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

2
00:00:00,310 --> 00:00:01,750
>> DAVID Malan: Ejja issa blow moħħok.

3
00:00:01,750 --> 00:00:06,500
Jirriżulta fid-dinja reali 1 maqsuma
b'10 huwa tabilħaqq 1/10, jew 0.1.

4
00:00:06,500 --> 00:00:10,370
Iżda fil-kompjuters li huma biss ikollhom finite
numru ta 'bits li biex

5
00:00:10,370 --> 00:00:14,290
jirrappreżentaw numri, inti ma tistax dejjem
jirrappreżentaw numri bħall 1/10 ma

6
00:00:14,290 --> 00:00:15,500
preċiżjoni perfetta.

7
00:00:15,500 --> 00:00:18,640
Fi kliem ieħor, kompjuters kultant ikollhom
biex jagħmlu sejħiet sentenza u ma

8
00:00:18,640 --> 00:00:22,740
neċessarjament jirrappreżentaw in-numru inti
jridu preċiżament kemm għandek il-ħsieb.

9
00:00:22,740 --> 00:00:27,020
>> Per eżempju, ejja ngħidu li mmur lura fil
dan il-programm u jibdlu l-0.1 u,

10
00:00:27,020 --> 00:00:32,073
oh, 0.28, u b'hekk jindika li
Nixtieq printf li printf li

11
00:00:32,073 --> 00:00:34,350
28 postijiet ta 'preċiżjoni.

12
00:00:34,350 --> 00:00:39,330
Ejja issa issalva u jikkumpilaw-programm,
din id-darba ma 'make floats2.

13
00:00:39,330 --> 00:00:41,910
Run ma floats2 dot slash.

14
00:00:41,910 --> 00:00:49,980
U, qalb Alla, din id-darba nara ma 0.1,
iżda 0.10000000, li huwa pjuttost

15
00:00:49,980 --> 00:00:51,070
tajba s'issa.

16
00:00:51,070 --> 00:00:57,830
Iżda mbagħad, 14901161193847656250.

17
00:00:57,830 --> 00:00:58,880
>> Well, x'inhu għaddej?

18
00:00:58,880 --> 00:01:02,280
Ukoll, jirriżulta li float huwa
tipikament maħżuna ġewwa ta 'kompjuter

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

20
00:01:03,500 --> 00:01:07,340
32 huwa ovvjament numru finit, li
jimplika li inti tista 'tirrappreżenta biss

21
00:01:07,340 --> 00:01:11,050
32 bits numru finit
ta 'floating valuri tal-punti.

22
00:01:11,050 --> 00:01:14,980
Sfortunatament, dan ifisser li l-
kompjuter ma jistax jirrappreżenta possibbli

23
00:01:14,980 --> 00:01:18,110
b'punt li jvarja numri, jew numri reali,
li jeżistu fid-dinja,

24
00:01:18,110 --> 00:01:19,980
minħabba li biss tant bits.

25
00:01:19,980 --> 00:01:23,940
>> U għalhekk dak li l-kompjuter apparentement
sar f'dan il-każ jirrappreżentaw 1/10 sa

26
00:01:23,940 --> 00:01:26,880
l Floating eqreb possibbli
valur tal-punt li jista '.

27
00:01:26,880 --> 00:01:31,050
Iżda jekk inħarsu, kif għandna hawnhekk, sa 28
postijiet deċimali, nibdew biex tara li

28
00:01:31,050 --> 00:01:31,970
impreċiżjoni.

29
00:01:31,970 --> 00:01:34,480
Allura dan huwa problema ma
ebda soluzzjoni perfetta.

30
00:01:34,480 --> 00:01:38,060
Nistgħu nużaw doppja minflok ta 'float,
li għandhom tendenza li jużaw 64 bits bħala

31
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kuntrarju 32.

32
00:01:39,410 --> 00:01:42,290
Iżda naturalment, 64 huwa wkoll finit,
għalhekk il-problema se

33
00:01:42,290 --> 00:01:43,630
jibqgħu anke ma jirdoppja.

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