Why is texture memory version of below program slower than global memory version

i am confused why my texture version is slower than my global memory version because the texture version should exploit spatial locality. I am trying to compute the dot product in the below case. Thus, if one thread accesses index i, its neighbour should access i+1. Thus, we see spatial locality.

Below is the texture memory version:

#include<cuda_runtime.h>
#include<cuda.h>
#include<stdio.h>
#include<stdlib.h>
#define intMin(a,b) ((a<b)?a:b)
//Threads per block
#define TPB 128
//blocks per grid
#define BPG intMin(128, ((n+TPB-1)/TPB))

texture<float> arr1;
texture<float> arr2;


const int n = 4;

__global__ void addVal( float *c){
    int tid = blockIdx.x * blockDim.x + threadIdx.x;

    //Using shared memory to temporary store results
    __shared__ float cache[TPB];
    float temp = 0;
    while(tid < n){
        temp += tex1Dfetch(arr1,tid) * tex1Dfetch(arr2,tid);
        tid += gridDim.x * blockDim.x;


    }
    cache[threadIdx.x] = temp;
    __sync开发者_StackOverflow社区threads();
    int i = blockDim.x/2;
    while( i !=0){
        if(threadIdx.x < i){
            cache[threadIdx.x] = cache[threadIdx.x] +cache[threadIdx.x + i] ;

        }
    __syncthreads();
    i = i/2;

    }
    if(threadIdx.x == 1){
        c[blockIdx.x ] = cache[0];
    }



}

int main(){

float a[n] , b[n] , c[BPG];
float *deva, *devb, *devc;
int i;
//Filling with random values to test
for( i =0; i< n; i++){
    a[i] = i;
    b[i] = i*2;
}
printf("Not using constant memory\n");
cudaMalloc((void**)&deva, n * sizeof(float));
cudaMalloc((void**)&devb, n * sizeof(float));
cudaMalloc((void**)&devc, BPG * sizeof(float));


cudaMemcpy(deva, a, n *sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(devb, b, n *sizeof(float), cudaMemcpyHostToDevice);
cudaBindTexture(NULL,arr1, deva,sizeof(float) * n); // note: deva shd be in gpu
cudaBindTexture(NULL,arr2, devb,sizeof(float) * n); // note: deva shd be in gpu
cudaEvent_t start, stop;
cudaEventCreate(&start);
cudaEventCreate(&stop);
cudaEventRecord(start, 0);

//Call function to do dot product
addVal<<<BPG, TPB>>>(devc);
cudaEventRecord(stop, 0);
cudaEventSynchronize(stop);
float time;
cudaEventElapsedTime(&time,start, stop);
printf("The elapsed time is: %f\n", time);


//copy result back
cudaMemcpy(c, devc, BPG * sizeof(float), cudaMemcpyDeviceToHost);
float sum =0 ;
for ( i = 0 ; i< BPG; i++){
    sum+=c[i];

}
//display answer
printf("%f\n",sum);
cudaUnbindTexture(arr1);
cudaUnbindTexture(arr2);
cudaFree(devc);

getchar();

return 0;
}

Global Memory version:

#include<cuda_runtime.h>
#include<cuda.h>
#include<stdio.h>
#include<stdlib.h>
#define intMin(a,b) ((a<b)?a:b)
//Threads per block
#define TPB 128
//blocks per grid
#define BPG intMin(128, ((n+TPB-1)/TPB))

const int n = 4;

__global__ void addVal(float *a, float *b, float *c){
    int tid = blockIdx.x * blockDim.x + threadIdx.x;

    //Using shared memory to temporary store results
    __shared__ float cache[TPB];
    float temp = 0;
    while(tid < n){
        temp += a[tid] * b[tid];
        tid += gridDim.x * blockDim.x;


    }
    cache[threadIdx.x] = temp;
    __syncthreads();
    int i = blockDim.x/2;
    while( i !=0){
        if(threadIdx.x < i){
            cache[threadIdx.x] = cache[threadIdx.x] +cache[threadIdx.x + i] ;

        }
    __syncthreads();
    i = i/2;

    }
    if(threadIdx.x == 1){
        c[blockIdx.x ] = cache[0];
    }



}

int main(){

float a[n] , b[n] , c[BPG];
float *deva, *devb, *devc;
int i;
//Filling with random values to test
for( i =0; i< n; i++){
    a[i] = i;
    b[i] = i*2;
}
printf("Not using constant memory\n");
cudaMalloc((void**)&deva, n * sizeof(float));
cudaMalloc((void**)&devb, n * sizeof(float));
cudaMalloc((void**)&devc, BPG * sizeof(float));
cudaMemcpy(deva, a, n *sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(devb, b, n *sizeof(float), cudaMemcpyHostToDevice);

cudaEvent_t start, stop;
cudaEventCreate(&start);
cudaEventCreate(&stop);
cudaEventRecord(start, 0);

//Call function to do dot product
addVal<<<BPG, TPB>>>(deva, devb, devc);
cudaEventRecord(stop, 0);
cudaEventSynchronize(stop);
float time;
cudaEventElapsedTime(&time,start, stop);
printf("The elapsed time is: %f\n", time);


//copy result back
cudaMemcpy(c, devc, BPG * sizeof(float), cudaMemcpyDeviceToHost);
float sum =0 ;
for ( i = 0 ; i< BPG; i++){
    sum+=c[i];

}
//display answer
printf("%f\n",sum);


getchar();

return 0;
}

---

While know your graphic device may help, for some type of problems, with compute capability 2.x the L1 and L2 cache works as good the texture cache.

In this case, you are not exploiting the texture cache, as you only read once value per thread. On the other chand, you are exploiting spatial locality in 1D what can be hide with global memory coalesced access.

I recommend you the book 'CUDA by Example: An Introduction to General-Purpose GPU Programming'. Great book for beginners. With graphics examples like JuliaSet or a very basic Raycasting (there are also the common add, reduce and dot product examples if you prefer thouse :).

Hope this help.

---

Further to pQB's answer, there is no data reuse in your program -- each input is read only once, and used only once. Memory indices are sequential across threads, and therefore perfectly coalesced. Because of these two reasons, there is no need for any device memory cacheing, so global memory access is more efficient than texture access. Add to this additional latency overhead in the texture cache (texture cache is designed to increase throughput, not decrease latency, unlike L1/L2 data caches), and the slowdown is explained.

BTW, what you are doing is a parallel reduction, so you may want to see the "reduction" example in the CUDA SDK for a fast implementation.