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>> [MUSIC PLAYING]

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>> DOUG LLOYD: Hi, bel蓹 ki, ed蓹k
C. operatorlar haqq谋nda dan谋艧maq

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Bel蓹 ki, biz art谋q bir g枚rd眉m, 蓹slind蓹,
tap艧谋r谋q operator b蓹rab蓹rdir.

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Bu, bizim yaln谋z qoymaq 眉莽眉n imkan verir
bir d蓹yi艧蓹n d蓹y蓹ri.

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Bu tap艧谋r谋q var
operator, bir b蓹rab蓹r i艧ar蓹si.

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Manipulyasiya 眉莽眉n v蓹 i艧
C d蓹y蓹rl蓹r v蓹 d蓹yi艧蓹nl蓹rin il蓹,

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biz operatorlar谋n bir s谋ra var
bizim s蓹r蓹ncam谋nda biz istifad蓹 ed蓹 bil蓹rsiniz.

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Nin bir n蓹z蓹r salaq
眉mumi olanlar谋 b蓹zi

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hesab operatorlar谋 il蓹 ba艧lay谋r.

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Siz g枚zl蓹y蓹 bil蓹r kimi, biz n蓹 ed蓹
C. olduqca 蓹sas riyaziyyat 蓹m蓹liyyatlar谋

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Biz, 蓹lav蓹 莽谋xmaq, 莽oxaltmaq v蓹 bil蓹r
plus istifad蓹 b枚lm蓹k n枚mr蓹l蓹ri, minus, ulduz,

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v蓹 m眉vafiq olaraq, do臒ramaq.

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Burada x蓹tl蓹ri bir ne莽蓹 var
code hans谋 biz bunu.

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Bel蓹 ki, biz int x y plus 1 b蓹rab蓹rdir var.

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Bir yerd蓹 ki, g眉man ed蓹k
kodu bu x蓹tt yuxar谋da

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biz int y 10 b蓹rab蓹rdir deyib.

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X d蓹y蓹ri I sonra n蓹
kodu bu ilk line icra?

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Siz 11 dem蓹k mi?

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Siz do臒ru had.

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Niy蓹 ki?

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Yax艧谋, y 10 idi.

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B蓹zi M蓹n deyir蓹m int x 10 plus 1 b蓹rab蓹rdir.

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10 plus 1 11 edir.

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Bel蓹 ki, d蓹y蓹ri 11 olur
d蓹yi艧蓹n x saxlan谋l谋r.

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脟ox pis deyil, sa臒?

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>> Nec蓹 bu n枚vb蓹ti x蓹tti haqq谋nda
code? x x d蓹f蓹 5 b蓹rab蓹rdir.

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B蓹li, 蓹vv蓹l biz icra
kodu bu x蓹tt x 11 idi.

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Bel蓹 ki, d蓹y蓹ri n蓹
x kodu bu x蓹tt sonra?

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Ikinci edin.

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Bel蓹 ki, x x d蓹f蓹 5 b蓹rab蓹rdir.

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x 11 idi.

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Bel蓹 ki, x 11 d蓹f蓹 5 b蓹rab蓹rdir.

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V蓹 ya 55.

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Siz 55 dedi Bel蓹 ki, siz do臒ru olard谋.

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>> 陌ndi, bir az 莽a艧d谋r谋c谋 ola bil蓹r, lakin
ki, tap艧谋r谋q C i艧l蓹yir yolu il蓹

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sa臒 d蓹y蓹ri olur
sol, d蓹y蓹ri t蓹yin.

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Bel蓹 ki, ilk biz x d蓹f蓹 5 qiym蓹tl蓹ndirm蓹k.

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Bel蓹 ki, 11 d蓹f蓹 5 55 edir.

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V蓹 sonra biz x ki, d蓹y蓹ri saxlamay谋n.

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Var idi 11
蓹vv蓹l 陌ndi 眉zerine yaz谋l谋r.

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Bel蓹 ki, x d蓹y蓹ri indi 55.

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脺mid edir蓹m ki, kifay蓹t q蓹d蓹r sad蓹 deyil.

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>> Siz var ba艧qa operator var
y蓹qin ki, m眉tl蓹q e艧itdim

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bu deyil蓹n, lakin var
蓹lb蓹tt蓹 ke莽mi艧d蓹 il蓹 i艧l蓹mi艧dir

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Siz uzun sizin g眉n xat谋rlay谋ram
geri grade m蓹kt蓹b b枚lg眉s眉 yol.

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Bu modulus operator deyirl蓹r.

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N蓹 modulus yoxdur bunu var
siz蓹 qalan verir

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Siz birlikd蓹 iki 蓹d蓹d b枚lm蓹k zaman.

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M蓹n dem蓹k Bel蓹 ki, 13 b枚l眉n眉r
4, qalan n蓹 var?

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V蓹 ki, d蓹y蓹ri hesablan谋r olunacaq
modulus operator t蓹r蓹find蓹n.

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>> Bel蓹 ki, m蓹n kodu bir x蓹tti
Burada int m 13 mod 4 b蓹rab蓹rdir.

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M蓹n bir comment burada dem蓹k
M d蓹y蓹ri indi 1.

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Niy蓹 ki, deyirsiniz?

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B蓹li, uzun b枚lg眉s眉 h蓹yata sizin
ba艧 bir ikinci m蓹niml蓹 da艧谋y谋rlar edin.

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Bel蓹 ki, m蓹n 4 13 oxuyasan.

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4 13 眉莽 d蓹f蓹 gider
1-qalan.

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Bel蓹 ki, 蓹sas蓹n, b眉t眉n
modulus operator yoxdur

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siz蓹 zaman deyir edir
b枚lm蓹k, siz qalan almaq.

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Siz, h蓹qiq蓹t蓹n, hesab bil蓹r
Biz olduqca faydal谋 艧ey,

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ancaq, h蓹qiq蓹t蓹n, heyran olar谋q
nec蓹 tez-tez ki, modulus il蓹

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operator laz谋ml谋 g蓹l蓹 bil蓹r.

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>> Probleml蓹rin bir ne莽蓹 var
biz il蓹 m蓹艧臒ul CS50 ed蓹c蓹yik.

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O, h蓹m莽inin bunu 眉莽眉n yax艧谋
t蓹sad眉fi say谋 kimi 艧eyl蓹r.

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Bel蓹 ki, m蓹s蓹l蓹n 茝g蓹r var 蓹g蓹r
bir t蓹sad眉fi say谋 generator e艧itdim,

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ki, bir s谋ra verm蓹k olacaq
0 b蓹zi b枚y眉k sayda.

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Amma b蓹lk蓹 siz yaln谋z h蓹qiq蓹t蓹n
0 20 bir s谋ra laz谋md谋r.

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Siz modulus operator istifad蓹 蓹g蓹r
ki, n蓹h蓹ng say谋 ki,

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t蓹r蓹find蓹n yarad谋lan 蓹vv蓹l
t蓹sad眉fi say谋 generator,

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sizi olacaq n蓹
Bu, b枚y眉k d蓹y蓹ri, 20 il蓹 b枚lm蓹k

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v蓹 qalan almaq.

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qalan yaln谋z
0 19 d蓹y蓹ri ola bil蓹r.

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Bel蓹 ki, modulus operator istifad蓹
Bu 莽ox etm蓹k

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v蓹 a艧a臒谋 bir 艧ey yava艧-yava艧 onu
bir az daha m蓹nal谋.

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M蓹n siz蓹 olacaq olduqca 蓹min蓹m
o, h蓹m d蓹 istifad蓹 ed蓹

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CS50 g蓹l蓹c蓹kd蓹 m眉蓹yy蓹n m蓹qamda.

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>> Bel蓹 ki, C d蓹 biz蓹 bir yol verir
bir hesab m眉raci蓹t etm蓹k

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bir d蓹yi艧蓹n operator
bir az daha stenoqrafiya 艧蓹kild蓹.

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Bel蓹 ki, 蓹vv蓹lki slayd,
biz x x d蓹f蓹 5 b蓹rab蓹rdir g枚rd眉m.

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Ki 莽al谋艧m谋艧d谋r.

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x d蓹f蓹 5 onda x geri saxlan谋l谋r olur.

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Bunu q谋sa yol fikir var,
v蓹 syntax x d蓹f蓹 5 b蓹rab蓹rdir var.

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Bu eyni d蓹qiq 艧ey
dey蓹r蓹k x x d蓹f蓹 5 b蓹rab蓹rdir.

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Bu, sad蓹c蓹 bir az var
bunu q谋sa yol.

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V蓹 b蓹zi g枚r蓹nd蓹
distribution kodu v蓹 ya

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bir 枚rnek kod oldu ki,
bu kimi 艧eyl蓹r yoxdur,

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yaln谋z il蓹 tan谋艧 olmaq
syntax n蓹 dem蓹kdir.

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Siz 蓹lb蓹tt蓹 ki, yoxdur
茝g蓹r istifad蓹, lakin,

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Bu kodu ed蓹 bil蓹r
bir az slicker baxmaq.

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V蓹 siz d蓹 h蓹r hans谋 bir istifad蓹 ed蓹 bil蓹rsiniz ki, bilirik
m眉xt蓹lif operatorlar谋 biz art谋q var

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蓹vv蓹l yerin蓹 d蓹f蓹 g枚r眉ld眉.

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Siz x plus 5, minus b蓹rab蓹rdir dey蓹 bil蓹r
5 d蓹f蓹, u莽urum, v蓹 mod b蓹rab蓹rdir.

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Bu i艧 b眉t眉n.

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>> Bir 艧ey d蓹 var
ki, C bel蓹 眉mumi var

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biz q蓹rar q蓹bul etdik ki,
daha ki hassasla艧t谋r谋n.

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1-d蓹yi艧蓹n incrementing
v蓹 ya 1 il蓹 d蓹yi艧蓹n decrementing

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Bel蓹 bir 眉mumi 艧ey edir x眉susil蓹
biz dan谋艧maq zaman haqq谋nda bir az sonra loops

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Us biz 蓹v蓹zin蓹 q蓹rar q蓹bul etdik ki,
, x plus kimi bir 艧ey 1 b蓹rab蓹rdir s枚yl蓹y蓹r蓹k

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v蓹 ya x x plus 1 b蓹rab蓹rdir, h蓹tta var
Q谋sa x plus Plus etdi.

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Bel蓹 ki, x x plus 1, x b蓹rab蓹rdir plus 1 b蓹rab蓹rdir,
v蓹 x plus plus b眉t眉n eyni 艧ey.

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Onlar 1 b眉t眉n art谋m x.

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Amma ki, incrementing
v蓹 1-decrementing

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biz ki, 眉mumi
plus plus v蓹 minus minus

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biz蓹 stenoqrafiya imkan verir
daha ki.

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>> Bel蓹 ki, ikinci di艧li ke莽id imkan
v蓹 Boolean ifad蓹l蓹r haqq谋nda dan谋艧maq.

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D蓹 c眉r olan b眉t眉n d眉艧m蓹k
operatorlar谋n 眉mumi kateqoriya.

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Amma Boolean ifad蓹l蓹r,
hesab operatorlar谋 f蓹rqli olaraq,

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d蓹y蓹rl蓹ri m眉qayis蓹 眉莽眉n istifad蓹 olunur.

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C Bel蓹 ki, daha, b眉t眉n Boolean ifad蓹l蓹r
iki m眉mk眉n d蓹y蓹rl蓹r s谋ras谋nda qiym蓹tl蓹ndirm蓹k,

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xat谋rlay谋ram.

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Do臒ru v蓹 ya yalan.

112
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Bu yaln谋z iki d蓹y蓹rl蓹r var ki,
Boolean d蓹yi艧蓹n ed蓹 bil蓹r.

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Biz n蓹tic蓹l蓹r istifad蓹 ed蓹 bil蓹rsiniz
bir Boolean ifad蓹

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proqramla艧d谋rma yollar谋 bir 莽ox.

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茝slind蓹, siz olacaq
bu bir 莽ox edir.

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>> M蓹s蓹l蓹n, biz q蓹rar q蓹bul ed蓹 bil蓹r,
Yax艧谋, bir 艧蓹rt do臒ru olduqda,

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b蓹lk蓹 m蓹n bu almaq laz谋md谋r
m蓹nim kodu a艧a臒谋 艧ax蓹l蓹nm蓹k.

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A 艧蓹rti, bel蓹 dan谋艧maq.

119
00:05:35,980 --> 00:05:37,970
Biz tezlikl蓹 d蓹 o 枚yr蓹nm蓹k laz谋md谋r.

120
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V蓹 ya b蓹lk蓹 kimi uzun kimi
m蓹n ist蓹yir蓹m, do臒rudur

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bunu saxlamaq 眉莽眉n
眉z蓹rind蓹 v蓹 眉z蓹rind蓹.

122
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A loop.

123
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H蓹r iki halda, biz istifad蓹 etdiyiniz bilirik ki,
do臒ru v蓹 ya yalan bir Boolean ifad蓹,

124
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v蓹 ya q蓹rar q蓹bul etm蓹k
x眉susi yol almaq.

125
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B蓹z蓹n biz 莽al谋艧谋r谋q
Boolean ifad蓹l蓹ri il蓹,

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biz n枚v眉 bool d蓹yi艧蓹nl蓹r istifad蓹 ed蓹c蓹k.

127
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Siz elan ed蓹 bil蓹r
bir Bool, d蓹yi艧蓹n tipli

128
00:05:58,461 --> 00:06:00,210
v蓹 istifad蓹 laz谋md谋r
M蓹ntiqi ifad蓹.

129
00:06:00,210 --> 00:06:02,130
Amma h蓹mi艧蓹 yoxdur.

130
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Bu, C, h蓹r bir qeyri-0 莽谋x谋r kimi
d蓹y蓹ri do臒ru dey蓹r蓹k kimi eyni.

131
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Siz b蓹yan 蓹g蓹r bir
n枚v眉 Boolean d蓹yi艧蓹n,

132
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v蓹 ki, do臒ru bu d蓹y蓹r verilir
bir tam elan eyni

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00:06:14,240 --> 00:06:17,410
v蓹 bu d蓹y蓹r t蓹yin
1, 2, 3, v蓹 ya, h蓹qiq蓹t蓹n, h蓹r hans谋 bir d蓹y蓹ri

134
00:06:17,410 --> 00:06:19,580
0 ba艧qa he莽.

135
00:06:19,580 --> 00:06:22,690
C 脟眉nki, h蓹r bir qeyri-0 d蓹y蓹ri do臒rudur.

136
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0, dig蓹r t蓹r蓹fd蓹n, yalan.

137
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Bu g蓹lm蓹k bil蓹r
bilm蓹k sonra laz谋ml谋,

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00:06:27,162 --> 00:06:28,620
lakin yaln谋z bir 艧ey unutmay谋n.

139
00:06:28,620 --> 00:06:31,890
Biz h蓹mi艧蓹 istifad蓹 etm蓹k yoxdur,
Boolean tipli d蓹yi艧蓹nl蓹r zaman biz

140
00:06:31,890 --> 00:06:34,980
Boolean ifad蓹l蓹r il蓹 i艧l蓹yir.

141
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>> Boolean iki 蓹sas n枚v眉 var
biz il蓹 i艧l蓹m蓹k laz谋md谋r ifad蓹l蓹ri.

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00:06:37,890 --> 00:06:40,640
M蓹ntiqi operatorlar v蓹
relational operatorlar谋.

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00:06:40,640 --> 00:06:42,640
var dil
olduqca 蓹h蓹miyy蓹tli deyil.

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M蓹n onlar谋n qrupla艧d谋r谋lmas谋 al谋ram nec蓹 h蓹qiq蓹t蓹n var.

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V蓹 蓹lb蓹tt蓹 ki, laz谋md谋r, m蓹n tez hesab edir蓹m ki,
bir relational operator n蓹 d蓹rk

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蓹sas谋nda onlar n蓹 zaman biz
ikinci onlar haqq谋nda dan谋艧maq.

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Amma m眉tl蓹q narahat etmir
m眉dd蓹tli m蓹ntiqi operator ezberlemeye

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v蓹 ya relational operator.

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M蓹n yaln谋z qrup 眉莽眉n istifad蓹 edir蓹m
Onlara bir m蓹ntiqi 艧蓹kild蓹.

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>> Bel蓹 ki, bir n蓹z蓹r salaq
眉莽 m蓹ntiqi operatorlar

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biz bir g枚r蓹c蓹ksiniz ki,
CS50 proqramla艧d谋rma bit

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v蓹 proqramla艧d谋rma daha 莽ox, 眉mumiyy蓹tl蓹.

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M蓹ntiqi v蓹 do臒ru 蓹g蓹r
yaln谋z h蓹r iki operands do臒ru 蓹g蓹r.

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茝ks halda yalan.

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Ki, harada dem蓹kdir?

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Bel蓹 ki, m蓹n bir am ki, bildirin
M蓹n m蓹nim kodu qeyd

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iki d蓹yi艧蓹nl蓹rin, x v蓹 y.

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M蓹n q蓹rar q蓹bul etm蓹k ist蓹yirsinizs蓹,
m蓹nim kodu bir 艧ey etm蓹k

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x do臒ru v蓹 y do臒ruysa 蓹saslan谋r.

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M蓹n yaln谋z bunu ist蓹yir蓹m
onlar谋n h蓹r ikisi, do臒ru

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蓹ks halda m蓹n a艧a臒谋 getm蓹k ist蓹mir蓹m
yol m蓹n蓹 k枚m蓹k etm蓹k niyy蓹tind蓹 deyil, 莽眉nki.

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N蓹 dey蓹 bil蓹r蓹m x & 蓹g蓹r y edir.

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Bu m蓹ntiqi Boolean olacaq
ifad蓹 m眉qayis蓹 x v蓹 y

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v蓹 m眉蓹yy蓹n bir yol alaraq
Onlar谋n d蓹y蓹rl蓹ri n蓹 蓹saslan谋r.

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X do臒ru v蓹 蓹g蓹r Bel蓹 ki, y do臒ru
Burada bu h蓹qiq蓹t masa 蓹sas谋nda,

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yaln谋z bundan sonra biz bu yolu en蓹c蓹k.

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X, & y varsa.

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Bu, yaln谋z true-- v蓹 yaln谋z var
x do臒ru v蓹 y do臒ru do臒ru 蓹g蓹r.

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Ya bir yalan varsa,
biz h蓹qiq蓹t masa g枚rm蓹k kimi,

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sonra x v蓹 y h蓹m do臒ru deyil.

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Bel蓹 ki, x & y yalan.

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>> M蓹ntiqi v蓹 ya yaln谋z do臒ru
蓹n az谋 bir operand do臒ru 蓹g蓹r.

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茝ks halda yalan.

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Bel蓹 ki, m蓹ntiqi v蓹 t蓹l蓹b
x v蓹 y h蓹m do臒ru olmaq.

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M蓹ntiqi v蓹 ya do臒ru v蓹 ya y olmaq x t蓹l蓹b edir
do臒ru olmaq v蓹 ya h蓹r iki x v蓹 y do臒ru olmaq.

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Bel蓹 ki, daha, biz n枚v tapmaq
bir v蓹ziyy蓹td蓹 枚z眉m眉z眉

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biz bizim kodu niyy蓹tind蓹 oldu臒unuz,
v蓹 biz yol bir 莽蓹ng蓹l 莽atd谋.

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V蓹 biz bir a艧a臒谋 getm蓹k ist蓹yir蓹m
X眉susil蓹 yol x do臒ru 蓹g蓹r

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v蓹 ya y do臒ru deyil,
m眉tl蓹q, h蓹m d蓹 do臒ru.

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Amma b蓹lk蓹 蓹g蓹r, h蓹m d蓹 do臒ru.

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X do臒ru v蓹 蓹g蓹r bel蓹 y
do臒ru, biz yol a艧a臒谋 getm蓹k laz谋md谋r.

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x do臒rudur.

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Onlardan biri, sa臒 do臒rudur?

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X do臒ru v蓹 y do臒rudur.

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X do臒ru v蓹 y yalan varsa,
onlardan biri h蓹l蓹 do臒rudur.

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Bel蓹 ki, x v蓹 ya y h蓹l蓹 do臒rudur.

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X yalan v蓹 y do臒ru deyil,
Onlardan biri sa臒, h蓹l蓹 d蓹 do臒rudur?

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y bu halda, do臒rudur.

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Bel蓹 ki, x v蓹 ya y do臒ru oldu臒unu do臒ru.

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X yalan v蓹 y yalan yaln谋z
ki, yol a艧a臒谋 getm蓹k yoxdur,

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x, n蓹 d蓹 y n蓹 do臒rudur, 莽眉nki.

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陌ndi siz arad谋臒谋n谋z 蓹g蓹r
ekran indi

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v蓹 n蓹 merak
simvolu m蓹ntiqi v蓹 ya 眉莽眉n,

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艧aquli bar deyirl蓹r.

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V蓹 蓹g蓹r klaviatura baxaraq
bir d蓹qiq蓹 眉莽眉n, m蓹n indi bunu al谋ram kimi,

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yaln谋z yuxar谋da ad蓹t蓹n var
Enter d眉ym蓹sini 蓹n klaviatura,

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backslash eyni d眉ym蓹d蓹.

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O, h蓹m莽inin ad蓹t蓹n do臒ru
kvadrat m枚t蓹riz蓹d蓹 yan谋nda.

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Bel蓹 ki, 蓹sas ola bil蓹r ki,
ke莽mi艧d蓹 莽ox 莽ap yoxdur.

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Lakin, 茝g蓹r bunu edirsinizs蓹
m蓹ntiqi m眉qayis蓹,

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Biz bunu olacaq kimi
zaman谋 莽ox, bu

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faydal谋 olacaq
ki, 蓹sas tapmaq v蓹 istifad蓹 edin.

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Bel蓹 ki, eyni d眉ym蓹d蓹 ad蓹t蓹n var
backslash kimi yuxar谋da daxil edin.

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>> final m蓹ntiqi operator deyil.

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V蓹 NOT olduqca sad蓹 var.

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Onun operand d蓹y蓹ri inverts.

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X do臒ru, onda deyil x yalan.

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X yalan, onda deyil x do臒rudur.

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B蓹z蓹n bu r蓹mzi duyars谋n谋z
Bang ya nida kimi t蓹l蓹ff眉z

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v蓹 ya deyil.

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Bu olduqca 莽ox eyni 艧ey.

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Halda ki, dan谋艧谋q e艧itm蓹k v蓹
Siz o dem蓹kdir ki, n蓹 蓹min deyilik

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yaln谋z nida var
point, lakin b蓹z蓹n var

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bir ne莽蓹 f蓹rqli 艧eyl蓹r deyilir.

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B眉t眉n h眉quqlar, bel蓹 ki, davam edir
m蓹ntiqi operatorlar qay臒谋.

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Bel蓹 ki, haqq谋nda dan谋艧maq ed蓹k
relational operatorlar谋.

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Yen蓹 bu il蓹 tan谋艧 de臒ilseniz
geri grade m蓹kt蓹b hesab,

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Siz y蓹qin ki, tan谋艧 oldu臒unuzda
il蓹 nec蓹 bu i艧 art谋q.

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Bekledi臒iniz kimi bu d蓹qiq davranmaq.

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Bel蓹 ki, az bu, do臒ru 莽ox
M蓹s蓹l蓹n, x y az olduqda.

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X 4 v蓹 蓹g蓹r bel蓹, y
6 x y azd谋r.

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D眉zd眉r.

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Daha az v蓹 ya b蓹rab蓹r
pretty eyni i艧l蓹yir.

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X 4, v蓹 y sonra, 4 蓹g蓹r
x y daha az v蓹 ya b蓹rab蓹rdir.

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Daha 莽ox. x y daha b枚y眉kd眉r.

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V蓹 ya daha 莽ox, x b蓹rab蓹r
y v蓹 ya daha 莽ox b蓹rab蓹rdir.

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Do臒ru, onda siz laz谋md谋r
ifad蓹 ke莽ir,

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v蓹 a艧a臒谋 getm蓹k laz谋md谋r
yolda ki yol.

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X y daha b枚y眉k olduqda, bir varsa,
v蓹 x, 蓹slind蓹, y daha b枚y眉k deyil,

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Siz n蓹 etm蓹k laz谋md谋r
艧蓹rti m枚vzu.

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>> Biz yoxdur ki, qeyd
az bir xarakter

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v蓹 ya b蓹rab蓹r, kimi ola bil蓹r
riyaziyyat d蓹rslikl蓹rd蓹n il蓹 tan谋艧.

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Bel蓹 ki, biz, r蓹mzi daha az
b蓹rab蓹r i艧ar蓹si sonra.

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Y蓹ni biz t蓹msil nec蓹
莽ox v蓹 ya b蓹rab蓹r daha az.

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V蓹 eyni, biz bunu etm蓹k
v蓹 ya daha 莽ox b蓹rab蓹r 眉莽眉n.

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>> son iki relational
vacibdir operatorlar谋

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b蓹rab蓹rlik v蓹 b蓹rab蓹rsizlik 眉莽眉n test.

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Bel蓹 ki, x b蓹rab蓹rdir 蓹g蓹r, y b蓹rab蓹rdir do臒rudur
X v蓹 Y d蓹y蓹ri eyni 蓹g蓹r.

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X 10 v蓹 y sonra 10 茝g蓹r
x b蓹rab蓹rdir y do臒ru b蓹rab蓹rdir.

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X 10 v蓹 y x 11, 蓹g蓹r
y do臒ru deyil b蓹rab蓹rdir b蓹rab蓹rdir.

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Biz d蓹 istifad蓹 b蓹rab蓹rsizlik 眉莽眉n test ed蓹 bil蓹rsiniz
nida point v蓹 ya bang v蓹 ya,

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yenid蓹n.

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茝g蓹r x 蓹g蓹r, y b蓹rab蓹r deyil
ki, biz burada istifad蓹 etdiyiniz test var

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biz getm蓹k 眉莽眉n yax艧谋 olard谋.

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Bel蓹 ki, x b蓹rab蓹r deyil
y, biz yol a艧a臒谋 getm蓹k laz谋md谋r.

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>> Burada h蓹qiq蓹t蓹n ehtiyatl谋 olun.

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Bu, h蓹qiq蓹t蓹n 眉mumi mistake-- v蓹
M蓹n, 蓹lb蓹tt蓹, q蓹bul olduqca 莽ox zaman

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M蓹n started-- 蓹ld蓹 edilib
t蓹sad眉f蓹n s蓹hv

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tap艧谋r谋q operator, bir b蓹rab蓹r,
b蓹rab蓹rlik m眉qayis蓹 operator 眉莽眉n,

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ikiqat b蓹rab蓹rdir.

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00:11:52,582 --> 00:11:54,540
B蓹zi q蓹rib蓹 s蓹b蓹b olacaq
Sizin kodu davran谋艧,

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v蓹 ad蓹t蓹n compiler ed蓹c蓹k
Siz c蓹hd zaman bu bar蓹d蓹 siz蓹 x蓹b蓹rdarl谋q

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v蓹 kodu t蓹rtib, lakin b蓹z蓹n
Siz onu x蓹b蓹r莽i ed蓹 bil蓹r.

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Bu, m眉tl蓹q yax艧谋 bir 艧ey deyil
Siz onu x蓹b蓹r莽i ki, baxmayaraq ki.

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Yaln谋z bunu 蓹g蓹r
bir b蓹rab蓹rsizlik test,

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Siz iki yoxlan谋lmas谋 edirsinizs蓹
m眉xt蓹lif d蓹yi艧蓹nl蓹rin eyni d蓹y蓹ri var

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onlar谋n i莽蓹risind蓹, istifad蓹 etm蓹k 眉莽眉n 蓹min olun
b蓹rab蓹r deyil, t蓹k b蓹rab蓹rdir.

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V蓹 sizin proqram olacaq
Siz niyy蓹tind蓹 davran谋艧 var.

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00:12:15,450 --> 00:12:18,400
M蓹n Doug Lloyd Ben v蓹 bu CS50 edir.

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