for loop range syntax - matlab

Question

I spent a couple of hours debugging a problem that I would have thought would have been a syntax error.
a = zeros(3);
for i=1:1size(a,2) % note the missing colon between 1 and size(a,2)
i
end
The following only displays
ans = 3
1
Essentially, it seems Matlab/Octave parses the above as:
for i=1:1
size(a,2)
i
end
Note however that
i=1:1size(a,2)
produces a syntax error. Is there a good reason that Matlab/Octave has this for loop syntax? Is there something that it's supposed to make easier? Just curious if anyone else has any thoughts about it. Thanks.

Answer 1

It is indeed a bit of a surprise that Matlab's syntax allows this. I don't know why this is allowed. One reason might be to allow for-loops on one line:
>> for i=1:3 disp(i);end
1
2
3
But interestingly, removing the space is not allowed:
>> for i=1:3disp(i);end
for i=1:3disp(i);end
|
Error: Unexpected MATLAB operator.
This reason for this is probably that a number followed by d is another way of writing a floating point number (3d10 == 3e10), so the parser/tokenizer initially thinks you define a number, but then gets confused when it sees the i. Daniel's example with fprintf does work, since a number followed by an f is not a valid number, so the tokenizer understands that you started a new token.
I guess that many years ago (>30?), when they defined matlab's syntax, they did not foresee that this could introduce this kind of hard-to-spot problems. I guess matlab was originally written by engineers for engineers, and not by someone who knows how to design a general purpose programming language. Other languages like C or Python use punctuation to separate loop conditions from loop body, so there is no ambiguity. I don't know if it is still possible to correct Matlab's syntax, since it might break old code that relies on the current behavior.
At least, if you use a recent version of Matlab, the editor warns for various problems in your code. Paying attention to the small red dashes in the right hand border could have saved you a few hours of debugging time (but maybe you were using octave). I try to make it a habit to fix all the warnings it indicates. For your code, it shows the following:

Answer 2

Your code is equivalent to
a = zeros(3);
for i=1:1
size(a,2)
i
end
There are some places where everyone would use newline or white space, but the parser itself does not require.
A minimal loop:
for i=1:3fprintf('%d',i),end
but I recommend to use at least a comma seperated version, everything else is horrible to read:
for i=1:3,fprintf('%d',i),end