Go: understanding strings

Question

I got slightly confused this morning when the following code worked.

// s points to an empty string in memory
s := new(string)

// assign 1000 byte string to that address
b := make([]byte, 0, 1000)
for i := 0; i < 1000; i++ {
    if i%100 == 0 {
        b = append(b, '\n')
    } else {
        b = append(b, 'x')
    }
}
*s = string(b)

// how is there room for it there?
print(*s)

http://play.golang.org/p/dAvKLChapd

I feel like I'm missing something obvious here. Some insight would be appreciated.

The line *s = string(b) assigns s to a new string. This discards the old object you made in your second line. — fuz
– fuz, Commented Apr 19, 2013 at 14:28
@FUZxxl I was thinking in terms of C strings, which seems obviously wrong in retrospect. — Ilia Choly
– Ilia Choly, Commented Apr 19, 2013 at 14:31
In Go, a string is already a reference type, just as a slice is. — fuz
– fuz, Commented Apr 19, 2013 at 14:34
To make it simpler, you can just consider t := ""; t = string(b); print(t) // how is there room for it there? where t is the *s — newacct
– newacct, Commented Apr 19, 2013 at 17:35

zzzz · Accepted Answer · 2013-04-19 14:23:56Z

8

I hope I understood the question...

An entity of type string is implemented by a run time struct, roughly

type rt_string struct {
        ptr *byte // first byte of the string
        len int   // number of bytes in the string
}

The line

*s = string(b)

sets a new value (of type rt_string) at *s. Its size is constant, so there's "room" for it.

More details in rsc's paper.

answered Apr 19, 2013 at 14:23

zzzz

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