Error in cudaMemcpyToSymbol using CUDA 5 - c

Question

The Problem
I have prepared one sample CUDA code using the constant memory. I can run this in cuda 4.2 successfully but I get "invalid device symbol" when I compile using the CUDA 5.
I have attached the sample code here.
The Code
#include <iostream>
#include <stdio.h>
#include <cuda_runtime.h>
#include <cuda.h>
struct CParameter
{
int A;
float B;
float C;
float D;
};
__constant__ CParameter * CONSTANT_PARAMETER;
#define PARAMETER "CONSTANT_PARAMETER"
bool ERROR_CHECK(cudaError_t Status)
{
if(Status != cudaSuccess)
{
printf(cudaGetErrorString(Status));
return false;
}
return true;
}
// Kernel that executes on the CUDA device
__global__ void square_array(float *a, int N)
{
int idx = blockIdx.x * blockDim.x + threadIdx.x;
if (idx<N)
{
a[idx] = CONSTANT_PARAMETER->A * a[idx];
}
}
////Main Function/////
int main(void)
{
/////Variable Definition
const int N = 10;
size_t size = N * sizeof(float);
cudaError_t Status = cudaSuccess;
CParameter * m_dParameter;
CParameter * m_hParameter;
float * m_D;
float * m_H;
//Memory Allocation Host
m_hParameter = new CParameter;
m_H = new float[N];
//Memory Allocation Device
cudaMalloc((void **) &m_D, size);
cudaMalloc((void**)&m_dParameter,sizeof(CParameter));
////Data Initialization
for (int i=0; i<N; i++)
m_H[i] = (float)i;
m_hParameter->A = 5;
m_hParameter->B = 3;
m_hParameter->C = 98;
m_hParameter->D = 100;
//Memory Copy from Host To Device
Status = cudaMemcpy(m_D, m_H, size, cudaMemcpyHostToDevice);
ERROR_CHECK(Status);
Status = cudaMemcpy(m_dParameter,m_hParameter,sizeof(CParameter),cudaMemcpyHostToDevice);
ERROR_CHECK(Status);
Status = cudaMemcpyToSymbol(PARAMETER, &m_dParameter, sizeof(m_dParameter));
ERROR_CHECK(Status);
// Do calculation on device:
int block_size = 4;
int n_blocks = N/block_size + (N%block_size == 0 ? 0:1);
square_array <<<n_blocks, block_size>>>(m_D,N);
// Retrieve result from device and store it in host array
cudaMemcpy(m_H, m_D, sizeof(float)*N, cudaMemcpyDeviceToHost);
// Print results
for (int i=0; i<N; i++)
printf("%d %f\n", i, m_H[i]);
// Cleanup
free(m_H);
free(m_hParameter);
cudaFree(m_dParameter);
cudaFree(m_D);
return 0;
}
I have tried WINDOWS: CUDA 5.0 Production Release and the Graphics card is GTX 590.
Any help will be appreciated.

Answer 1

In an effort to avoid being "Stringly Typed", the use of character strings to refer to device symbols was deprecated in CUDA runtime API functions in CUDA 4.1, and removed in CUDA 5.0.
The CUDA 5 release notes read:
** The use of a character string to indicate a device symbol, which was possible
with certain API functions, is no longer supported. Instead, the symbol should be
used directly.
If you change your code to the following, it should work.
Status = cudaMemcpyToSymbol(CONSTANT_PARAMETER, &m_dParameter, sizeof(m_dParameter));
ERROR_CHECK(Status);

Answer 2

From the CUDA 5.0 Release Notes:
** The use of a character string to indicate a device symbol, which was possible with certain API functions, is no longer supported. Instead, the symbol should be used directly. "
These API functions still exist, but they accept the target symbol argument only as a bare identifier now, not as either a bare identifier or a string literal naming an ident. E.g.
__ device__ __ constant__ type ident;
main() { cudaMemcpyToSymbol("ident", ...); } // no longer valid, returns cudaErrorInvalidSymbol
main() { cudaMemcpyToSymbol(ident, ...); } // valid
So get rid of this:
#define PARAMETER "CONSTANT_PARAMETER"
And change this:
Status = cudaMemcpyToSymbol(PARAMETER, &m_dParameter, sizeof(m_dParameter));
To this:
Status = cudaMemcpyToSymbol(CONSTANT_PARAMETER, &m_dParameter, sizeof(m_dParameter));
And I think it will work.