問題描述

我有一道考試題要復(fù)習(xí)，題目是 4 分.

I have an exam question I am revising for and the question is for 4 marks.

在 java 中，我們可以將 int 分配給 double 或 float".這會(huì)丟失信息嗎?為什么?

"In java we can assign a int to a double or a float". Will this ever lose information and why?

我之所以這么說是因?yàn)檎麛?shù)通常是固定長(zhǎng)度或大小的——存儲(chǔ)數(shù)據(jù)的精度是有限的，而以浮點(diǎn)數(shù)存儲(chǔ)信息可以是無限的，本質(zhì)上我們會(huì)因此而丟失信息

I have put that because ints are normally of fixed length or size - the precision for storing data is finite, where storing information in floating point can be infinite, essentially we lose information because of this

現(xiàn)在我有點(diǎn)粗略地知道我是否在這里擊中了正確的區(qū)域.我很確定它會(huì)失去精確度，但我無法確切說明原因.請(qǐng)問我能得到一些幫助嗎?

Now I am a little sketchy as to whether or not I am hitting the right areas here. I very sure it will lose precision but I can't exactly put my finger on why. Can I get some help, please?

推薦答案

在 Java 中 Integer 使用 32 位來表示它的值.

In Java Integer uses 32 bits to represent its value.

在 Java 中，F(xiàn)LOAT 使用 23 位尾數(shù)，因此大于 2^23 的整數(shù)將被截?cái)嗥渥畹陀行?例如 33554435(或 0x200003)將被截?cái)酁榇蠹s 33554432 +/- 4

In Java a FLOAT uses a 23 bit mantissa, so integers greater than 2^23 will have their least significant bits truncated. For example 33554435 (or 0x200003) will be truncated to around 33554432 +/- 4

在 Java 中，DOUBLE 使用 52 位尾數(shù)，因此能夠表示 32 位整數(shù)而不會(huì)丟失數(shù)據(jù).

In Java a DOUBLE uses a 52 bit mantissa, so will be able to represent a 32bit integer without lost of data.

另請(qǐng)參閱維基百科上的浮點(diǎn)"

See also "Floating Point" on wikipedia

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