safety and auto vectorization

Chad Joan chadjoan at
Mon Aug 3 18:42:30 UTC 2020

On Sunday, 2 August 2020 at 17:31:45 UTC, Bruce Carneal wrote:
> import std;
> void f0(int[] a, int[] b, int[] dst) @safe {
>     dst[] = a[] + b[];
> }
> void f1(int[] a, int[] b, int[] dst) @trusted {
>     const minLen = min(a.length, b.length, dst.length);
>     dst[0..minLen] = a[0..minLen] + b[0..minLen];
>     assert(dst.length == minLen);
> }
> I was surprised that f0 ran just fine with a.length and 
> b.length geq dst.length.  Is that a bug or a feature?
> Assuming it's a feature, are f0 and f1 morally equivalent?  I 
> ask because f1 auto-vectorizes in ldc while f0 does not.  Not 
> sure why.  As a guess I'd say that the front end doesn't hoist 
> bounds checks in f0 or at least doesn't convey the info to the 
> back end in a comprehensible fashion.  Non-guesses welcome.

I don't know what's going on auto-vectorization-wise, but to 
address the behavioral issues, the next thing I would do if I 
were in your shoes is something like this:

import std.stdio;
int[100]  a, b, dst;
a[]   = 2;
b[]   = 3;
dst[] = 42;
f0(a[0..$], b[0..$], dst[0..50]); // Notice: dst is a smaller 
writefln("dst[49] == %d", dst[49]); // Should be 5.
writefln("dst[50] == %d", dst[50]); // 42 or 5?

On my machine (Linux 64-bit DMD v2.093.0) it prints this:
dst[49] == 5
dst[50] == 42

Which suggests that it is doing the minimum-length calculation, 
as the dst[] values outside of the lesser-sized slice were 

This was DMD, so it's going to be worth trying on your compiler 
to see what you get.

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