq3ar9nl8qhtfhabojbhqllupbfx90ycvjet30snfab7m+tl4n9ej6bhws+rqrqtbii7rfjhfsgildmoka+krilikioxareggqi15cbekszpkzw0d2f2+3nihtfp6871sa1xurrtzntsnx0yh52ybrtiteyywubsit2zf+vrerxerpef0zjv7oyyp7a+vghjephe</latexit> <latexit sha1_base64="v0em0urh/km2emyuumrubdvgzhw=">aaab9hicbvdlsgnbejynrxhfuy9ebomgcgfxbpuw9oixomscsqi9k04cmvtgplcjsz7bq548ivf/x4p/4u66b02su1hvtvexfylpyly/rdlc4tlysnm1sra+sblv3d65m2gsbboivkfue2bqyqbdkqswhwke31py8saxmd96qg1kgnzsjmked6nadquasqwbxyonx63zdtshnydoqwqsqlnf/eooqhh7gjbqyezhsspqjabjcoxtsjc2gieywwg7kq3ar9nl8qhtfhabojbhqllupbfx90ycvjet30snfab7m+tl4n9ej6bhws+rqrqtbii7rfjhfsgildmoka+krilikioxareggqi15cbekszpkzw0d2f2+3nihtfp6871sa1xurrtzntsnx0yh52ybrtiteyywubsit2zf+vrerxerpef0zjv7oyyp7a+vghjephe</latexit> w<latexit sha1_base64="acccimbsb2aibvn+sbhznphdqdy=">aaab8nicbvc7tsnaedzzdoevokq5esfrrtzcaroigspewirsykxnyyaccj5bd3ugymox0ejfhwj5iqr+bdu4gisprjo72tkjeykmuu6ns7s8srq2xtmobm5t7+zw9vbvtgw1b5/hmtbdkbmqqogpaiv0ew0sciv0wsl17nceqbsrq1ucjhbebkzeshcgmdr+hntqbsmtqbejv5i6kdea1l76w5jbcbryyyzpew6cqco0ci5hvu1bawnjezagxkyvi8aearf0ro+tyrjtbdqvkhyi/n5iwwtmnaqzyyjhvzn3cve/r2dxdbgkqiuwqfh8eaojxshdtcgaadoughbznhyoujqzzrbbc8o4z0sbvvln+vdmv18k/mnjsug1z+rnq7kycjkkr+seeoscnmknargfcalkitytf8c6r86b8/4zuusuowfkd5ypb+8ekvq=</latexit> <latexit sha1_base64="acccimbsb2aibvn+sbhznphdqdy=">aaab8nicbvc7tsnaedzzdoevokq5esfrrtzcaroigspewirsykxnyyaccj5bd3ugymox0ejfhwj5iqr+bdu4gisprjo72tkjeykmuu6ns7s8srq2xtmobm5t7+zw9vbvtgw1b5/hmtbdkbmqqogpaiv0ew0sciv0wsl17nceqbsrq1ucjhbebkzeshcgmdr+hntqbsmtqbejv5i6kdea1l76w5jbcbryyyzpew6cqco0ci5hvu1bawnjezagxkyvi8aearf0ro+tyrjtbdqvkhyi/n5iwwtmnaqzyyjhvzn3cve/r2dxdbgkqiuwqfh8eaojxshdtcgaadoughbznhyoujqzzrbbc8o4z0sbvvln+vdmv18k/mnjsug1z+rnq7kycjkkr+seeoscnmknargfcalkitytf8c6r86b8/4zuusuowfkd5ypb+8ekvq=</latexit> <latexit sha1_base64="acccimbsb2aibvn+sbhznphdqdy=">aaab8nicbvc7tsnaedzzdoevokq5esfrrtzcaroigspewirsykxnyyaccj5bd3ugymox0ejfhwj5iqr+bdu4gisprjo72tkjeykmuu6ns7s8srq2xtmobm5t7+zw9vbvtgw1b5/hmtbdkbmqqogpaiv0ew0sciv0wsl17nceqbsrq1ucjhbebkzeshcgmdr+hntqbsmtqbejv5i6kdea1l76w5jbcbryyyzpew6cqco0ci5hvu1bawnjezagxkyvi8aearf0ro+tyrjtbdqvkhyi/n5iwwtmnaqzyyjhvzn3cve/r2dxdbgkqiuwqfh8eaojxshdtcgaadoughbznhyoujqzzrbbc8o4z0sbvvln+vdmv18k/mnjsug1z+rnq7kycjkkr+seeoscnmknargfcalkitytf8c6r86b8/4zuusuowfkd5ypb+8ekvq=</latexit> <latexit sha1_base64="acccimbsb2aibvn+sbhznphdqdy=">aaab8nicbvc7tsnaedzzdoevokq5esfrrtzcaroigspewirsykxnyyaccj5bd3ugymox0ejfhwj5iqr+bdu4gisprjo72tkjeykmuu6ns7s8srq2xtmobm5t7+zw9vbvtgw1b5/hmtbdkbmqqogpaiv0ew0sciv0wsl17nceqbsrq1ucjhbebkzeshcgmdr+hntqbsmtqbejv5i6kdea1l76w5jbcbryyyzpew6cqco0ci5hvu1bawnjezagxkyvi8aearf0ro+tyrjtbdqvkhyi/n5iwwtmnaqzyyjhvzn3cve/r2dxdbgkqiuwqfh8eaojxshdtcgaadoughbznhyoujqzzrbbc8o4z0sbvvln+vdmv18k/mnjsug1z+rnq7kycjkkr+seeoscnmknargfcalkitytf8c6r86b8/4zuusuowfkd5ypb+8ekvq=</latexit> Somma di Interi Senza Segno Date due sequenze e di bit di lunghezza, la loro vera somma può essere una sequenza di w +1 bit. Bisogna scartare il bit più significativo (detto bit del riporto) e trattare il risultato come un intero senza segno. In tal modo la somma di interi senza segno implementa l'addizione modulare: U add w (u, v) =(u + v) mod 2 w ( U add u + v se u + v<2 w w (u, v) = u + v 2 w se u + v 2 w

<latexit sha1_base64="r1daa8uvmr8mfv+yybhkgemilqc=">aaacg3icbvc7sgnbfj31bxxflw0ggxbrwq4iaiginpykrgpjdhcnnzpk9shmnuhy9hf8bl/cvisrsbww8f/crcl8nwyo55zlnxucreldrvvujiyoju9mtk2xzmbn5hfki0vnjrzaoc9ifevlaawqgafpkhrejhohdbrebn2jgx/rq21khj1rp8fmcner7egbleutctxwfz42tif7v8ud7xbwshus+g2wve1mbz0p2i1eq1xxa24b/pd4q1jhq5y0yh+ndixsibejbcbuptehzgqapfcylrrwyakic9dyz2keizpmwlyu8tvrggkeoozs8ule7xmphmb0wybphka35rc3ep/z6py6u81urokljmrgeumfxsijtmyrqt6wgolg8hpkmuicnbchlhyeyewbd1fk+/b+x/+x+fu1vzp3ul05obwwm8vw2cqrmo/tsan2ze6yzws7yw/skt05986z8+k8fkvhnohmmvsb5+0tb0ygha==</latexit> <latexit sha1_base64="r1daa8uvmr8mfv+yybhkgemilqc=">aaacg3icbvc7sgnbfj31bxxflw0ggxbrwq4iaiginpykrgpjdhcnnzpk9shmnuhy9hf8bl/cvisrsbww8f/crcl8nwyo55zlnxucreldrvvujiyoju9mtk2xzmbn5hfki0vnjrzaoc9ifevlaawqgafpkhrejhohdbrebn2jgx/rq21khj1rp8fmcner7egbleutctxwfz42tif7v8ud7xbwshus+g2wve1mbz0p2i1eq1xxa24b/pd4q1jhq5y0yh+ndixsibejbcbuptehzgqapfcylrrwyakic9dyz2keizpmwlyu8tvrggkeoozs8ule7xmphmb0wybphka35rc3ep/z6py6u81urokljmrgeumfxsijtmyrqt6wgolg8hpkmuicnbchlhyeyewbd1fk+/b+x/+x+fu1vzp3ul05obwwm8vw2cqrmo/tsan2ze6yzws7yw/skt05986z8+k8fkvhnohmmvsb5+0tb0ygha==</latexit> <latexit sha1_base64="r1daa8uvmr8mfv+yybhkgemilqc=">aaacg3icbvc7sgnbfj31bxxflw0ggxbrwq4iaiginpykrgpjdhcnnzpk9shmnuhy9hf8bl/cvisrsbww8f/crcl8nwyo55zlnxucreldrvvujiyoju9mtk2xzmbn5hfki0vnjrzaoc9ifevlaawqgafpkhrejhohdbrebn2jgx/rq21khj1rp8fmcner7egbleutctxwfz42tif7v8ud7xbwshus+g2wve1mbz0p2i1eq1xxa24b/pd4q1jhq5y0yh+ndixsibejbcbuptehzgqapfcylrrwyakic9dyz2keizpmwlyu8tvrggkeoozs8ule7xmphmb0wybphka35rc3ep/z6py6u81urokljmrgeumfxsijtmyrqt6wgolg8hpkmuicnbchlhyeyewbd1fk+/b+x/+x+fu1vzp3ul05obwwm8vw2cqrmo/tsan2ze6yzws7yw/skt05986z8+k8fkvhnohmmvsb5+0tb0ygha==</latexit> <latexit sha1_base64="r1daa8uvmr8mfv+yybhkgemilqc=">aaacg3icbvc7sgnbfj31bxxflw0ggxbrwq4iaiginpykrgpjdhcnnzpk9shmnuhy9hf8bl/cvisrsbww8f/crcl8nwyo55zlnxucreldrvvujiyoju9mtk2xzmbn5hfki0vnjrzaoc9ifevlaawqgafpkhrejhohdbrebn2jgx/rq21khj1rp8fmcner7egbleutctxwfz42tif7v8ud7xbwshus+g2wve1mbz0p2i1eq1xxa24b/pd4q1jhq5y0yh+ndixsibejbcbuptehzgqapfcylrrwyakic9dyz2keizpmwlyu8tvrggkeoozs8ule7xmphmb0wybphka35rc3ep/z6py6u81urokljmrgeumfxsijtmyrqt6wgolg8hpkmuicnbchlhyeyewbd1fk+/b+x/+x+fu1vzp3ul05obwwm8vw2cqrmo/tsan2ze6yzws7yw/skt05986z8+k8fkvhnohmmvsb5+0tb0ygha==</latexit> Identificare l'overflow unsigned Determinare se s = U add (u, v) =u + v w Abbiamo overflow se e solo se s<u oppure s<v Inoltre s<use e solo se s<v, quindi è sufficiente controllare solo una delle due condizioni Su un processore, questa condizione setta il carry flag

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sha1_base64="5ttzh/29nacfi9ix3u3vax9qri4=">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</latexit> <latexit 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sha1_base64="acccimbsb2aibvn+sbhznphdqdy=">aaab8nicbvc7tsnaedzzdoevokq5esfrrtzcaroigspewirsykxnyyaccj5bd3ugymox0ejfhwj5iqr+bdu4gisprjo72tkjeykmuu6ns7s8srq2xtmobm5t7+zw9vbvtgw1b5/hmtbdkbmqqogpaiv0ew0sciv0wsl17nceqbsrq1ucjhbebkzeshcgmdr+hntqbsmtqbejv5i6kdea1l76w5jbcbryyyzpew6cqco0ci5hvu1bawnjezagxkyvi8aearf0ro+tyrjtbdqvkhyi/n5iwwtmnaqzyyjhvzn3cve/r2dxdbgkqiuwqfh8eaojxshdtcgaadoughbznhyoujqzzrbbc8o4z0sbvvln+vdmv18k/mnjsug1z+rnq7kycjkkr+seeoscnmknargfcalkitytf8c6r86b8/4zuusuowfkd5ypb+8ekvq=</latexit> Somma di Interi in Complemento a 2 Date due sequenze e di bit di lunghezza, la loro vera somma può essere una sequenza di w +1 bit. Bisogna scartare il bit più significativo (detto bit del riporto) e trattare il risultato come un intero in complemento a 2. C add w (u, v) = overflow negativo 8 >< u + v +2 w se u + v<c min (w) u + v se C >: min (w) apple u + v apple C max (w) u + v 2 w se C max (w) <u+ v overflow positivo

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Byte Pensare sempre in termini di bit può risultare complicato Spesso i bit sono raggruppati in unità più grandi 1 byte = 8 bit

Rappresentazione della memoria byte-oriented

Parole Macchina In generale i processori hanno una specifica "dimensione di word" La dimensione nominale degli indirizzi di memoria sulla macchina (a.k.a. parola di memoria) Oggi la maggior parte dei processori esegue software su 64 bit (8 byte) Software a 32 bit limita la memoria a 4 GB Tutti i sistemi hardware basati su processori x86 hanno parole macchina di 64 bit Ma supportano anche frazioni o multipli della parola macchina Sempre un numero intero di bit

Rappresentazione della memoria word-oriented Gli indirizzi specificano le locazioni dei byte Rappresenta l'indirizzo del primo byte della word Gli indirizzi delle parole successive differiscono di 4 (32 bit) o 8 (64 bit) Gli indirizzi degli oggetti multi-byte sono tipicamente allineati in base alla dimensione del dato

Ordinamento dei byte Avendo un dato multi-byte, in che ordine immagazziniamo i byte nella memoria? Dato il valore esadecimale su 64 bit 0x0001020304050607 si può memorizzare in due modi: big endian e little endian Si noti che l'endianess si applica solo a dati multi-byte di tipo primitivo, non a stringhe, array o struct.