Result of reduction#1 is wrong - cuda

Question

I implemented the reduction#1 form the well-known slides by Mark Harris, but I obtain 0 as result. I filled the input array with the same values shown in the slides. I compiled with cuda 7.0 using the command nvcc reduction1.cu -o red1. Where is the mistake? Thanks.
#include <stdio.h>
#include <cuda_runtime.h>
#define THREADS_PER_BLOCK 16
__global__ void reduce1(int *g_idata, int *g_odata) {
extern __shared__ int sdata[];
// each thread loads one element from global to shared mem
unsigned int tid = threadIdx.x;
unsigned int i = blockIdx.x*blockDim.x + threadIdx.x;
sdata[tid] = g_idata[i];
__syncthreads();
// do reduction in shared mem
for(unsigned int s=1; s < blockDim.x; s *= 2)
{
if (tid % (2*s) == 0) sdata[tid] += sdata[tid + s];
__syncthreads();
}
// write result for this block to global mem
if (tid == 0) g_odata[blockIdx.x] = sdata[0];
}
int main()
{
int inputLength=16;
int hostInput[16]={10,1,8,-1,0,-2,3,5,-2,-3,2,7,0,11,0,2};
int hostOutput=0;
int *deviceInput;
int *deviceOutput;
cudaMalloc((void **)&deviceInput, inputLength * sizeof(int));
cudaMalloc((void **)&deviceOutput, sizeof(int));
cudaMemcpy(deviceInput, hostInput, inputLength * sizeof(int),cudaMemcpyHostToDevice);
reduce1<<<1,THREADS_PER_BLOCK>>>(deviceInput, deviceOutput);
cudaDeviceSynchronize();
cudaMemcpy(&hostOutput, deviceOutput,sizeof(int), cudaMemcpyDeviceToHost);
printf("%d\n",hostOutput);
cudaFree(deviceInput);
cudaFree(deviceOutput);
return 0;
}

Answer 1

As talonmies said, you are using dynamic shared memory, but you are not allocating any memory space for it. You have to specify the size of this memory as the third argument of your kernel execution configuration.
reduce1<<<1, THREADS_PER_BLOCK, 64>>>(deviceInput, deviceOutput);
^^
Another way to fix this code is to use static shared memory. Declare your shared memory like this:
__shared__ int sdata[16];
Please read this before asking questions for CUDA.