1
00:00:00,000 --> 00:00:00,310

2
00:00:00,310 --> 00:00:01,750
>> DAVID Malan: Hebu sasa pigo akili yako.

3
00:00:01,750 --> 00:00:06,500
Ni zamu nje katika ulimwengu halisi 1 kugawanywa
na 10 ni kweli 1/10, au 0.1.

4
00:00:06,500 --> 00:00:10,370
Lakini katika kompyuta tu na finite
idadi ya bits ambayo kwa

5
00:00:10,370 --> 00:00:14,290
kuwakilisha idadi, unaweza daima
kuwakilisha idadi kama 1/10 na

6
00:00:14,290 --> 00:00:15,500
usahihi kamilifu.

7
00:00:15,500 --> 00:00:18,640
Kwa maneno mengine, kompyuta wakati mwingine kuwa
kupiga simu hukumu na si

8
00:00:18,640 --> 00:00:22,740
lazima kuwakilisha idadi wewe
wanataka kama just kama unakusudia.

9
00:00:22,740 --> 00:00:27,020
>> Kwa mfano, tuseme mimi kurudi nyuma katika
mpango huu na mabadiliko ya 0.1 kwa,

10
00:00:27,020 --> 00:00:32,073
oh, 0.28, na hivyo kuonyesha kwamba
Ningependa printf kwa printf kwa

11
00:00:32,073 --> 00:00:34,350
28 maeneo ya usahihi.

12
00:00:34,350 --> 00:00:39,330
Hebu sasa ila na mpango wa kukusanya,
wakati huu na kufanya floats2.

13
00:00:39,330 --> 00:00:41,910
Kuendesha na dot slash floats2.

14
00:00:41,910 --> 00:00:49,980
Na, mpendwa wa Mungu, wakati huu naona si 0.1,
lakini 0.10000000, ambayo ni pretty

15
00:00:49,980 --> 00:00:51,070
nzuri hadi sasa.

16
00:00:51,070 --> 00:00:57,830
Lakini basi, 14901161193847656250.

17
00:00:57,830 --> 00:00:58,880
>> Naam, ni nini kinaendelea?

18
00:00:58,880 --> 00:01:02,280
Naam, zinageuka kuwa kuelea ni
kawaida kuhifadhiwa ndani ya kompyuta

19
00:01:02,280 --> 00:01:03,500
na 32 bits.

20
00:01:03,500 --> 00:01:07,340
32 ni wazi finite idadi, ambayo
ina maana kwamba unaweza tu kuwakilisha

21
00:01:07,340 --> 00:01:11,050
na 32 bits idadi mahususi
ya yaliyo maadili ya uhakika.

22
00:01:11,050 --> 00:01:14,980
Kwa bahati mbaya, hiyo ina maana kwamba
kompyuta hawezi kuwakilisha wote iwezekanavyo

23
00:01:14,980 --> 00:01:18,110
idadi yaliyo uhakika, au idadi halisi,
ambazo zipo katika dunia,

24
00:01:18,110 --> 00:01:19,980
kwa sababu tu ina bits wengi.

25
00:01:19,980 --> 00:01:23,940
>> Na hivyo kile kompyuta ni inaonekana
kufanyika katika kesi hii ni kuwakilisha 1/10 hadi

26
00:01:23,940 --> 00:01:26,880
floating karibu iwezekanavyo
thamani uhakika kwamba wanaweza.

27
00:01:26,880 --> 00:01:31,050
Lakini kama sisi kuangalia, kama sisi hapa, 28
maeneo decimal, sisi kuanza kuona kwamba

28
00:01:31,050 --> 00:01:31,970
kutokuwa sahihi.

29
00:01:31,970 --> 00:01:34,480
Hivyo hili ni tatizo kwa
hakuna suluhisho kamili.

30
00:01:34,480 --> 00:01:38,060
Tunaweza kutumia mara mbili badala ya kuelea,
ambao huelekea kutumia bits 64 kama

31
00:01:38,060 --> 00:01:39,410
kinyume na 32.

32
00:01:39,410 --> 00:01:42,290
Lakini bila shaka, 64 pia ni finite,
hivyo tatizo mapenzi

33
00:01:42,290 --> 00:01:43,630
kubaki hata kwa mara mbili.

34
00:01:43,630 --> 00:01:46,323