switch (x)
   {
   case 1:
             {
             z++;
             ...
             break;
             }
...

Yes, those { } delimiting the case-labeled statement sequence. A quick check of my C source benchmarks showed this usage occurring in around 1% of case-labels. Panic over.

What is the statistical significance, i.e., variance, of that 1%? Have I simply measured an unrepresentative sample, what would be a representative sample and what would be the expected variance within a representative sample?

I am interested in commercial software development and so I have selected half a dozen or so largish code bases as my source benchmark, preferably written in a commercial environment even if currently available as Open source. I would prefer this benchmark to be an order of magnitude larger and perhaps I will get around to adding more programs soon.

My if/switch measurements were aimed at finding usage characteristics that varied between the two kinds of selection statements. One characteristic measured was the number of equality tests in the associated controlling expression. For instance, in:

if (x == 1 || x == 2)
   z--;
else if (x == 3)
   z++;

the first controlling expression contains two equality tests and the second one equality test.

Plotting the percentage of equality tests that occur in the controlling expressions of if-if/if-else-if sequences and switch statements we get the following:

Do these results indicate that if-if/if-else-if sequences and switch statements differ in the number of equality tests contained in their controlling expressions? If I measured a completely different set of source code, would the results be very different?

To answer this question a probability model is needed. Take as an example the controlling expressions present in an if-if sequence. If each controlling expression is independent of the others, then the probability of two equality tests, for instance, occurring in any of these expressions is constant and thus given a large sample the distribution of two equality tests in the source has a binomial distribution. The same argument can be applied to other numbers of equality tests and other kinds of sequence.

For each measurement point in the above plot the associated error bars span the square-root of the variance of that point (assuming a binomial distribution, for a normal distribution the length of this span is known as the standard deviation). The error bars overlap suggesting that the apparent difference in percentage of equality tests in each kind of sequence is not statistically significant.

The existence of some dependency between controlling expression equality tests would invalidate this simply analysis, or at least reduce its reliability. I did notice that in a sequence that containing two equality tests, the controlling expression that contained it tended to appear later in the sequence (the reverse of the example given above). Did I notice this because I tend to write this way? A question for another day.

For example, a struct containing five fields has four field pairs having distance 1 from each other, three distance 2, two distance 2, and one field pair having distance 4; the average is 2.

The surprising answer, to me at least, is (N+1)/3.

Proof: The average distance can be obtained by summing the distances between all possible field pairs and dividing this value by the number of possible different pairs.

                  Distance 1  2  3  4  5  6
Number of fields
            4              3  2  1
            5              4  3  2  1
            6              5  4  3  2  1
            7              6  5  4  3  2  1

The above table shows the pattern that occurs as the number of fields in a definition increases.

In the case of a definition containing five fields the sum of the distances of all field pairs is: (4*1 + 3*2 + 2*3 + 1*4) and the number of different pairs is: (4+3+2+1). Dividing these two values gives the average distance between two randomly chosen fields, e.g., 2.

Summing the distance over every field pair for a definition containing 3, 4, 5, 6, 7, 8, … fields gives the sequence: 1, 4, 10, 20, 35, 56, … This is sequence A000292 in the On-Line Encyclopedia of Integer sequences and is given by the formula n*(n+1)*(n+2)/6 (where n = N − 1, i.e., the number of fields minus 1).

Summing the number of different field pairs for definitions containing increasing numbers of fields gives the sequence: 1, 3, 6, 10, 15, 21, 28, … This is sequence A000217 and is given by the formula n*(n + 1)/2.

Dividing these two formula and simplifying yields (N + 1)/3.