I just got started with fotran95; I was given a code and I am studying it; I came across a subroutine that calls a function but I don't understand what the output is:
here is the subroutine:
SUBROUTINE collisione(vga, ga, vgb, gb)
IMPLICIT NONE
DOUBLE PRECISION, DIMENSION(3), INTENT(INOUT) :: ga, gb
DOUBLE PRECISION, DIMENSION(3), INTENT(INOUT) :: vga, vgb
DOUBLE PRECISION, DIMENSION(3) :: r, ra, rb, r_r
ra = pos_ini(ga)
rb = pos_ini(gb)
END SUBROUTINE
here is the function:
FUNCTION pos_ini(g)
IMPLICIT NONE
DOUBLE PRECISION, DIMENSION(3) :: pos_ini
DOUBLE PRECISION, DIMENSION(3), INTENT(IN) :: g
DOUBLE PRECISION, DIMENSION(3) :: es, eg, es_p
DOUBLE PRECISION :: rnd, b, kos, ang
CALL RANDOM_NUMBER(rnd)
b = 1.D0 - 2.D0*rnd
kos = SQRT(1.D0 - b*b)
CALL RANDOM_NUMBER(rnd)
ang = pi2 * rnd
es(1) = kos * COS(ang)
es(2) = kos * SIN(ang)
es(3) = b
eg = g / SQRT(DOT_PRODUCT(g,g))
es_p = es - DOT_PRODUCT(es,eg) * eg
pos_ini = d * es_p/SQRT(DOT_PRODUCT(es_p,es_p)) ! IS THIS THE OUTPUT THAT GIVES THE VALUE OF ra and rb???????
END FUNCTION
(read the comment). So here comes my question: in the subroutine I see that the variables ra and rb are defined after using the function pos_ini where the input is the vector ga or gb.
In the function pos_ini however I don't understand what the output is; is it pos_ini = d * es_p/SQRT(DOT_PRODUCT(es_p,es_p)) ?? if so why? and how come there is no intent(OUT) in function pos_ini?
In Fortran the returned function value is connected to the function name or a variable in the optional result clause. Therefore the function value will be whatever happened to be the value of pos_ini at the end of the function.
In your case it is the espression d * es_p/SQRT(DOT_PRODUCT(es_p,es_p)).
This is covered by any Fortran tutorial.