In this particular case, cumsum does exactly this,
julia> cumsum([1:10])
10-element Array{Int64,1}:
1
3
6
10
15
21
28
36
45
55
I guess that would be idiomatic Julia ;)
An equivalent foldl would be
foldl((a, b) -> push!(a, a[end]+b), Int[1], [2:10])
10-element Array{Int64,1}:
1
3
6
10
15
21
28
36
45
55
A generic cumfoldl would be
julia> cumfoldl(f, x0, itr) = foldl((a, b) -> push!(a, f(a[end], b)),
[itr[1]], itr[2:end])
cumfoldl (generic function with 1 method)
julia> cumfoldl(*, 0, map(string, 1:10))
10-element Array{ASCIIString,1}:
"1"
"12"
"123"
"1234"
"12345"
"123456"
"1234567"
"12345678"
"123456789"
"12345678910"
I often find it useful to remind myself that foldl(push!, eltype(list)[],
list) constructs the same list and take it from there.
