print on diag operator appears broken [BUG]

[luitjens]

Here is a trivial reproducer:

`
#include "matx.h"
#include
#include
#include <math.h>

using namespace matx;

int main([[maybe_unused]] int argc, [[maybe_unused]] char **argv)
{
MATX_ENTER_HANDLER();
auto A = matx::make_tensor({4,4});

A.SetVals({ {0,1,2,3},{4,5,6,7},{8,9,10,11},{12,13,14,15}});
cudaDeviceSynchronize();

printf("A:\n"); print(A);
printf("diag A:\n"); print(diag(A));
printf("sqrt diag A:\n"); print(sqrt(diag(A)));

CUDA_CHECK_LAST_ERROR();
MATX_EXIT_HANDLER();
}`

output:
A: Tensor{float} Rank: 2, Sizes:[4, 4], Strides:[4,1] 000000: 0.0000e+00 1.0000e+00 2.0000e+00 3.0000e+00 000001: 4.0000e+00 5.0000e+00 6.0000e+00 7.0000e+00 000002: 8.0000e+00 9.0000e+00 1.0000e+01 1.1000e+01 000003: 1.2000e+01 1.3000e+01 1.4000e+01 1.5000e+01 diag A: Operator{float} Rank: 1, Sizes:[4] 000000: 0.0000e+00 000001: 0.0000e+00 000002: 0.0000e+00 000003: 0.0000e+00 sqrt diag A: Operator{float} Rank: 1, Sizes:[4] 000000: 0.0000e+00 000001: 2.2361e+00 000002: 3.1623e+00 000003: 3.8730e+00

Notice the diag output is all zeros. In my full use case I get the first row of A. This indicates something is going wrong to me. Note that once you wrap diag in a square root you get the right values printed. In my real use case the device side has the right answers but the prints that I was debugging were wrong.

[luitjens]

@cliffburdick reverting this commit fixes the issue: 59bcbb7

We will need to debug why that patch doesn't fully work for the diag operator.

[luitjens]

Fixed in #552

[cliffburdick]

I copied that code from another place we print. Will take a look at it when back