Tutorial:JMathLab/Programming

Programming

Programs can be created and run interactively. Functions can be either built-in or user defined.

Branches

Branches can be defined as in any language:

if test1
    code1;
else 
    code2;
end;

"if x A else B end"

As an example a function with the if statement

<jc lang="math"> function y = H2(f, e)

 if (f==1)
   y = e+f;
 else
   y = e+2*f;
 end

end y= H2(1,2) printf('%f\n',y) y=H2(2,2) printf('%f',y) </jc>

Loops

Loops with condition x and statement(s) A: while x A end The while-loop is repeated until x becomes false (0).

<jc lang="math"> x=1;y=1; while(x<10) y=x+y; x++; end printf('%f',y) </jc>

Loops with counter z and statement(s) A:

"for z = vector A end"

In the for-loop the counter is formally initialized by a vector. In each execution of the loop the counter takes on the value of the next element of vector.

<jc lang="math"> x=1;y=1; for(x=1:0.1:100) y=x^2+y; end printf('%f',y) </jc>

Jumps

return, continue, break

A function may be part of the loop, and begins another cycle. break permanently leaves the loop.

<jc lang="math"> x=1; while( 1 )

   if(x>1000) 
      break; 
   end 
   x++; 

end printf('%f',x) </jc>

Summary

Definition	Example
Branch	if x==0 xp=1; end if x==0 xp=1; else xp=2; end
Loop	while x<17 x=x+1; end for i=0:100 x=x+i; end
Jump	return, continue, break
Function	function y=ttwo(x) y=2*x; end
Evaluate	eval('x=24')
Text output	printf('Number=%f', x)
Error message	error('Test')

Benchmarking

You can benchmark your code to ensure that your code is optimized. Here is a small example to see how fast jMathLab is:

<jc lang="math"> a1=time(0) x=0 n=0 for i=0:1000000

    x=x+cos(i)*sin(i)*sqrt(i*10); 
    n=n+1;
    end

printf('%f\n',x) a2=time(0) printf('iterations=%f\n',n) printf('processed for %f (sec)',(a2-a1)/1000) </jc>

Here "time(0)" return the number of milliseconds since 1970. Run this code and compare with any other similar code.

Let us compare performance of 2 codes: one is written in jMathLab and one in Jython/Python. Here what we have:

<jc lang="math"> d1=time(0) s=1; n=0 for(x=0:1:1000000)

       s=s+cos(x)*sin(x^2);
      n=n+1;

end d2=time(0); printf('\nNr of iterations=%f\n',n) ; printf('Time=%f',(d2-d1)/1000) </jc>

No let's rewrite the same program in Jython:

<jc lang="math"> import math import time start = time.clock() n=0 s=0 for x in range(1000000):

    s=s+math.cos(x)*math.sin(x*x)
    n=n+1

print ' CPU time (s)=',time.clock()-start </jc>