`@allocated` and `@allocations` are giving the wrong result.

[sadish-d]

julia> @time Int[]
  0.000002 seconds (1 allocation: 64 bytes)
Int64[]

julia> @allocated Int[]
0

julia> @allocations Int[]
0

julia> @time Array{Int}(undef)
  0.000003 seconds (1 allocation: 64 bytes)
0-dimensional Array{Int64, 0}:
0

julia> @allocated Array{Int}(undef)
0

julia> @allocations Array{Int}(undef)
0

julia> @time println("")

  0.000194 seconds (8 allocations: 160 bytes)

julia> @allocated println("")

160

julia> @allocations println("")

8

versioninfo

julia> versioninfo()
Julia Version 1.10.3
Commit 0b4590a5507 (2024-04-30 10:59 UTC)
Build Info:
  Official https://julialang.org/ release
Platform Info:
  OS: macOS (arm64-apple-darwin22.4.0)
  CPU: 8 × Apple M1
  WORD_SIZE: 64
  LIBM: libopenlibm
  LLVM: libLLVM-15.0.7 (ORCJIT, apple-m1)
Threads: 1 default, 0 interactive, 1 GC (on 4 virtual cores)
Environment:
  JULIA_EDITOR = code
  JULIA_NUM_THREADS = 

[Keno]

@time returns the result. @allocated does not, so the system didn't need to allocate it.

[sadish-d]

From the documentation of @allocations:

A macro to evaluate an expression, discard the resulting value, and instead return the total number of allocations during evaluation of the expression.

Say I have a function f() = Int[].

If I do a = @time f(), I know that the value of a will be Int[]. If I do b = @allocations f(), I know from the docstring that b is not going to be Int[]-- the return value of f().

But why does @allocations tell me that f() it makes no allocations when @time` tells me it makes one? Shouldn't the answer both be the same?

I believe @allocations compiles the function before evaluating, but I'm not sure that explains the difference.

(Happy to move this discussion elsewhere if that's more appropriate.)