Conditional execution is done using the
else statements in the following construct:
if (logical_expression1) then ! Block of code else if (logical_expression2) then ! Block of code else ! Block of code end if
You may have as many
else if statements as you desire.
The following operators can be used when making expressions:
|Operation||Modern Fortran||Old FORTRAN|
|Logical not equivalent||.NEQV.|
Note: The Fortran standard mandates
.NEQ. cannot be used with logicals but some compilers will not enforce the standard
To check more than one statement, use parentheses.
if ((a .gt. b) .and. .not. (a .lt. c)) then
The following program generates a random number between 0 and 1 and tests if it is between 0 and 0.3, 0.3 and 0.6, or between 0.6 and 1.0.
program xif implicit none real :: x real, parameter :: x1 = 0.3, x2 = 0.6 call random_seed() call random_number(x) if (x < x1) then print *, x, "<",x1 else if (x < x2) then print *, x, "<", x2 else print *, x, ">=", x2 end if end program xif
There are two interesting archaic forms of
IF (<logical_expression>) GOTO <statement_label> IF (<arithmetic_expression>) <first_label>, <second_label>, <third_label>
In the first form, things are pretty straightforward. In the second form, the arithmetic expression is evaluated. If the expression evaluates to a negative number, then execution continues at the first line number. If the expression evaluates to zero, then execution continues at the second line number. Otherwise, execution continues at the third line number.
- select case(...) case (...); ... end select
If an if block consists of repeated tests on a single variable, it may be possible to replace it with a select case construct. For example, the code
if (month=="January" .or. month=="December") then num_days = 31 else if (month=="February") then num_days = 28 else if (month=="March") then num_days = 31 else num_days = 30 end if
can be replaced by
select case (month) case ("January", "December") num_days = 31 case ("February") num_days = 28 case ("March") num_days = 31 case default num_days = 30 end select
Fortran does not need a break statement.
- do i=1,10 ... end do
To iterate, Fortran has a do loop. The following loop prints the squares of the integers from 1 to 10:
do i=1,10 print *, i**2 end do
One can exit a loop early using exit, as shown in the code below, which prints the squares of integers until one of the squares exceeds 25.
do i=1,10 isquare = i**2 if (isquare > 25) exit print *, isquare end do
Loops can be nested. The following code prints powers 2 through 4 of the integers from 1 to 10
do i=1,10 do ipower=1,3 print *, i, ipower, i**ipower end do end do
In an archaic form of
DO, a line number on which the loop(s) end is used. Here's the same loop, explicitly stating that label
1 is the last line of each loop:
DO 1 i=1,10 DO 1 ipower=1,3 1 PRINT *, i, ipower, i**ipower
If using the archaic form, the loop must not end on an
GO TO statement. You may use a
CONTINUE statement as an anchor for a the
There is also an optional increment argument when declaring a do loop. The following will count up by two's. 2, 4, 6, ...
do i=2,10,2 write (*,*) i end do
Arguments to the do loop don't have to be numbers, they can be any integer that is defined elsewhere in the program. first, last, and increment can be any variable name.
do i=first,last,increment ! Code goes here end do
goto statement_label will jump to the specified statement number.
stop exit_code will stop with the specified condition code or exit code.
stop may be coded without an argument. Note that on many systems,
stop 0 is still a failure. Also note that pre-Fortran 2008, the condition code must be a constant expression and not a variable.
exit will leave a loop.
continue can be used to end an archaic
DO loop when it would otherwise end on an
cycle will transfer the control of the program to the next
end do statement.
return leaves a subroutine or function.