The documentation is correct, and your interpretation of what cudaMemset
does is wrong. The function really does set byte values. Your example sets the first 32 bytes to 0x12
, not all 32 integers to 0x12
, viz:
#include <cstdio>
int main(void)
{
const int n = 32;
const size_t sz = size_t(n) * sizeof(int);
int *dJunk;
cudaMalloc((void**)&dJunk, sz);
cudaMemset(dJunk, 0, sz);
cudaMemset(dJunk, 0x12, 32);
int *Junk = new int[n];
cudaMemcpy(Junk, dJunk, sz, cudaMemcpyDeviceToHost);
for(int i=0; i<n; i++) {
fprintf(stdout, "%d %x\n", i, Junk[i]);
}
cudaDeviceReset();
return 0;
}
produces
$ nvcc memset.cu
$ ./a.out
0 12121212
1 12121212
2 12121212
3 12121212
4 12121212
5 12121212
6 12121212
7 12121212
8 0
9 0
10 0
11 0
12 0
13 0
14 0
15 0
16 0
17 0
18 0
19 0
20 0
21 0
22 0
23 0
24 0
25 0
26 0
27 0
28 0
29 0
30 0
31 0
ie. all 128 bytes set to 0, then first 32 bytes set to 0x12
. Exactly as described by the documentation.