cuda kernel warning : expression has no effect

Question

I'm new to Cuda programming and trying my luck with Particle in Cell Code. But the first Problem is to build a particle mover. But when I'm trying to compile this code i get error messages like this:

error : expression must have integral or enum type / warning : expression has no effect.

My code:

__global__ void kernel(int* x, int* x_1, int* E_x, int* t, int* m)
{
    int idx = 0;
    if (idx < N)
        // move particles
        x_1[idx] = (E_x[idx] / m[1]) * t[1] * t[1] + x[idx];
}

kernel<<1,1>>( dev_x , dev_x_1, dev_E_x , dev_t, dev_m );

The integers defined as follows:

int x[N], x_1[N], v_x[N], v_y[N], v_z[N], E_x[N], m[1], t[1];
int *dev_x, *dev_v_x, *dev_x_1, *dev_v_y, *dev_v_z, *dev_E_x, *dev_m, *dev_t;

Look [here](http://stackoverflow.com/questions/359732/why-is-it-considered-a-bad-practice-to-omit-curly-braces) - first answer. — user2398029, Dec 26 '12 at 22:13
Perhaps you could show the *actual code* which produces that error message, because what you have posted clearly isn't it. — talonmies, Dec 26 '12 at 22:21
And it would help to know which line is producing the error. — Barry Brown, Dec 27 '12 at 02:10
The incorrect double chevron syntax on the kernel calls is producing the errors and warnings indicated by the post. — Robert Crovella, Dec 27 '12 at 02:20

score 4 · Accepted Answer · answered Dec 26 '12 at 22:53

4

One problem is you are using a double-chevron syntax instead of the proper triple-chevron syntax on your kernel launch parameters. Instead of this:

kernel<<1,1>>( dev_x , dev_x_1, dev_E_x , dev_t, dev_m );

Do this:

kernel<<<1,1>>>( dev_x , dev_x_1, dev_E_x , dev_t, dev_m );

answered Dec 26 '12 at 22:53

Robert Crovella

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Oooh sorry!!! It was 2 o´clock in the morning when i answerd this. THANKS!!! Now it´s working!!! – suehprom Dec 27 '12 at 11:02

cuda kernel warning : expression has no effect

1 Answers1