[kstenerud]

I think you may have misread the code:

    while (howmany != 0) {
        val = random();
        if( val is odd) {
          out[index] =  val;
          index += 1;
        }
        howmany--;
    }

vs

    while (howmany != 0) {
        val = random();
        out[index] = val;
        index += (val bitand 1);
        howmany--;
    }

Both of these store a list of odd numbers in out[], with "index" containing the resulting count of how many numbers are in out[]. Both will have an "index" (count) value of 0 if all inputs were even, and neither attempts to access out[index-1].

[XJ6]

You were right I misinterpreted "out[0] to out[index-1]" but your next statement:

> count of how many numbers are in out[]

is not true, in the latter case it's a count of how many numbers you want to be in out[].

Consider what happens if howmany is 1 and it generates a single even number. In the original you have an empty array and index 0, in the newer one you have an array e.g. [2] and an index 0.

Yes, you can solve this by 'manually' only returning the first "index" elements but it's just asking for trouble and a source of bugs as seen in your attempt to describe it have a difference between the reality and your description.

You might not consider that to matter, but the point I'm trying to make isn't just nitpicking, it's that there are certain expected behaviours from code, and an array which is "all odd numbers, except maybe all odd but one last even number" is bound to lead to broken expectations.

[kstenerud]

> > count of how many numbers are in out[]

> is not true, in the latter case it's a count of how many numbers you want to be in out[].

I'm failing to see how "index" is the count of how many numbers you want to be in out[] rather than the count of how many numbers actually ARE in out[], or why this would be different between code examples.

> Consider what happens if howmany is 1 and it generates a single even number. In the original you have an empty array and index 0, in the newer one you have an array e.g. [2] and an index 0.

Yes, that's the point, as was explicitly stated in the article.

> Yes, you can solve this by 'manually' only returning the first "index" elements but it's just asking for trouble and a source of bugs as seen in your attempt to describe it have a difference between the reality and your description.

What is there to solve? The end result in both cases is that variable "index" is 0, because 0 of the values in out[] are valid.

> You might not consider that to matter, but the point I'm trying to make isn't just nitpicking, it's that there are certain expected behaviours from code, and an array which is "all odd numbers, except maybe all odd but one last even number" is bound to lead to broken expectations.

Yes, and the expectation is that "index" tells you where the first garbage element is in the array (in both examples). You're either going to have an array with all garbage (index = 0), all garbage except the first element (index = 1), all garbage except the first and second elements (index = 2), and so on. What actual values are in a garbage index (even or odd or 0 or 1 or 2 or Martin Luther's birth year), are irrelevant because you're not going to read from those indices.

[couchand]

Granted, the examples lack enough detail to know precisely how `out` is allocated. However, it seems safe to assume from the context that `out` is allocated by the client and `index` is returned to indicate the total number. There is no case where `out` is an "empty array", it always must have enough space for `howmany` integers.

[gpderetta]

You are reading too much in a minimalistic example to show some specific detail.