I'm trying to write a histogram kernel in OpenCL to compute the R, G, and B histograms of an input image. My kernel looks like this:

Code :
const sampler_t mSampler = CLK_NORMALIZED_COORDS_FALSE |
__kernel void computeHistogram(read_only image2d_t input, __global int* rOutput,
                               __global int* gOutput, __global int* bOutput)
    int2 coords = {get_global_id(0), get_global_id(1)};
    float4 sample = read_imagef(input, mSampler, coords);
    uchar rbin = floor(sample.x * 255.0f);
    uchar gbin = floor(sample.y * 255.0f);
    uchar bbin = floor(sample.z * 255.0f);

When I run it on an 2100 x 894 image (1,877,400 pixels) i tend to only see in or around 1,870,000 total values being recorded when I sum up the histogram values for each channel. It's also a different number each time. I did expect this since once in a while two kernels probably grab the same value from the output array and increment it, effectively cancelling out one increment operation (I'm assuming?).

The 1,870,000 output is for a {1,1} workgroup size (which is what seems to get set by default if I don't specify otherwise). If I force a larger workgroup size like {10,6}, I get a drastically smaller sum in my histogram (proportional to the change in workgroup size). This seemed strange to me, but I'm guessing what happens is that all of the work items in the group increment the output array value at the same time, and so it just counts as a single increment?

Anyways, I've read in the spec that OpenCL has no global memory syncronization, only syncronization within local workgroups using their __local memory. The histogram example by nVidia breaks up the histogram workload into a bunch of subproblems of a specific size, computes their partial histograms, then merges the results into a single histogram after. This doesn't seem like it'll work all that well for images of arbitrary size. I suppose I could pad the image data out with dummy values...

Being new to OpenCL, I guess I'm wondering if there's a more straightforward way to do this (since it seems like it should be a relatively straightforward GPGPU problem).