JAVA BufferOverflowException，同时保持缓冲区容量

Question

所以我有一个方法，从一个ByteBuffer创建一个IntBuffer:

public static IntBuffer directIntBuffer(int[]buffer){
        ByteBuffer bb = ByteBuffer.allocateDirect(4*buffer.length);
        bb.order(ByteOrder.nativeOrder());
        IntBuffer ib = bb.asIntBuffer();
        ib.put(buffer).flip();
        return ib;
    }

在另一个类的构造函数中，我初始化这些intbuffer之一，并填充它嵌套的for循环:

EDIT:所以这是我的完整构造函数。我填充FloatBuffer FBVertices与实际顶点坐标和IntBuffer IBVerticesIndex与索引。后来，我用GL11. gldrawelements (GL11)绘制了这些。GL_TRIANGLES IBVerticesIndex);

public sphereHelper(float fRadius, int iSlices, int iStacks)
{  
    FBVertices = GLDrawHelper.directFloatBuffer(new float[iSlices*iStacks*4*3]); 
    FBVertices.clear();
    IBVerticesIndex = GLDrawHelper.directIntBuffer(new int[iSlices*iStacks*4]);
    IBVerticesIndex.clear();
    int index=0;


    float verticalDegreePerStack= (float)Math.PI/iStacks; //1*Pi because we only want the height of the sphere once (from -PI/2 to PI/2)
    float horizontalDegreePerSlice = 2.0f* (float)Math.PI/iSlices; //2*Pi because of unit circle/polar coordinates going once around the sphere
    for (int k=0; k< iStacks; k++)
    {
        //overall vertical angle
        float rho= (float)k*verticalDegreePerStack;
        //sinus of that angle
        float srho =(float) (Math.sin(rho));
        float crho = (float) (Math.cos(rho));
        //sinus and cosinus of vertical angle+ angle of 1 stack
        float srhodrho= (float) (Math.sin(rho + verticalDegreePerStack));
        float crhodrho = (float) (Math.cos(rho + verticalDegreePerStack));
        for (int j=0; j<iSlices;j++)
        {
            //overall angle along the horizontal border (0 degree at full 2*PI circle)
            float theta = (j==iSlices)? 0.0f : j*horizontalDegreePerSlice;
            //sinus and cosinus
            float stheta = (float)(-Math.sin(theta));
            float ctheta = (float)(Math.cos(theta));
            //coordinates of first vertex of current triangle
            float x=stheta*srho;
            float z=ctheta*srho;
            float y=crho;

            //put coordinates of first vertex into buffer
            FBVertices.put(new float[]{x*fRadius,y*fRadius,z*fRadius});
            //put index of first vertex of polygon in index array, increase index counter by one
            if (IBVerticesIndex.position()<IBVerticesIndex.capacity())
            {IBVerticesIndex.put(index);}
            index++;
            //second vertex
            x = stheta*srhodrho;
            z = ctheta*srhodrho;
            y = crhodrho;
            //put these coordinates into Buffer
            FBVertices.put(new float[] {x*fRadius,y*fRadius,z*fRadius});

             if (IBVerticesIndex.position()+2<IBVerticesIndex.capacity())
            {
            IBVerticesIndex.put(new int[] {index,index+1,index+2});}
            index++;
            System.out.println(IBVerticesIndex.position()+"/"+IBVerticesIndex.capacity());
        }
        IBVerticesIndex.flip();
        FBVertices.flip();
       }
    }

我得到一个BufferOverflowException，在这个输出:

4/400 8/400 12/400 16/400 20/400 24/400 28/400 32/400 36/400 40/400

线程"main"出现异常java.nio.BufferOverflowException

所以它在缓冲区容量的十分之一溢出?我在一开始就清除了缓冲区，所以位置应该被重置。

感谢您的输入!

Answer 1

EDIT: +1对Arjan对原始问题的评论。我把Don的代码复制到我的机器上，它在Java 8上工作，也得到400/400作为输出。您是否在注释代码的其他地方设置了limit ?

你读过Buffer超类是如何工作的吗?具体地说，这个类有一个limit属性，这个属性没有一个记录良好的初始值。写入超过limit值会抛出一个BufferOverflowException

在我看来，你想使用这个IntBuffer像一个原始数组，在这种情况下，在你的初始化例程中，你需要在调用flip()后将limit设置为capacity，根据文档将limit重置为mark。