[ffriend]

Even better would be to write (see dot vectorization [1]):

    C .= A .+ B

Benchmarks for 3 matrices of size 1000x1000:

    julia> using BenchmarkTools

    julia> @benchmark C = A + B
    BenchmarkTools.Trial: 
      memory estimate:  7.63 MiB
      allocs estimate:  2
      --------------
      minimum time:     2.359 ms (0.00% GC)
      median time:      2.713 ms (0.00% GC)
      mean time:        3.794 ms (28.81% GC)
      maximum time:     62.708 ms (95.27% GC)
      --------------
      samples:          1314
      evals/sample:     1

    julia> @benchmark C .= A .+ B
    BenchmarkTools.Trial: 
      memory estimate:  128 bytes
      allocs estimate:  4
      --------------
      minimum time:     1.232 ms (0.00% GC)
      median time:      1.320 ms (0.00% GC)
      mean time:        1.356 ms (0.00% GC)
      maximum time:     2.572 ms (0.00% GC)
      --------------
      samples:          3651
      evals/sample:     1

Note that memory usage dropped from 7.63MiB to 128 bytes.

[1]: https://docs.julialang.org/en/stable/manual/functions/#man-v...

[vanderZwan]

Thanks! Like I said: I never truly dove into it, although I loved reading about the approach to the type system, and the multiple dispatch.

> Note that memory usage dropped from 7.63MiB to 128 bytes.

Which is important if you're working with large data-sets. Both for performance and for being able to run the calculations at all.

[stabbles]

A little bit late to the party here, but the number of allocations is really just 0 bytes. It shows 128 bytes because the benchmark is creating new references to A, B and C. To correct this use either interpolation with $A, $B and $C or initialize A, B and C in the setup phase:

    > @benchmark C .= A .+ B setup = (A = rand(1000, 1000); B = rand(1000, 1000); C = rand(1000, 1000))
    BenchmarkTools.Trial:
      memory estimate:  0 bytes
      allocs estimate:  0
      --------------
      minimum time:     2.048 ms (0.00% GC)

This is showing 0 bytes indeed.