""" Returns a resource path based on the call stack.

This type of resource path is normally requested from the constructor for an
object whose resources are relative to the module constructing the object.
"""

# Standard library imports.
import sys

from inspect import stack
from os.path import dirname, exists
from os      import getcwd
    
def resource_path ( level = 2 ):
    """Returns a resource path calculated from the caller's stack.
    """
    module = stack( level )[ level ][0].f_globals[ '__name__' ]
    
    if module != '__main__':
        # Return the path to the module:
        return dirname( getattr( sys.modules.get( module ), '__file__' ) )
        
    # '__main__' is not a real module, so we need a work around:
    for path in [ dirname( sys.argv[0] ), getcwd() ]:
        if exists( path ):
            break
            
    return path
    
#### EOF ######################################################################