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C 0 0000 +0 1 0000 -0 0 0001 +1 1 0001 -1 0 0010 +2 1 0010 -2 0 0011 +3 1 0011 -3 0 0100 +4 1 0100 -4 0 0101 +5 1 0101 -5 0 0110 1 0110 0 0111 1 0111 0 1000 1 1000 0 1001 1 1001 0 1010 1 1010 0 1011 1 1011 0 1100 1 1100 0 1101 1 1101 0 1110 +14 1 1110 -14 0 1111 +15 1 1111 -15DPDv8Binrn zporn sla obecnv~dy musme vdt, kolikabitov je kd kolik Ydo/bito mme k dispozici => kter bit je znamnkov (MSB) napY. v pYmm kdu 1+3 bity: 1 101(2) = -5(10) 1+4 bity: 0 1101(2) = +13(10) stejn (+13d) i pro 1+5, 1+6, ... bito nebo pro 0+4 bity '!!<9'B         - n(Od tn  pYm kdjako v destkov ode te se po Ydech zprava pYi ode tn  0-1 se vypoj   1 zleva ode t se mena od vtaho, pYi zmn menaence a menaitele se zmn znamnkov bit 26-9=17 26(10)=16+8+0+2+0(10)=11010(2) 9(10)=8+0+0+1(10)=1001(2) 11010-01001=10001(2) 1101-0101; 1011-1001; 1001-0011; 1000-0011 od tn vce sel ? postupn; nebo nejdYve vaechny menaitele se stPPPNPAP1PN  A1t Inverzn kdfjako pYm, jen zporn sla jsou inverz kladnho sla (logick negace bit po bitu) +5(10)=0 101(2) -5(10)=1 010(2) rozsah jako u pYmho (nula dvakrt), jen jin zporn sla$W =p@DoplHkov kd (dvojkov doplnk)!!(bkladn sla jsou v nm stejn jako v pYmm (nebo inverznm) kdu zporn slo jako v inverznm, ale navc se pYi t  1 =>nula jen jednou => -1 + 1 = 0 => -1 + 2 = 1 funguje matematika jedin reln vyu~van kd pro prci se zpornmi sly navc se rozsah datovho typu zva o 1 (v zporn oblasti)|.9<X    9<x DDoplHkov kd pYklady pro 1+4bity##(-3(10): +3(10)=0 0011(2) -3(10)=1 1100(2) + 1 -3(10)=1 1101(2) +15(10): +15(10)=0 1111(2) -16(10): -15(10)=1 0001(2) -16(10)=1 0000(2) pYete en: 15(10)+1=? 0 1111(2)+1=1 0000(2) 1 0000(2)=-16(10) 15(10)+1=-16(10) => rozsah -16 a~ +15P7P PP P$P PDPP7    $ D  |"8PYete en (registru, pamti)zsituace, kdy je snaha o pYenos do neexistujcho bitu (nesta  bity/Ydy) maximum+1 = minimum plat i u datovch typo pro kladn sla: 1111 + 1 = 10000 ale pYi 4 bitech se zapae jen 0000N_*$_*$  z!.DoplHkov kd  rozsah(Jnula jen jednou => rozsah o 1 vta ne~ u pYm.a in. +0: 0 0000 -0: 1 1111 + 1 = 0 0000 jazyk Pascal  nkter datov typy Shortint: -128..+127 (8b: 1MSB, zbyde 7; 27=128; -128 a~ -1 a 0 a~ 127) 127+1 = -128  pYete en Byte: 0..+255 (8b) Word: 0..65535 (16b) Integer: -32768..+32767 (16b) dYve se datov typy volily odpovdn, snahou byla nenro nost programu na RAM  640kB mus sta it ka~dmu  autor vroku = ? Bill Gates o zkladn pamti v OS MS-DOS pro pYstup do zbytku RAM (napY. 8MB, 32MB) slo~itja postup dnes napY. 2GB, 4GB RAM5Z#Z#ZHZZZ/Z)ZVZ5##*      /)V  P{j{$: Od tn s doplHkovm kdem8-6=2 8-6 = 8 + (-6) 8(10) = 0 1000(2) 6(10) = 0 0110(2) -6(10) = 1 1001(2) + 1 = 1 1010 (2) 0 1000 + 1 1010 = 0 0010 (2) = 2 po ta  od tn obejde s tnm ! a nejen on ! 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Martin Admek594Microsoft Office PowerPoint@0M@WS,@`X#koxGhg  ,T%&" WMFC <8  0AA @8  fQ ʚ;y8ʚ;g4YdYd} 0ppp@ <4ddddă&) 0 <4!d!dăH* 0g4SdSd` 08p@ pp<4BdBdă&) 0t0___PPT10 pp4___PPT9n3n)/ypf}I[ePNG  IHDR asRGB`PLTEkjjjUU@@@??@??+++**+**#g cmPPJCmp07128I tRNS@fAIDAT 0 ~{[8([Pz2S82(\dXP< zIENDB`0n&WvlHMU+vnPNG  IHDR asRGB`PLTEkjjkjjUUTU@@@??@++*+**M#A cmPPJCmp07128I tRNS@f>IDAT1W}{{J %s4!u61QJuRW&x&vxIENDB` ne!(mPɈPNG  IHDRysRGBPLTEffffffff̙̙O) cmPPJCmp0712 jtRNS ADIDATcP, SVA%6AC!P.iuD!".l`)SbPbL@]KIENDB`B&?v, bPRIPO, cvi en  Ing. Martin Admek (UHK-FIM-KIT)O  ='.PRIPO Principy po ta o$20.10.  cvi en . 5 Od tn v binrn soustav DoplHkov kd Procvi en pYevodo mezi soustavami *dZb  *Organiza n drobnosti dochzka F  $fTPYipomenut: S tn ve dvojkov soustav++("pln stejn jako destkov pYi psemnm s tn pod sebou jasn&nzorn Jen mme k dispozici jen dv slice jsme nuceni dYve pou~t vyaa Yd n-%#-%#  @ Qr6Zporn sla v pYmm kduPYidn zvlatn znamnkov bit 0 ~ + 1 ~ - most-significant bit (MSB) Nula redundantn zabr dv seln kombinace 8 bito: MSB=0: 0 a~ 127 MSB=1: -127 a~ -0 celkov rozsah: -127..+127 (msto 0 a~ 255 u pYirozench sel) (tabulka pYevzata z pYednaky)P'P6PbP!P'6b    B!    C 0 0000 +0 1 0000 -0 0 0001 +1 1 0001 -1 0 0010 +2 1 0010 -2 0 0011 +3 1 0011 -3 0 0100 +4 1 0100 -4 0 0101 +5 1 0101 -5 0 0110 1 0110 0 0111 1 0111 0 1000 1 1000 0 1001 1 1001 0 1010 1 1010 0 1011 1 1011 0 1100 1 1100 0 1101 1 1101 0 1110 +14 1 1110 -14 0 1111 +15 1 1111 -15DPDv8Binrn zporn sla obecnv~dy musme vdt, kolikabitov je kd kolik Ydo/bito mme k dispozici => kter bit je znamnkov (MSB) napY. v pYmm kdu 1+3 bity: 1 101(2) = -5(10) 1+4 bity: 0 1101(2) = +13(10) stejn (+13d) i pro 1+5, 1+6, ... bito nebo pro 0+4 bity '!!<9'B         - n(Od tn  pYm kdjako v destkov ode te se po Ydech zprava pYi ode tn  0-1 se vypoj   1 zleva ode t se mena od vtaho, pYi zmn menaence a menaitele se zmn znamnkov bit 26-9=17 26(10)=16+8+0+2+0(10)=11010(2) 9(10)=8+0+0+1(10)=1001(2) 11010-01001=10001(2) 1101-0101; 1011-1001; 1001-0011; 1000-0011 od tn vce sel ? postupn; nebo nejdYve vaechny menaite  !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~      "#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuxyz{|}~wvRoot EntrydO)p*ko@Pictures Current UserVSummaryInformation(!PowerPoint Document(^DocumentSummaryInformation8 . @n?" dd@  @@`` @ / /))*-+E\]hiqrstuvxyz{|}~,b$#̒͡daݾC>  0AA @8  fQ ʚ;y8ʚ;g4YdYd} 0ppp@ <4ddddă&) 0 <4!d!dăH* 0g4SdSd` 08p@ pp<4BdBdă&) 0t0___PPT10 pp4___PPT9n3n)/ypf}I[ePNG  IHDR asRGB`PLTEkjjjUU@@@??@??+++**+**#g cmPPJCmp07128I tRNS@fAIDAT 0 ~{[8([Pz2S82(\dXP< zIENDB`0n&WvlHMU+vnPNG  IHDR asRGB`PLTEkjjkjjUUTU@@@??@++*+**M#A cmPPJCmp07128I tRNS@f>IDAT1W}{{J %s4!u61QJuRW&x&vxIENDB` ne!(mPɈPNG  IHDRysRGBPLTEffffffff̙̙O) cmPPJCmp0712 jtRNS ADIDATcP, SVA%6AC!P.iuD!".l`)SbPbL@]KIENDB`B&?v, bPRIPO, cvi en  Ing. Martin Admek (UHK-FIM-KIT)O  ='.PRIPO Principy po ta o$20.10.  cvi en . 5 Od tn v binrn soustav DoplHkov kd Procvi en pYevodo mezi soustavami *dZb  *Organiza n drobnosti dochzka F  $fTPYipomenut: S tn ve dvojkov soustav++("pln stejn jako destkov pYi psemnm s tn pod sebou jasn&nzorn Jen mme k dispozici jen dv slice jsme nuceni dYve pou~t vyaa Yd n-%#-%#  @ Qr6Zporn sla v pYmm kduPYidn zvlatn znamnkov bit 0 ~ + 1 ~ - most-significant bit (MSB) Nula redundantn zabr dv seln kombinace 8 bito: MSB=0: 0 a~ 127 MSB=1: -127 a~ -0 celkov rozsah: -127..+127 (msto 0 a~ 255 u pYirozench sel) (tabulka pYevzata z pYednaky)P'P6PbP!P'6b    B!    C 0 0000 +0 1 0000 -0 0 0001 +1 1 0001 -1 0 0010 +2 1 0010 -2 0 0011 +3 1 0011 -3 0 0100 +4 1 0100 -4 0 0101 +5 1 0101 -5 0 0110 1 0110 0 0111 1 0111 0 1000 1 1000 0 1001 1 1001 0 1010 1 1010 0 1011 1 1011 0 1100 1 1100 0 1101 1 1101 0 1110 +14 1 1110 -14 0 1111 +15 1 1111 -15DPDv8Binrn zporn sla obecnv~dy musme vdt, kolikabitov je kd kolik Ydo/bito mme k dispozici => kter bit je znamnkov (MSB) napY. v pYmm kdu 1+3 bity: 1 101(2) = -5(10) 1+4 bity: 0 1101(2) = +13(10) stejn (+13d) i pro 1+5, 1+6, ... bito nebo pro 0+4 bity '!!<9'B         - n(Od tn  pYm kdjako v destkov ode te se po Ydech zprava pYi ode tn  0-1 se vypoj   1 zleva ode t se mena od vtaho, pYi zmn menaence a menaitele se zmn znamnkov bit 26-9=17 26(10)=16+8+0+2+0(10)=11010(2) 9(10)=8+0+0+1(10)=1001(2) 11010-01001=10001(2) 1101-0101; 1011-1001; 1001-0011; 1000-0011 od tn vce sel ? postupn; nebo nejdYve vaechny menaitele se stPPPNPAP1PN  A1t Inverzn kdfjako pYm, jen zporn sla jsou inverz kladnho sla (logick negace bit po bitu) +5(10)=0 101(2) -5(10)=1 010(2) rozsah jako u pYmho (nula dvakrt), jen jin zporn sla$W =p@DoplHkov kd (dvojkov doplnk)!!(bkladn sla jsou v nm stejn jako v pYmm (nebo inverznm) kdu zporn slo jako v inverznm, ale navc se pYi t  1 =>nula jen jednou => -1 + 1 = 0 => -1 + 2 = 1 funguje matematika jedin reln vyu~van kd pro prci se zpornmi sly navc se rozsah datovho typu zva o 1 (v zporn oblasti)|.9<X    9<x DDoplHkov kd pYklady pro 1+4bity##(-3(10): +3(10)=0 0011(2) -3(10)=1 1100(2) + 1 -3(10)=1 1101(2) +15(10): +15(10)=0 1111(2) -16(10): -15(10)=1 0001(2) -16(10)=1 0000(2) pYete en: 15(10)+1=? 0 1111(2)+1=1 0000(2) 1 0000(2)=-16(10) 15(10)+1=-16(10) => rozsah -16 a~ +15P7P PP P$P PDPP7    $ D  |"8PYete en (registru, pamti)zsituace, kdy je snaha o pYenos do neexistujcho bitu (nesta  bity/Ydy) maximum+1 = minimum plat i u datovch typo pro kladn sla: 1111 + 1 = 10000 ale pYi 4 bitech se zapae jen 0000N_*$_*$  z!.DoplHkov kd  rozsah(Jnula jen jednou => rozsah o 1 vta ne~ u pYm.a in. +0: 0 0000 -0: 1 1111 + 1 = 0 0000 jazyk Pascal  nkter datov typy Shortint: -128..+127 (8b: 1MSB, zbyde 7; 27=128; -128 a~ -1 a 0 a~ 127) 127+1 = -128  pYete en Byte: 0..+255 (8b) Word: 0..65535 (16b) Integer: -32768..+32767 (16b) dYve se datov typy volily odpovdn, snahou byla nenro nost programu na RAM  640KB mus sta it ka~dmu  autor vroku = ? Bill Gates o zkladn pamti v OS MS-DOS pro pYstup do zbytku RAM (napY. 8MB, 32MB) slo~itja postup dnes napY. 2GB, 4GB RAM5Z#Z#ZHZZZ/Z)ZVZ5##*      /)V  b{jg){$: Od tn s doplHkovm kdem8-6=2 8-6 = 8 + (-6) 8(10) = 0 1000(2) 6(10) = 0 0110(2) -6(10) = 1 1001(2) + 1 = 1 1010 (2) 0 1000 + 1 1010 = 0 0010 (2) = 2 po ta  od tn obejde s tnm ! a nejen on ! (a nejen od tn)  # ~#Procvi it sipYevody mezi selnmi soustavami (cv. 4) doplHkov kd (cv. 5) vpo et Un a In v srioparalelnm zapojen rezistoro (cv. 2-3)6I0P# ,Z PYat'logick funkce vod do Booleovy algebry/[goqsuwy {!}"#$  ! 0 ,$(  x  c $p*0 !     c $p+0x<$D 0  X  0A?!H  0޽h ? 3f{FEɄJ B ___PPT10" .P`C+YD ' z= @B D ' = @BA?%,( < +O%,( < +D4' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*x%(D4' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D4' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*%(D4' =%(D' =%(D' =4@BBBB%(D' =1:Bvisible*o3>+B#style.visibility<*&%(+r2z(j! 1v 7( D/ 0|DArialr Newman$ 0DTimes New R ՜.+,0p    Pedvdn na obrazovceNchod^' ArialTimes New Roman WingdingsSymbol Courier NewPaprskyPRIPO Principy potaOrganizan drobnosti+Pipomenut: Stn ve dvojkov soustavZporn sla v pmm kduBinrn zporn sla obecnOdtn pm kd Inverzn kd!Doplkov kd (dvojkov doplnk)#Doplkov kd pklady pro 1+4bityPeteen (registru, pamti)Doplkov kd rozsahOdtn s doplkovm kdem Procviit siPt Pouit psmaablona nvrhuNadpisy snmk*_:Ing. Martin AdmekIng. Male se stPPPNPAP1PN  A1t Inverzn kdfjako pYm, jen zporn sla jsou inverz kladnho sla (logick negace bit po bitu) +5(10)=0 101(2) -5(10)=1 010(2) rozsah jako u pYmho (nula dvakrt), jen jin zporn sla$W =p@DoplHkov kd (dvojkov doplnk)!!(bkladn sla jsou v nm stejn jako v pYmm (nebo inverznm) kdu zporn slo jako v inverznm, ale navc se pYi t  1 =>nula jen jednou => -1 + 1 = 0 => -1 + 2 = 1 funguje matematika jedin reln vyu~van kd pro prci se zpornmi sly navc se rozsah datovho typu zva o 1 (v zporn oblasti)|.9<X    9<x DDoplHkov kd pYklady pro 1+4bity##(-3(10): +3(10)=0 0011(2) -3(10)=1 1100(2) + 1 -3(10)=1 1101(2) +15(10): +15(10)=0 1111(2) -16(10): -15(10)=1 0001(2) -16(10)=1 0000(2) pYete en: 15(10)+1=? 0 1111(2)+1=1 0000(2) 1 0000(2)=-16(10) 15(10)+1=-16(10) => rozsah -16 a~ +15P7P PP P$P PDPP7    $ D  |"8PYete en (registru, pamti)zsituace, kdy je snaha o pYenos do neexistujcho bitu (nesta  bity/Ydy) maximum+1 = minimum plat i u datovch typo pro kladn sla: 1111 + 1 = 10000 ale pYi 4 bitech se zapae jen 0000N_*$_*$  z!.DoplHkov kd  rozsah(Jnula jen jednou => rozsah o 1 vta ne~ u pYm.a in. +0: 0 0000 -0: 1 1111 + 1 = 0 0000 jazyk Pascal  nkter datov typy Shortint: -128..+127 (8b: 1MSB, zbyde 7; 27=128; -128 a~ -1 a 0 a~ 127) 127+1 = -128  pYete en Byte: 0..+255 (8b) Word: 0..65535 (16b) Integer: -32768..+32767 (16b) dYve se datov typy volily odpovdn, snahou byla nenro nost programu na RAM  640KB mus sta it ka~dmu  autor vroku = ? Bill Gates o zkladn pamti v OS MS-DOS pro pYstup do zbytku RAM (napY. 8MB, 32MB) slo~itja postup dnes napY. 2GB, 4GB RAM5Z#Z#ZHZZZ/Z)ZVZ5##*      /)V  P{j{$: Od tn s doplHkovm kdem8-6=2 8-6 = 8 + (-6) 8(10) = 0 1000(2) 6(10) = 0 0110(2) -6(10) = 1 1001(2) + 1 = 1 1010 (2) 0 1000 + 1 1010 = 0 0010 (2) = 2 po ta  od tn obejde s tnm ! a nejen on ! (a nejen od tn)  # ~#Procvi it sipYevody mezi selnmi soustavami (cv. 4) doplHkov kd (cv. 5) vpo et Un a In v srioparalelnm zapojen rezistoro (cv. 2-3)6I0P# ,Z PYat'logick funkce vod do Booleovy algebry/[goqsuwy {!}"#$X$ 0 Ph(  X  C       S ۷  0   j<s tn je snaza ne~ od tn H  0޽h ? 3380___PPT10.PС.r:w9$ w 8( D/ 0|DArialr Newman$ 0DTimes New Roman$ 0 DWingdingsRoman$ 00DSymbolgsRoman$ 0@DCourier Newman$ 01@ . @n?" dd@  @@`` @ / /))*-+E\]hiqrstuvxyz{|}~,b$#̒͡daݾC>  0AA @8  fQ ʚ;y8ʚ;g4YdYd} 0ppp@ <4dddd&) 0 <4!d!dH* 0g4SdSd` 08p@ pp<4BdBd&) 0t0___PPT10 pp4___PPT9n3n)/ypf}I[ePNG  IHDR asRGB`PLTEkjjjUU@@@??@??+++**+**#g cmPPJCmp07128I tRNS@fAIDAT 0 ~{[8([Pz2S82(\dXP< zIENDB`0n&WvlHMU+vnPNG  IHDR asRGB`PLTEkjjkjjUUTU@@@??@++*+**M#A cmPPJCmp07128I tRNS@f>IDAT1W}{{J %s4!u61QJuRW&x&vxIENDB` ne!(mPɈPNG  IHDRysRGBPLTEffffffff̙̙O) cmPPJCmp0712 jtRNS ADIDATcP, SVA%6AC!P.iuD!".l`)SbPbL@]KIENDB`B&?v, bPRIPO, cvi en  Ing. Martin Admek (UHK-FIM-KIT)O  =x(.PRIPO Principy po ta o$20.10.  cvi en . 5 Od tn v binrn soustav DoplHkov kd Procvi en pYevodo mezi soustavami *dZb  *Organiza n drobnosti dochzka F  $fTPYipomenut: S tn ve dvojkov soustav++("pln stejn jako destkov pYi psemnm s tn pod sebou jasn&nzorn Jen mme k dispozici jen dv slice jsme nuceni dYve pou~t vyaa Yd n-%#-%#  @ Qr6Zporn sla v pYmm kduPYidn zvlatn znamnkov bit 0 ~ + 1 ~ - most-significant bit (MSB) Nula redundantn zabr dv seln kombinace 8 bito: MSB=0: 0 a~ 127 MSB=1: -127 a~ -0 celkov rozsah: -127..+127 (msto 0 a~ 255 u pYirozench sel) (tabulka pYevzata z pYednaky)P'P6PbP!P'6b    B!    C 0 0000 +0 1 0000 -0 0 0001 +1 1 0001 -1 0 0010 +2 1 0010 -2 0 0011 +3 1 0011 -3 0 0100 +4 1 0100 -4 0 0101 +5 1 0101 -5 0 0110 1 0110 0 0111 1 0111 0 1000 1 1000 0 1001 1 1001 0 1010 1 1010 0 1011 1 1011 0 1100 1 1100 0 1101 1 1101 0 1110 +14 1 1110 -14 0 1111 +15 1 1111 -15DPDv8Binrn zporn sla obecnv~dy musme vdt, kolikabitov je kd kolik Ydo/bito mme k dispozici => kter bit je znamnkov (MSB) napY. v pYmm kdu 1+3 bity: 1 101(2) = -5(10) 1+4 bity: 0 1101(2) = +13(10) stejn (+13d) i pro 1+5, 1+6, ... bito nebo pro 0+4 bity '!!<9'B         - n(Od tn  pYm kdjako v destkov ode te se po Ydech zprava pYi ode tn  0-1 se vypoj   1 zleva ode t se mena od vtaho, pYi zmn menaence a menaitele se zmn znamnkov bit 26-9=17 26(10)=16+8+0+2+0(10)=11010(2) 9(10)=8+0+0+1(10)=1001(2) 11010-01001=10001(2) 1101-0101; 1011-1001; 1001-0011; 1000-0011 od tn vce sel ? postupn; nebo nejdYve vaechny menaitele se stPPPNPAP1PN  A1t Inverzn kdfjako pYm, jen zporn sla jsou inverz kladnho sla (logick negace bit po bitu) +5(10)=0 101(2) -5(10)=1 010(2) rozsah jako u pYmho (nula dvakrt), jen jin zporn sla$W =p@DoplHkov kd (dvojkov doplnk)!!(bkladn sla jsou v nm stejn jako v pYmm (nebo inverznm) kdu zporn slo jako v inverznm, ale navc se pYi t  1 =>nula jen jednou => -1 + 1 = 0 => -1 + 2 = 1 funguje matematika jedin reln vyu~van kd pro prci se zpornmi sly navc se rozsah datovho typu zva o 1 (v zporn oblasti)|.9<X    9<x DDoplHkov kd pYklady pro 1+4bity##(-3(10): +3(10)=0 0011(2) -3(10)=1 1100(2) + 1 -3(10)=1 1101(2) +15(10): +15(10)=0 1111(2) -16(10): -15(10)=1 0001(2) -16(10)=1 0000(2) pYete en: 15(10)+1=? 0 1111(2)+1=1 0000(2) 1 0000(2)=-16(10) 15(10)+1=-16(10) => rozsah -16 a~ +15P7P PP P$P PDPP7    $ D  |"8PYete en (registru, pamti)zsituace, kdy je snaha o pYenos do neexistujcho bitu (nesta  bity/Ydy) maximum+1 = minimum plat i u datovch typo pro kladn sla: 1111 + 1 = 10000 ale pYi 4 bitech se zapae jen 0000N_*$_*$  z!.DoplHkov kd  rozsah(Jnula jen jednou => rozsah o 1 vta ne~ u pYm.a in. +0: 0 0000 -0: 1 1111 + 1 = 0 0000 jazyk Pascal  nkter datov typy Shortint: -128..+127 (8b: 1MSB, zbyde 7; 27=128; -128 a~ -1 a 0 a~ 127) 127+1 = -128  pYete en Byte: 0..+255 (8b) Word: 0..65535 (16b) Integer: -32768..+32767 (16b) dYve se datov typy volily odpovdn, snahou byla nenro nost programu na RAM  640KB mus sta it ka~dmu  autor vroku = ? Bill Gates o zkladn pamti v OS MS-DOS pro pYstup do zbytku RAM (napY. 8MB, 32MB) slo~itja postup dnes napY. 2GB, 4GB RAM5Z#Z#ZHZZZ/Z)ZVZ5##*      /)V  P{j{$: Od tn s doplHkovm kdem8-6=2 8-6 = 8 + (-6) 8(10) = 0 1000(2) 6(10) = 0 0110(2) -6(10) = 1 1001(2) + 1 = 1 1010 (2) 0 1000 + 1 1010 = 0 0010 (2) = 2 po ta  od tn obejde s tnm ! a nejen on ! (a nejen od tn)  # ~#Procvi it sipYevody mezi selnmi soustavami (cv. 4) doplHkov kd (cv. 5) vpo et Un a In v srioparalelnm zapojen rezistoro (cv. 2-3)6I0P# ,Z PYatProcvi en probranch tmat PYespYat logick funkce vod do Booleovy algebry( (P )/[goqsuwy {!}"#$}  0 t$(  tr t S *0 `   r t S |+0 `  H t 0޽h ? 3f{FEɄ___PPT10i.K`3+D=' z= @B +rձZ "