I have to dynamically accumulate strings and for each string simply accumulate characters in an array
push!(word, c)
and then using this nifty one-liner to convert to a string
foldl(*, word)
I’m just wondering if that’s the best way to solve the problem ?
julia> word = ['a','b','c'];
julia> join(word)
"abc"
No, write to a buffer: create buf = IOBuffer(), accumulate characters as print(buf, c), and then do String(take!(buf)) at the end.
To be clear, this is exactly what join does internally, right? And the advantage of this over using join is just that you also skip the unnecessary collection into a Vector{Char} ?
Yes.
@stevengj is the winner ! 
@yha comes in second place
and I came in last 
The foldl method is slowest because it does the most allocations. What’s particularly interesting is that the IOBuffer method uses about 1/3 the allocations of using join. And join uses about 1/2 the allocations of foldl.
Thanks everyone