parseInt 在 rhino 中返回 double

Question

我有这个 JavaScript 文件（Rhino 1.7R4）。

importPackage(java.io);
importPackage(java.lang);
importPackage(java.util);

var reader = new BufferedReader( new InputStreamReader(System['in']) );

var line = reader.readLine();
var tok = new java.util.StringTokenizer(line);

var A = Integer.parseInt(tok.nextToken());
var B = Integer.parseInt(tok.nextToken());
var C = Integer.parseInt(tok.nextToken());

// System.out.printf( "A=%d, B=%d, C=%d\n", A, B, C );
System.out.printf( "A=%f, B=%f, C=%f\n", A, B, C );

当我首先取消评论时printf- 我得到

A=Exception in thread "main" org.mozilla.javascript.WrappedException: Wrapped java.util.IllegalFormatConversionException: d != java.lang.Double
        at org.mozilla.javascript.Context.throwAsScriptRuntimeEx(Context.java:1754)
        at org.mozilla.javascript.MemberBox.invoke(MemberBox.java:148)
        at org.mozilla.javascript.NativeJavaMethod.call(NativeJavaMethod.java:225)
        at org.mozilla.javascript.optimizer.OptRuntime.callN(OptRuntime.java:52)
        at test._c_script_0(Unknown Source)
        at test.call(Unknown Source)
        at org.mozilla.javascript.ContextFactory.doTopCall(ContextFactory.java:394)
        at org.mozilla.javascript.ScriptRuntime.doTopCall(ScriptRuntime.java:3091)
        at test.call(Unknown Source)
        at test.exec(Unknown Source)
        at org.mozilla.javascript.optimizer.OptRuntime$1.run(OptRuntime.java:218)
        at org.mozilla.javascript.Context.call(Context.java:489)
        at org.mozilla.javascript.ContextFactory.call(ContextFactory.java:504)
        at org.mozilla.javascript.optimizer.OptRuntime.main(OptRuntime.java:206)
        at test.main(Unknown Source)
Caused by: java.util.IllegalFormatConversionException: d != java.lang.Double
        at java.util.Formatter$FormatSpecifier.failConversion(Unknown Source)
        at java.util.Formatter$FormatSpecifier.printInteger(Unknown Source)
        at java.util.Formatter$FormatSpecifier.print(Unknown Source)
        at java.util.Formatter.format(Unknown Source)
        at java.io.PrintStream.format(Unknown Source)
        at java.io.PrintStream.printf(Unknown Source)
        at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
        at sun.reflect.NativeMethodAccessorImpl.invoke(Unknown Source)
        at sun.reflect.DelegatingMethodAccessorImpl.invoke(Unknown Source)
        at java.lang.reflect.Method.invoke(Unknown Source)
        at org.mozilla.javascript.MemberBox.invoke(MemberBox.java:126)
        ... 13 more

Answer 1

这是 JavaScript 类型系统的结果。JavaScript 中的所有数字都是双精度数，因此虽然 `Integer.parseInt("4.5") 返回 4，但 JavaScript 需要将其转换为 4.0。

幸运的是，64 位双精度数有足够的精度来精确表示任何 32 位 int，因此在 JavaScript 方面，您通常不需要担心它。但值后面的 Java 对象始终是 a Double，并且会在需要时自动转换。但由于System.out.printf采用类型参数，Object因此必须按原样传递。