关于cudaError_t的返回值为77，怎么解决？

[帅气的小王子]

代码如下所示，

	
	cudaMalloc((unsigned char **)&dev_bdecoded, CodeLength * sizeof(unsigned char));
	cudaMalloc((int **)&dev_jc, (n + 1) * sizeof(int));
	cudaMalloc((double **)&dev_LLR_extrinsic, nzmax * sizeof(double));
	cudaMalloc((double **)&dev_LLR_intrinsic, CodeLength * sizeof(double));

	cudaMemcpy(dev_bdecoded, bdecoded, CodeLength * sizeof(unsigned char), cudaMemcpyHostToDevice);
	cudaMemcpy(dev_jc, jc, (n + 1) * sizeof(int), cudaMemcpyHostToDevice);
	cudaMemcpy(dev_LLR_intrinsic, LLR_intrinsic, CodeLength * sizeof(double), cudaMemcpyHostToDevice);

        threadsPerBlock = 256;
	blocksPerGrid = (n + threadsPerBlock - 1) / threadsPerBlock;
	Kernel1<<<blocksPerGrid,threadsPerBlock>>>(dev_bdecoded, dev_jc, dev_LLR_extrinsic, dev_LLR_intrinsic);
	cudaStatus = cudaDeviceSynchronize();
	if (cudaStatus != cudaSuccess) {
		fprintf(stderr, "cudaDeviceSynchronize returned error code %d after launching addKernel!\n", cudaStatus);
	}

其中bdecoded,jc,LLR_intrinsic是作为参数传递进来的指针。
内核函数如下

__global__ void Kernel1(unsigned char *dev_bdecoded, int *dev_jc, double *dev_LLR_extrinsic, double *dev_LLR_intrinsic)
{
	int i = blockDim.x * blockIdx.x + threadIdx.x;
	int j;
	dev_bdecoded[i] = 0;
	for (j = dev_jc[i]; j < dev_jc[i + 1]; j++)
		dev_LLR_extrinsic[j] = dev_LLR_intrinsic[i];
}

在上面cudaStatus的返回值为77，

    /**
     * The device encountered a load or store instruction on an invalid memory address.
     * The context cannot be used, so it must be destroyed (and a new one should be created).
     * All existing device memory allocations from this context are invalid
     * and must be reconstructed if the program is to continue using CUDA.
     */
    cudaErrorIllegalAddress               =     77,

去查了下解释说是内存问题，如上面所示，我想知道该如何解决这个问题？

[帅气的小王子]

引用 1 楼 u014693181 的回复:

cudaMalloc( (void**)&d_Mat_A , size ) ; 应该是 这样的形式。内存分配出错。

改过之后依然不行。

[帅气的小王子]

引用 2 楼 u010333076 的回复:

[quote=引用 1 楼 u014693181 的回复:] cudaMalloc( (void**)&d_Mat_A , size ) ; 应该是 这样的形式。内存分配出错。

改过之后依然不行。[/quote]

cudaMalloc((void **)&dev_jc, (n + 1) * sizeof(int));

这样子改的

[YCMyTot]

cudaMalloc( (void**)&d_Mat_A , size ) ; 应该是 这样的形式。内存分配出错。

[帅气的小王子]

找到问题了，内核函数中dev_jc[i]，i的取值超过了dev_jc的范围，数组越界了。

[wishkind]

还没有开始cuda编程，先看看