CODEWITHSUNDEEP
cuda
user3910910
user3910910

Reputation: 99

Why is the first cudaMalloc the only bottleneck?

I defined this function :

void cuda_entering_function(...)
{
    StructA *host_input, *dev_input;
    StructB *host_output, *dev_output;

    host_input = (StructA*)malloc(sizeof(StructA));
    host_output = (StructB*)malloc(sizeof(StructB));
    cudaMalloc(&dev_input, sizeof(StructA));
    cudaMalloc(&dev_output, sizeof(StructB));

    ... some more other cudaMalloc()s and cudaMemcpy()s ...

    cudaKernel<< ... >>(dev_input, dev_output);

    ...
}

This function is called several times (about 5 ~ 15 times) throughout my program, and I measured this program's performance using gettimeofday().

Then I found that the bottleneck of cuda_entering_function() is the first cudaMalloc() - the very first cudaMalloc() throughout my whole program. Over 95% of the total execution time of cuda_entering_function() was consumed by the first cudaMalloc(), and this also happens when I changed the size of first cudaMalloc()'s allocating memory or when I changed the executing order of cudaMalloc()s.

What is the reason and is there any way to reduce the first cuda allocating time?

Upvotes: 3

Views: 639

Answers (1)

Etienne Pellegrini
Etienne Pellegrini

Reputation: 748

The first cudaMalloc is responsible for the initialization of the device too, because it's the first call to any function involving the device. This is why you take such a hit: it's overhead due to the use of CUDA and your GPU. You should make sure that your application can gain a sufficient speedup to compensate for the overhead.

In general, people use a call to an initialization function in order to setup their device. In this answer, you can see that apparently a call to cudaFree(0) is the canonical way to do so. This sample shows the use of cudaSetDevice, which could be a good habit if you ever work on machines with several CUDA-ready devices.

Upvotes: 9

Related Questions