C++ syntax, nested loops and continue — an embarrassing experiment?

For whatever reason, the C++ continue keyword is not used often in the portions of the code I normally deal with.  I was moving code in a function that had gotten waaay to big into a helper function. This changed the scope of some continue and break usage, and thought I’d double check that these only impact the inner loop:

#include <stdio.h>

int main()
{
   for ( int i = 0 ; i < 5 ; i++ )
   {
      for ( int j = 0 ; j < 5 ; j++ )
      {
         if ( j % 2 )
         {
            continue ;
         }

         printf( "i,j = %d, %d\n", i, j ) ;
      }
   }

   printf( "\n\n" ) ;
   for ( int i = 0 ; i < 5 ; i++ )
   {
      for ( int j = 0 ; j < 5 ; j++ )
      {
         if ( 3 == j )
         {
            break ;
         }

         printf( "i,j = %d, %d\n", i, j ) ;
      }
   }

   return 0 ;
}

I see:

i,0 = 0, 0
i,2 = 0, 2
i,4 = 0, 4
i,0 = 1, 0
i,2 = 1, 2
i,4 = 1, 4
i,0 = 2, 0
i,2 = 2, 2
i,4 = 2, 4
i,0 = 3, 0
i,2 = 3, 2
i,4 = 3, 4
i,0 = 4, 0
i,2 = 4, 2
i,4 = 4, 4


i,0 = 0, 0
i,1 = 0, 1
i,2 = 0, 2
i,0 = 1, 0
i,1 = 1, 1
i,2 = 1, 2
i,0 = 2, 0
i,1 = 2, 1
i,2 = 2, 2
i,0 = 3, 0
i,1 = 3, 1
i,2 = 3, 2
i,0 = 4, 0
i,1 = 4, 1
i,2 = 4, 2

This makes logical sense, and protects me from breaking code containing the inner loop in question should I move it to a helper function (where there will no longer be an outer loop in scope).

Should it be embarrassing that I actually did this experiment after so many years of programming?

Scotiabank iTrade RESP … they are sneaky little buggers with their hidden fees.

Watch out if you are considering opening a Scotiabank iTrade RESP account.  I got the email below from Scotiabank where I currently have a very minimal RESP account.  I’d opened this account with the intention of increasing it, assuming I’d eventually have the funds to do so, but will now likely close it out of principle and look elsewhere:

“As a valued Scotia iTRADE® client, your continued satisfaction is important to us.
That’s why we’re writing to you today to remind you that your combined assets with Scotia iTRADE need to reach certain levels as of August 31st, 2012 for you to avoid an annual fee charged to your registered account(s) on September 15th, 2012 (covering a one year period to September 14th, 2013).
Here is the breakdown for meeting asset levels to avoid the fee:
• If you hold an RESP account with Scotia iTRADE, you need to have $15,000 or greater in combined assets with Scotia iTRADE to avoid a $25 annual fee (plus HST) charged to your RESP account(s) on September 15th.
• For registered accounts, other than RESP accounts, you need to have $25,000 or greater in combined assets with Scotia iTRADE to avoid a $100 annual fee (plus HST) charged to your registered account(s) on September 15th.
With our powerful trading platforms, easy-to-use trading and investment tools, and friendly customer service, consolidating more of your assets with Scotia iTRADE is a wise decision.
We appreciate your business and will continue to offer you robust investment solutions to help you achieve your investment needs.”

Somewhere in the fine print of the Scotia iTrade agreement must be some statement of fees for less than this minimum “$15,000″.  When I read the fine print, I saw the fees schedule for the per trade fees, but not this one.

I can summarize this email from Scotiabank, “As a valued iTrade customer, we will charge you fees that will encourage you to look into our competitors banking products.”

Geometry of general Jones vector (problem 2.8)

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Another problem from [1].

Problem

The general case is represented by the Jones vector

Show that this represents elliptically polarized light in which the major axis of the ellipse makes an angle

with the axis.

Solution

Prior to attempting the problem as stated, let’s explore the algebra of a parametric representation of an ellipse, rotated at an angle as in figure (1). The equation of the ellipse in the rotated coordinates is

Figure 1: Rotated ellipse

 

which is easily seen to have the required form

We’d like to express and in the “fixed” frame. Consider figure (2) where our coordinate conventions are illustrated. With

Figure 2: 2d rotation of frame

 

and we find

so that the equation of the ellipse can be stated as

or

Observing that

we have, with a Jones vector representation of our rotated ellipse

Since we can absorb a constant phase factor into our argument, we can write this as

This has the required form once we make the identifications

What isn’t obvious is that we can do this for any , , and . Portions of this problem I tried in Mathematica starting from the elliptic equation derived in section 8.1.3 of [2]. I’d used Mathematica since on paper I found the rotation angle that eliminated the cross terms to always be 45 degrees, but this turns out to have been because I’d first used a change of variables that scaled the equation. Here’s the whole procedure without any such scaling to arrive at the desired result for this problem. Our starting point is the Jones specified field, again as above I’ve using

We need our cosine angle addition formula

Using this and writing we have

Subtracting from we have

Squaring this and using , and 1.2.17 we have

which expands and simplifies to

which is an equation of a rotated ellipse as desired. Let’s figure out the angle of rotation required to kill the cross terms. Writing , and rotating our primed coordinate frame by degrees

we have

To kill off the cross term we require

or

This yields 1.1.2 as desired. We also end up with expressions for our major and minor axis lengths, which are respectively for

which completes the task of determining the geometry of the elliptic parameterization we see results from the general Jones vector description.

References

[1] G.R. Fowles. Introduction to modern optics. Dover Pubns, 1989.

[2] E. Hecht. Optics. 1998.

Complex form of Poynting relationship

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This is a problem from [1], something that I’d tried back when reading [2] but in a way that involved Geometric Algebra and the covariant representation of the energy momentum tensor. Let’s try this with plain old complex vector algebra instead.

Question: Average Poynting flux for complex 2D fields (problem 2.4)

Given a complex field phasor representation of the form

Here we allow the components of and to be complex. As usual our fields are defined as the real parts of the phasors

Show that the average Poynting vector has the value

Answer

While the text works with two dimensional quantities in the plane, I found this problem easier when tackled in three dimensions. Suppose we write the complex phasor components as

and also write , and , then our (real) fields are

and our Poynting vector before averaging (in these units) is

We are tasked with computing the average of cosines

So, our average Poynting vector is

We have only to compare this to the desired expression

This proves the desired result.

References

[1] G.R. Fowles. Introduction to modern optics. Dover Pubns, 1989.

[2] JD Jackson. Classical Electrodynamics Wiley. John Wiley and Sons, 2nd edition, 1975.

Ease of screwing up C string operations.

Here’s a new way to mess up a strncpy that I hadn’t seen before. I found about 50 instances of this in the component I was working on today:

   ossStrNCopy( pszTarget, pszSource, strlen( pszSource ) + 1 ) ;

Our ossStrNCopy function is a lot like strncpy, but unlike the library interface, ours is explicitly truncating and won’t leave a dangling non-null terminated string. It’s supposed to be called like:

   ossStrNCopy( pszTarget, pszSource, sizeof(pszTarget) ) ;

(where pszTarget is a char [] array and not a char *). The idea is to avoid buffer overflows that would occur if you were to copy more bytes than the target can hold.

This incorrect use is an excellent example of what was probably clueless cut and paste. Perhaps the thought process was: “I don’t know what this last parameter is for. It’s too much work to read the documentation … I’ll just put something.”

The caller is requesting to have one plus the null terminator of the source string copied into the target. If you are going to do that, why not just call strcpy?