# How can I implement this?
**URL:** https://discourse.myhdl.org/t/how-can-i-implement-this/416
**Category:** Support
**Created:** [March 2, 2020, 12:11pm UTC](https://discourse.myhdl.org/t/how-can-i-implement-this/416 "2020-03-02T12:11:37Z")
**Posts on this page:** 3
**Page:** 1
### Author:  [@krs](https://discourse.myhdl.org/u/krs)
#### Post date: [March 2, 2020, 12:11pm UTC](https://discourse.myhdl.org/t/how-can-i-implement-this/416/1 "2020-03-02T12:11:37Z")
Hello I would like to implement the following logic with a for loop in myhdl.
Can anyone tell me how to do it ? Thanks
```
intMux = [Signal(intbv(0)[16:]) for i in range(8)]
@always(reset.posedge, clk.posedge)
def statsGen():
if reset == 1 :
sumInt.next = 0
elif (vld == 1):
sumInt.next = sumInt + intMux[0] + intMux[1] + intMux[2] + \
intMux[3] + intMux[4] + intMux[5] + intMux[6] + intMux[7]
```
---
### Author:  [@JanCoombs](https://discourse.myhdl.org/u/JanCoombs)
#### Post date: [March 2, 2020, 5:49pm UTC](https://discourse.myhdl.org/t/how-can-i-implement-this/416/2 "2020-03-02T17:49:47Z")
> [@krs](#):
>
> Hello I would like to implement the following logic with a for loop in myhdl.
> Can anyone tell me how to do it ? Thanks
```
@always(reset.posedge, clk.posedge)
def statsGen():
if (reset == 1): sumInt.next = 0
elif (vld == 1):
sumVar = 0
for ii in range( AddendCount ):
sumVar = sumVar + intMux[ii]
sumInt.next = sumVar
```
Jan Coombs
---
### Author:  [@josyb](https://discourse.myhdl.org/u/josyb)
#### Post date: [March 3, 2020, 8:57pm UTC](https://discourse.myhdl.org/t/how-can-i-implement-this/416/3 "2020-03-03T20:57:07Z")
My take on this:
```python
def sum(a):
if len(a) > 2:
# split in two
n = len(a) // 2
return sum(a[:n]) + sum(a[n:])
elif len(a) == 2:
return a[0] + a[1]
else:
return a[0]
intMux = [Signal(intbv(0)[16:]) for __ in range(8)]
@always_seq(clk.posedge, reset)
def statsgen():
if vld:
sumInt.next = sum(intMux)
```
This implements a _binary tree_-like circuit, and will give you the fastest implementation in hardware.