Question

我对于找到无符号幅度和2恭维的最小位数感到有些困惑。到目前为止，这是我的理由：例如，

a) 243 decimal
Since 2^8 = 256, Unsigned and 2's compliment would both need a minimum 8 bits.
b) -56 decimal
This is impossible for unsigned.
2^6 = 64. One more bit is needed to show it is negative, so minimum 7 bits.

我的推理是否正确？

Answer 1

需要＆＃34;比特＆＃34;对于无符号，只是最重要的位（+1，取决于MSB的定义），对于2的补码，你可以只是否定该值并减去1使其为正，然后为符号标志添加另一位

Exception in thread "main" com.github.axet.wget.info.ex.DownloadRetry: no video with required quality found
    at com.github.axet.vget.info.VGetParser.getVideo(VGetParser.java:38)
    at com.github.axet.vget.info.YouTubeParser.extract(YouTubeParser.java:388)
    at com.github.axet.vget.info.VGetParser.extract(VGetParser.java:15)
    at com.github.axet.vget.info.VideoInfo.extract(VideoInfo.java:100)
    at com.github.axet.vget.VGet.download(VGet.java:224)
    at com.github.axet.vget.VGet.download(VGet.java:68)
    at Test.main(Test.java:10)
    at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:57)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
    at java.lang.reflect.Method.invoke(Method.java:606)
    at com.intellij.rt.execution.application.AppMain.main(AppMain.java:140)

这基于你对问题的定义，至少是这样。对我而言，似乎+243需要9位用于2的补码，因为符号位的0仍然是相关的。

如何找到无符号幅度和2恭维的最小位数？

1 个答案: