Look at the following statement

lzp961

Look at the following statement [Copy link]

 

A series of statements:
begin
a <= 1;
end
begin
b <= 2;
end
is followed by:

begin
a = 1;
b = 2;

end
and

begin
a <=1;
b <=2;

end
? Do the two groups have the same semantics?

littleshrimp

Is it the one at the bottom?

wangerxian

And the one at the bottom, a=1 and a<=1 do not mean the same thing.

lzp961

wangerxian posted on 2021-12-7 15:46 And the one at the bottom, a=1 and a<=1 do not mean the same thing.

Not to say

Are begin end and begin end executed sequentially?

wangerxian

lzp961 posted on 2021-12-7 16:13 Doesn't it mean that begin end and begin end are executed sequentially?

begin
a <= 1;
end
begin
b <= 2;
end

It should be executed first a<=1 and then b<=2. The bottom one is also executed in this order.

lzp961

wangerxian published on 2021-12-7 16:18 begin a <= 1; end begin ...

begin

a <= 1;

b <= 2;

end

Aren't a and b assigned at the same time here?

wangerxian

lzp961 posted on 2021-12-7 16:21 begin a <= 1; b <= 2; end Aren't a and b assigned at the same time?

No, is this C language? If so, [<=] is a judgment statement.

wsmysyn

wangerxian posted on 2021-12-7 17:26 No, is this C language? If so, [<=] is a judgment statement.

This is Verilog HDL, non-blocking assignment

LuJianchang

1 and 3 are the same, both are non-blocking assignment statements, begin end is the same as () in C language, and 2 is a blocking assignment statement.

heningbo

This is what non-blocking means.

In fact, these sentences cannot be seen

It should be

begin

a<= 1;

b <= a;

end

and

begin

a<= 1;

b <= a;

end

To be able to see it.

It is generally recommended to use <= in the always statement.

Fred_1977

I just learned a little bit, these two usages are not easy to understand, I need to study them repeatedly.

forest2000

Blocking assignment statements and non-blocking assignment statements, the difference between sequential logic and combinational logic