Error (status) codes

In all FINUFFT interfaces, the returned value ier is a status indicator. It is 0 if successful, otherwise the error code has the following meanings (see include/finufft_errors.h):

1  requested tolerance epsilon too small to achieve (warning only)
2  attemped to allocate internal array larger than MAX_NF (defined in defs.h)
3  spreader: fine grid too small compared to spread (kernel) width
4  spreader: if chkbnds=1, a nonuniform point coordinate is out of input range [-3pi,3pi]^d
5  spreader: array allocation error
6  spreader: illegal direction (should be 1 or 2)
7  upsampfac too small (should be >1.0)
8  upsampfac not a value with known Horner poly eval rule (currently 2.0 or 1.25 only)
9  ntrans not valid in "many" (vectorized) or guru interface (should be >= 1)
10 transform type invalid
11 general allocation failure
12 dimension invalid
13 spread_thread option invalid
14 invalid mode array (more than ~2^31 modes, dimension with 0 modes, etc)
15 cuda failure (failure to call any cuda function/kernel)
16 attempt to destroy an uninitialized plan
17 invalid spread/interp method for dim (attempt to blockgather in 1D, e.g.)
18 size of bins for subprob/blockgather invalid
19 GPU shmem too small for subprob/blockgather parameters

When ier=1 (warning only) the transform(s) is/are still completed, at the smallest epsilon achievable, so, with that caveat, the answer should still be usable.

For any other nonzero values of ier the transform may not have been performed and the output should not be trusted. However, we hope that the value of ier will help to narrow down the problem.

FINUFFT sometimes also sends error text to stderr if it detects faulty input parameters.

If you are getting error codes, please reread the documentation for your language, then see our troubleshooting advice.

Large internal arrays

In case your input parameters demand the allocation of very large arrays, an internal check is done to see if their size exceeds a rather generous internal limit, set in defs.h as MAX_NF. The current value in the source code is 1e11, which corresponds to about 1TB for double precision. Allocations beyond this cause a graceful exit with error code 2 as above. Such a large allocation can be due to enormous N (in types 1 or 2), or M, but also large values of the space-bandwidth product (loosely, range of \(\mathbf{x}_j\) points times range of \(\mathbf{k}_j\) points) for type 3 transforms; see Remark 5 in reference FIN. Note that mallocs smaller than this, but which still exceed available RAM, may cause segfaults as usual. For simplicity of code, we do not do error checking on every malloc or STL vector creation in the code, and neither is this recommended in modern style guides. If you have a large-RAM machine and want to exceed the above hard-coded limit, you will need to edit defs.h and recompile.