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I am trying to understand how the vec function is used.

I have produced the following code, but the behaviour is not as expected.

$binstring = '@' ; # 01000000 binary ascii code
print vec($binstring,0,1); # returns 0, the binary value of bits 0 and 1
vec($binstring,0,1) = 3; # bits 0 and 1
print $binstring; # returns A, I am expecting C (binary ascii code 01000011)
vec($binstring,2,1) = 3; # bits 2 and 3
print $binstring; # returns E, I am expecting O (binary ascii code 01001111)

Documentation tells me that vec has the syntax:

vec EXPR, OFFSET, BITS

where BITS is a power of 2 from 1 to 32.

I presumed offset of zero would give me the leftmost (most significant) bit.
It actually gives me the rightmost bit.

Is this expected behaviour?

The number of bits is a power of two ranging from 1 to 32. Presumably this
means that minimum element size (BITS = 1) is 2^1 = 2 bits, and therefore I
cannot use single bit elements.

Is it correct that if BITS = 1, my elements should be 2 bits wide?

I presumed that the offset was in number of elements, so an offset of 1, would
point to bits two and three. It actually appears to point to bit 1, which is
part of the first element.

Presumably then, offset is in number of bits, not number of elements. Is this
correct?

When I set the value of my element to three, indicating both bits set to 1, I
only appear to be getting a single bit set.

What is going on here?

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Mark Hobley
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(Mark Hobley) wrote:


> The number of bits is a power of two ranging from 1 to 32. Presumably
> this means that minimum element size (BITS = 1) is 2^1 = 2 bits, and
> therefore I cannot use single bit elements.

Yes you can, but you can't use 3 or 5, the *number* has to be a power of
two, so 2^0, 2^1, 2^2, 2^3, 2^4.

> Is it correct that if BITS = 1, my elements should be 2 bits wide?

No, it's 1, but you can't use 3 bits wide, since 3 is not a power of 2 (if
I read the documentation correctly).


--
John Bokma Freelance software developer
&
Experienced Perl programmer: http://castleamber.com/