Loops and loop control

Last updated on 2026-01-27 | Edit this page

Estimated time: 20 minutes

Overview

Questions

  • How can I make the program do work repeatedly?
  • How can I have the program only do as much work as is required?

Objectives

  • Learn how to use a do construct to execute a single block of code many times.
  • See how to have a do construct skip a loop with cycle or end early with exit.
  • Learn how to use loop control variables with do constructs.

Uncontrolled do construct

A simple iteration is provided by the do statement. For example, …

  do
    ! ... go around for ever ...
  end do

A slightly more useful version requires some control (see example1.f90):

   integer :: i = 0
   do
     i = i + 1
     if (mod(i, 2) == 0) cycle    ! go to the next iteration
     if (i >= 10) exit            ! exit loop completely
     ! ... some computation ...
   end do

   ! ... control continues here after exit ...

Loop constructs may be nested, and may also be named (see example2.f90):

  some_outer_loop: &
  do
    some_inner_loop: &
    do
      if (i >= 10) exit some_outer_loop  ! exit belongs to outer loop
      ! ... continues ...
    end do some_inner_loop
  end do some_outer_loop

If the control statements cycle or exit do not have a label, they belong to the innermost construct in which they appear.

Loop control

More typically, one encounters controlled iterations with an integer loop control variable. E.g.,

  integer :: i
  do i = 1, 10, 2
    ! ... perform some computation ...
  end do

Formally, we have

[do-name:] do [loop-control]
             do-block
	   end do [do-name]

with loop-control of the form:

  do-variable = int-expr-lower, int-expr-upper [, int-expr-stride]

and where the number of iterations will be

  max(0, (int-expr-upper - int-expr-lower + int-expr-stride)/int-expr-stride)

in the absence of any subsequent control. The number of iterations may be zero. Any of the expressions may be negative. If the stride is not present, it will be 1 (unity); if the stride is present, it may not be zero.

Exercise (2 minutes)

Challenge

How many iterations make a do loop?

What is the number of iterations in the following cases?

   do i = 1, 10
     print *, "i is  ", i
   end do

   do i = 1, 10, -2
     print *, "i is ", i
   end do

   do i = 10, 1, -2
     print *, "i is ", i
   end do

You can confirm your answers by running example3.f90. Note there is no way (cf. C for) to limit the scope of the loop variable to the loop construct itself; the variable will then have a final value after exit from the loop.

The following output is produced:

OUTPUT 

 First loop: 1, 10
 i is  1
 i is  2
 i is  3
 i is  4
 i is  5
 i is  6
 i is  7
 i is  8
 i is  9
 i is  10
 Second loop: 1, 10, -2
 Third loop: 10, 1, -2
 i is  10
 i is  8
 i is  6
 i is  4
 i is  2

The first loop has ten iterations with i running from 0 to 10 in steps of 1.

The second loop runs zero times as, with a stride of -2, there is no way to iterate from 1 to 10.

The third loop has five iterations, from 10 to 2 in steps of -2. The stated end value of 1 can’t be reached, and 0 would go too far, so it finishes at 2.

Exercise (5 minutes)

Challenge

Calculating pi with loops

We return the exercises discussed in the earlier episode on variables. You can use your own solutions, or the new template here.

For exercise1.f90 which computes an approximation to the constant pi using the Gauss-Legendre algorithm, introduce an iteration to compute a fixed number of successive improvements. How many iterations are required to converge when using kind(1.d0)? Would you be able to adjust the program to halt the iteration if the approximation is within a given tolerance of the true answer?

Sample code implementing loops with this problem is used as a template to the exercise in the episode on arrays.

You should be able to observe that with kind(1.d0) the value of pi converges after only three iterations. You also can have the code exit the loop and halt iteration by storing the previous estimate of pi if the difference between the old and new values is less than the desired tolerance.

Exercise (5 minutes)

Challenge

Conductance of a channel with loops

exercise2.f90 returns to the calculation of the conductance in the narrow channel and will need similar work. Use a loop to compute a fixed number of terms in the sum over index k (use real type real64 for the sum). Here you should find convergence is much slower (you may need about 1000 terms); check by printing out the current value every so many terms (say 20).

Expert question: What happens if you accumulate the sum in the reverse order in this case? Can you think why this happens?

Answer to the expert question: floating point numbers of a given format (such as real64) all have the same relative error, no matter how big or small they are. The absolute error on the other hand is the produce of the number’s value and the relative error; thus, a small real has a small absolute error, and a large real has a large absolute error. If the large numbers are summed first, the sum’s absolute error quickly becomes large, and the smaller numbers are subsumed into it. If the small numbers are summed first, the sum’s absolute error remains small enough that the small numbers are able to contribute. Consider further how in floating point maths it is true that (x + y) + z /= x + (y + z), if you are interested look up Kahan’s algorithm.

Key Points
  • Iteration in Fortran is based around the do construct (somewhat analogous to C for construct). There is no equivalent of the C++ iterator class.
  • Without any control, a do loop will execute forever.
  • A loop iteration can be skipped with a cycle statement.
  • A loop can be ended if an exit statement is encountered.
  • It is very common to control the execution of a loop with an integer variable.