Two comparators in ISE synthesis report

[hithesh123]

I have a 4bit counter in my vhdl code. when the count<8, I load data to a MS 8-bits of a registers and when the count>8, I load data to LS 8-bits of the register.
In the synthesis report, why are 2 5-bit comparators inferred - one for less than and the other for greater/equal.

 

[barry]

Wait. You have a 4-bit counter but a 16-bit register? Without your code it's impossible to tell what's going on here.

 

[mrflibble]

I have a 4bit counter in my vhdl code. when the count<8, I load data to a MS 8-bits of a registers and when the count>8, I load data to LS 8-bits of the register.
In the synthesis report, why are 2 5-bit comparators inferred - one for less than and the other for greater/equal.

I bet one meeeelion dollahs that you actually have behavioral code for two comparators in there. Combine that with the fact that sometimes the synth tools are not all that clever and ta-da! two comparators.

Generally speaking, what you could do is make a signal, and assign that the value of ONE of the comparison functions. Then you use that as the input of a mux that selects between the values.

Specifically speaking: 4-bit counter, 8 is a threshold value. Oh whatever shall we do? Hey look, the most significant bit of the counter. It wants some attention, I wonder why that is...

Get it? that's a hint. several times! About 2^3 times.

 

[hithesh123]

I bet one meeeelion dollahs that you actually have behavioral code for two comparators in there. Combine that with the fact that sometimes the synth tools are not all that clever and ta-da! two comparators.

Generally speaking, what you could do is make a signal, and assign that the value of ONE of the comparison functions. Then you use that as the input of a mux that selects between the values.

Specifically speaking: 4-bit counter, 8 is a threshold value. Oh whatever shall we do? Hey look, the most significant bit of the counter. It wants some attention, I wonder why that is...

Get it? that's a hint. several times! About 2^3 times.

Are you saying, just check for the MSbit. If its '0' then (count<8), if it's '1' then (count>8)
In this case how do I make the count=0, after it reaches 0xF.

 

[mrflibble]

Are you saying, just check for the MSbit. If its '0' then (count<8), if it's '1' then (count>8)

No. I am saying, just check for the MSbit. If its '0' then (count<8), if it's '1' then (count>=8)

In this case how do I make the count=0, after it reaches 0xF.

As what bit sequence is 0xF + 1 represented on a 4-bit counter?