java 将 C++ long 类型转换为 JNI jlong
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Casting a C++ long type to a JNI jlong
提问by lost_bits1110
I am using JNI to pass data between C++ and Java. I need to pass a 'long' type, and am doing so using something like:
我正在使用 JNI 在 C++ 和 Java 之间传递数据。我需要传递一个“长”类型，并且正在使用以下内容：
long myLongVal = 100;
jlong val = (jlong)myLongVal;
CallStaticVoidMethod(myClass, "(J)V", (jvalue*)val);
However in Java, when the 'long' parameter is retrieved, it gets retrieved as some very large negative number. What am I doing wrong?
然而，在 Java 中，当检索“long”参数时，它会被检索为某个非常大的负数。我究竟做错了什么？
回答by Seva Alekseyev
When you pass a jlong (which is 64 bit) as a pointer (which is, most likely, 32-bit) you necessarily lose data. I'm not sure what's the convention, but try either this:
当您将 jlong（64 位）作为指针（最有可能是 32 位）传递时，您必然会丢失数据。我不确定约定是什么，但请尝试以下任一方法：
CallStaticVoidMethodA(myClass, "(J)V", (jvalue*)&val); //Note address-of!
or this:
或这个：
CallStaticVoidMethod(myClass, "(J)V", val);
It's ...Amethods that take a jvalue array, the no-postfix methods take C equivalents to scalar Java types.
它...A是采用 jvalue 数组的方法，无后缀方法采用 C 等效于标量 Java 类型。
The first snippet is somewhat unsafe; a better, if more verbose, alternative would be:
第一个片段有点不安全；更好的，如果更详细的话，替代方法是：
jvalue jv;
jv.j = val;
CallStaticVoidMethodA(myClass, "(J)V", &jv);
On some exotic CPU archtectures, the alignment requirements for jlongvariables and jvalueunions might be different. When you declare a union explicitly, the compiler takes care of that.
在一些奇特的 CPU 架构上，jlong变量和jvalue联合的对齐要求可能不同。当您显式声明联合时，编译器会处理它。
Also note that C++ longdatatype is often 32-bit. jlong is 64 bits, on 32-bit platforms the nonstandard C equivalent is long longor __int64.
另请注意，C++long数据类型通常为 32 位。jlong 是 64 位，在 32 位平台上，非标准 C 等效项是long longor __int64。
回答by Puppy
CallStaticVoidMethod(myClass, "(J)V", (jvalue*)val);
This is undefined behaviour. You are casting an integer to be a pointer. It is not a pointer. You need, at the very least, to pass the address. This code would on most platforms instantly crash.
这是未定义的行为。您正在将整数转换为指针。它不是指针。您至少需要传递address。此代码在大多数平台上会立即崩溃。
