問題描述

我知道 Java 不允許無(wú)符號(hào)類型，所以我想知道它是如何將整數(shù)轉(zhuǎn)換為字節(jié)的.假設(shè)我有一個(gè)值為 255 的整數(shù) a 并將整數(shù)轉(zhuǎn)換為一個(gè)字節(jié).該值是否以字節(jié) 11111111 表示?換句話說(shuō)，該值是更多地被視為一個(gè)有符號(hào)的 8 位整數(shù)，還是直接復(fù)制整數(shù)的最后 8 位?

I know Java doesn't allow unsigned types, so I was wondering how it casts an integer to a byte. Say I have an integer a with a value of 255 and I cast the integer to a byte. Is the value represented in the byte 11111111? In other words, is the value treated more as a signed 8 bit integer, or does it just directly copy the last 8 bits of the integer?

推薦答案

這叫做 縮小基元轉(zhuǎn)換.根據(jù)規(guī)范:

有符號(hào)整數(shù)到整數(shù)類型的窄化轉(zhuǎn)換T 只丟棄除n 個(gè)最低位之外的所有位，其中n 是用于表示類型 T 的位數(shù).除了可能丟失有關(guān)數(shù)值大小的信息外，這還可能導(dǎo)致結(jié)果值的符號(hào)與輸入值的符號(hào)不同.

A narrowing conversion of a signed integer to an integral type T simply discards all but the n lowest order bits, where n is the number of bits used to represent type T. In addition to a possible loss of information about the magnitude of the numeric value, this may cause the sign of the resulting value to differ from the sign of the input value.

所以這是您列出的第二個(gè)選項(xiàng)(直接復(fù)制最后 8 位).

So it's the second option you listed (directly copying the last 8 bits).

我不確定你是否知道有符號(hào)整數(shù)值是如何表示的，所以為了安全起見，我會(huì)指出字節(jié)值 1111 1111 在 二的補(bǔ)碼 系統(tǒng)(Java 使用).

I am unsure from your question whether or not you are aware of how signed integral values are represented, so just to be safe I'll point out that the byte value 1111 1111 is equal to -1 in the two's complement system (which Java uses).

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