Alex Rivera | Logout

How to efficiently concatenate strings in go

Asked 2009-11-19T03:44:10.347
1022

In Go, a string is a primitive type, which means it is read-only, and every manipulation of it will create a new string.

So if I want to concatenate strings many times without knowing the length of the resulting string, what's the best way to do it?

The naive way would be:

var s string
for i := 0; i < 1000; i++ {
    s += getShortStringFromSomewhere()
}
return s

but that does not seem very efficient.

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1 Answer

49

I just benchmarked the top answer posted above in my own code (a recursive tree walk) and the simple concat operator is actually faster than the BufferString.

func (r *record) String() string {
    buffer := bytes.NewBufferString("");
    fmt.Fprint(buffer,"(",r.name,"[")
    for i := 0; i < len(r.subs); i++ {
        fmt.Fprint(buffer,"\t",r.subs[i])
    }
    fmt.Fprint(buffer,"]",r.size,")\n")
    return buffer.String()
}

This took 0.81 seconds, whereas the following code:

func (r *record) String() string {
    s := "(\"" + r.name + "\" ["
    for i := 0; i < len(r.subs); i++ {
        s += r.subs[i].String()
    }
    s += "] " + strconv.FormatInt(r.size,10) + ")\n"
    return s
} 

only took 0.61 seconds. This is probably due to the overhead of creating the new BufferString.

Update: I also benchmarked the join function and it ran in 0.54 seconds.

func (r *record) String() string {
    var parts []string
    parts = append(parts, "(\"", r.name, "\" [" )
    for i := 0; i < len(r.subs); i++ {
        parts = append(parts, r.subs[i].String())
    }
    parts = append(parts, strconv.FormatInt(r.size,10), ")\n")
    return strings.Join(parts,"")
}
answered 2012-04-29T01:15:00.597

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