可能的重复：
Java 整数除法，如何产生双精度数？ https://stackoverflow.com/questions/3144610/java-integer-division-how-do-you-produce-a-double
double wang = 3 / 2;
Log.v("TEST", "Wang: " + Double.toString(wang));
Logcat 输出...
07-04 09:01:03.908: VERBOSE/TEST(28432): Wang: 1.0
我确信这个问题有一个明显的答案，可能我只是因为整夜编码而感到疲倦，但这让我难住了。
In many languages, Java being one of them, the way you write a number in an expression decides what type it gets. In Java, a few of the common number types behave like this1:
// In these cases the specs are obviously redundant, since all values will be
// cast correctly anyway, but it was the easiest way to show how to get to the
// different data types :P
int i = 1;
long l = 1L;
float f = 1.0f; // I believe the f and d for float and double are optional, but
double d = 1.0d; // I wouldn't bet on what the default is if they're omitted...
因此，当你声明3 / 2，你确实是在说(the integer 3) / (the integer 2)。 Java 执行除法，结果为1 (i.e. the integer 1...) 因为这是 3 和 2 除以整数的结果。最后，the integer 1被投射到the double 1.0d它存储在您的变量中。
要解决这个问题，您应该（正如许多其他人所建议的那样）计算
(the double 3) / (the double 2)
或者，在 Java 语法中，
double wang = 3.0 / 2.0;
1 Source: The Java Tutorial http://download.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html from Oracle
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