# block.gp.pos

Syntax

v/f = block.gp.pos(bgpp<,i>)

Get the position of a gridpoint.

Returns: v or f - position vector or component bgpp - block gridpoint pointer i - optional index of component

## Component Access

f = block.gp.pos.x(bdpp)

Get the $$x$$-component of the position.

Returns: f - $$x$$-component of the position vector bggp - gridpoint pointer
f = block.gp.pos.y(bdpp)

Get the $$y$$-component of the position.

Returns: f - $$y$$-component of the position vector bggp - gridpoint pointer
f = block.gp.pos.z(bdpp)

Get the $$z$$-component of the position.

Returns: f - $$z$$-component of the position vector bggp - gridpoint pointer