Pass arguments and take sum - fortran

Question

I am passing two values to my Fortran program, I need to get the sum of those arguments and print it as result:
I have the program for reading arguments as follows:
PROGRAM Argtest
IMPLICIT NONE
integer*4 nargs,i
character arg*80
nargs = iargc()
do i = 0,nargs
call getarg(i, arg)
print '(a)', arg
end do
END
I am passing the values 10 and 20.
I tried like this:
PROGRAM Argtest
IMPLICIT NONE
integer:: nargs,i
character:: arg
integer:: num1
integer:: num2
integer:: result
nargs = iargc()
do i = 1,nargs
call getarg(i, arg)
!print *, arg
IF( i == 1) THEN
num1 = ichar(arg)
ELSE IF(i == 2) THEN
num2 = ichar(arg)
ELSE
end IF
end do
result = num1+num2
print *, num1
print*,num2
END
I need to print the answer as 30. But I am getting values 49 and 50 instead of getting 10 and 30. Please help me.

Answer 1

Here is a very simple version: It reads the arguments as strings, converts them into ints one after the other, and adds them all up.
PROGRAM Argtest
IMPLICIT NONE
integer*4 nargs,i
character arg*80
integer :: total, int_arg
nargs = iargc()
total = 0
do i = 1,nargs
call getarg(i, arg)
read(arg, *) int_arg
total = total + int_arg
end do
print *, "total is ", total
END
Note that I am starting from argument 1, not 0 (as that is your program name, and can't be converted into a number).
You have now updated your question: ichar converts a single character into the integer that corresponds to that character's ASCII code.
You need to use read(ch_num, '(I)') int_num to convert a string like "10" to the integer number 10.