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Home/ Questions/Q 3211284
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Editorial Team
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Editorial Team
Asked: May 17, 20262026-05-17T14:43:02+00:00 2026-05-17T14:43:02+00:00

say I have a kernel foo(int a, int b) { __shared__ int array[a]; }

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say I have a kernel

foo(int a, int b)
{
    __shared__ int array[a];
}

it seems a has to be a constant value, I added const in front of int. It sill didn’t work out,
any idea?

foo(const int a, const int b)
{
    __shared__ int array[a];
}
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  1. Editorial Team
    Editorial Team
    2026-05-17T14:43:02+00:00Added an answer on May 17, 2026 at 2:43 pm

    While you can’t have a dynamically-sized array because of the constraints of the C language (as mentioned in other answers), what you can do in CUDA is something like this:

    extern __shared__ float fshared[];
    
    __global__ void testShmem( float * result, unsigned int shmemSize ) {
        // use fshared - shmemSize tells you how many bytes
        // Note that the following is not a sensible use of shared memory!
        for( int i = 0; i < shmemSize/sizeof(float); ++i ) {
           fshared[i] = 0;
        }
    }
    

    providing you tell CUDA how much shared memory you want during kernel invocation, like so:

    testShmem<<<grid, block, 1024>>>( pdata, 1024 );
    
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