Comparing integer numbers

[Deepu]

Hi All,

I had a question on how to compare 5-6 integer numbers.

always @ (a or b)
begin
if ( a == 1)
begin
if (b == 1000 || 2000 || 3000 || 4000 || 5000 || 6000)
begin
$display ("All is Fine");
end
end
end

In the above example 'a' will be '1' and then when it enters the if
loop, it doesnot enter the 2nd 'if' loop even though b is 1000 or 2000
so on, Is this way of comparing allowed? How else i can do the
comparison of integer numbers.

Thanks

 

[John_H]

Deepu wrote:

Hi All,

I had a question on how to compare 5-6 integer numbers.

always @ (a or b)
begin
if ( a == 1)
begin
if (b == 1000 || 2000 || 3000 || 4000 || 5000 || 6000)
begin
$display ("All is Fine");
end
end
end

In the above example 'a' will be '1' and then when it enters the if
loop, it doesnot enter the 2nd 'if' loop even though b is 1000 or 2000
so on, Is this way of comparing allowed? How else i can do the
comparison of integer numbers.

Thanks

Wrap your head around "operator precedence" and find it in a Verilog
reference (or C manual for that matter).

If b==1000, the if executes. Next, if 2000 is logically true, the if
executes. It was my recollection - flawed if you never entered the loop
- that nonzero integers evaluated to TRUE meaning the second loop should
have always been executed.

What you probably want is

if( b==1000 || b==2000 || b==3000 || b==4000 || b==5000 || b==6000 )

The Verilog language treats each logical entity as an entity onto its
own, not as part of a list to use in a previous evaluation as you
assumed. Since any operation can produce a result and those results are
used to determine your logic, you need to understand how to properly
compartmentalize your evaluations to achieve your goals.

Good luck.

 

[mk]

On Sun, 16 Mar 2008 17:21:55 -0700 (PDT), Deepu <pradeep.bg@gmail.com>
wrote:

Hi All,

I had a question on how to compare 5-6 integer numbers.

always @ (a or b)
begin
if ( a == 1)
begin
if (b == 1000 || 2000 || 3000 || 4000 || 5000 || 6000)
begin
$display ("All is Fine");
end
end
end

In the above example 'a' will be '1' and then when it enters the if
loop, it doesnot enter the 2nd 'if' loop even though b is 1000 or 2000
so on, Is this way of comparing allowed? How else i can do the
comparison of integer numbers.

It's allowed but it doesn't do what you want. The expression of the if
condition is equivalent to ((b==1000)||2000|| ...) so it's always
true. What you want seems to be
if (b==1000 || b == 2000 || b==3000 ...).

HTH.

 

[Jonathan Bromley]

On Sun, 16 Mar 2008 17:21:55 -0700 (PDT),
Deepu <pradeep.bg@gmail.com> wrote:

Hi All,

I had a question on how to compare 5-6 integer numbers.

always @ (a or b)
begin
if ( a == 1)
begin
if (b == 1000 || 2000 || 3000 || 4000 || 5000 || 6000)
begin
$display ("All is Fine");
end
end
end

As others have said, this isn't what you think.

A nice way to code it is...

case (b)
1000, 2000, 3000, 4000, 5000, 6000:
$display("It's one of the six choices");
endcase

although the *meaning* of this case statement is exactly
the same as a series of if...else if.... tests.

Alternatively, if you are using SystemVerilog,
you can write this:

if (b inside {1000, 2000, 3000, 4000, 5000, 6000} )
$display("b is a member of the set of values");

--
Jonathan Bromley, Consultant

DOULOS - Developing Design Know-how
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