编译时如何在路径中包含库？

Question

I'm reading this post about go and was trying to compile the source code found here

I downloaded the source code, compiled the first file with make and I can see the object is generated:

$pwd
/Users/oscarryz/code/go/rsc/rosetta/graph

$ls -ltR
total 136
-rw-r--r--  1 oscarryz  staff  61295 Sep 17 16:20 _go_.6
drwxr-xr-x  3 oscarryz  staff    102 Sep 17 16:20 _obj
-rw-r--r--  1 oscarryz  staff    126 Sep 17 16:17 Makefile
-rw-r--r--  1 oscarryz  staff   2791 Sep 17 16:17 graph.go

./_obj:
total 0
drwxr-xr-x  3 oscarryz  staff  102 Sep 17 16:20 rsc.googlecode.com

./_obj/rsc.googlecode.com:
total 0
drwxr-xr-x  3 oscarryz  staff  102 Sep 17 16:20 hg

./_obj/rsc.googlecode.com/hg:
total 0
drwxr-xr-x  3 oscarryz  staff  102 Sep 17 16:20 rosetta

./_obj/rsc.googlecode.com/hg/rosetta:
total 136
-rw-r--r--  1 oscarryz  staff  68486 Sep 17 16:20 graph.a

No my question is, how do I refer to that compiled code from the maze directory:

/Users/oscarryz/code/go/rsc/rosetta/maze/maze.go

Whose import declarations are:

import (
    "bytes"
    "fmt"
    "rand"
    "time"

    "rsc.googlecode.com/hg/rosetta/graph"
)

And right now is failing to compile with the error message:

6g  -o _go_.6 maze.go 
maze.go:20: can't find import: rsc.googlecode.com/hg/rosetta/graph
make: *** [_go_.6] Error 1

Answer 1

Ok, I found it, wasn't that hard.

6g flags: -I DIR search for packages in DIR

I have to specify the -I option like this:

6g -I ../graph/_obj/ -o _go_.6 maze.go