方法中的局部变量如何计算Java字节码

Question

I have been learning about java byte-code recently, and i have been understanding most of it, but i am confused about how the local variable count for example is counted. I thought it would just be the total of the local variables, but this code generates 1 local variable when looking through the bytecode

public int testFail()
{
    return 1;
}

But i thought it should be zero local variables because no local variable are defined.

Additionally this method also generates one local variable but it has more local variables than the previous example.

Finally this method

public static int testFail(int a, int b)
{
    return a+b;
}

gnerates two local variable in the bytecode.

public static int testFail(int a)
{
    return a;
}

Answer 1

Non-static methods use a local variable slot for this . Another complication is that long s and double s count as 2 each. Also, depending on your compiler and settings, you may not see a one-to-one mapping between local variables in the source code and local variables in the byte code. For example, if debug information is left out, the compiler may eliminate unnecessary local variables.

Edit:

I just remembered: compilers may also re-use local variable slots. For example, given this code:

public static void test() {
    for(int i = 0; i < 100; i++) {
        ...
    }
    for(int j = 0; j < 100; j++) {
    }
}

the same slot can be used for i and j because their scopes don't overlap.

Answer 2

第一个具有局部变量的原因是因为它是一种非静态方法，所以存在一个隐式的this参数。