20 Jun
2016
20 Jun
'16
9:06 p.m.
On Jun 20, 2016, at 1:56 PM, Swift Griggs
<swiftgriggs at gmail.com> wrote:
On Mon, 20 Jun 2016, Paul Koning wrote:
The best way to think about the two different assignment operators (<= is an
assignment *not* less-than-or-equal) is that one happens ?now? and the other
is resolved on clock edges.
So assuming that a & b are 1 bit wide and that initially both a and b are 0 (or
in verilog terms 1b0).
In the first example, both a and b will be 1b1. In the second example, at the end
of the first clock cycle, a will be 1b1 and b will be 1b0. At the end of the next
clock cycle, a will be 1b1 and b will be 1b1 also.
TTFN - Guy
used to how C or similar languages work. For
example, in this C code:
a = 1;
b = a;
a and b will both equal 1 at the end. But in the VHDL code:
a <= 1;
b <= a;
Whoa. That makes total sense, though. In the real world, I'm guessing the
"less than" just reflects that a signal might not have the level you
expect.