+ Post New Thread
Results 1 to 6 of 6

6th December 2018, 23:06 #1
 Join Date
 May 2013
 Location
 Mordor
 Posts
 421
 Helped
 96 / 96
 Points
 3,148
 Level
 13
Verilog Interview Question
Hi,
I was recently asked this question in a Logic Design/DV interview,
Suppose you have the following, what should be the bit width of c,d,e ? The interviewer said that the answers of my d & e were wrong. For e my answer was based on 3 as 2'b11 which would give a width of 18. What do you guys think ?
Code Verilog  [expand] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
// The code is my interpretation of his question. logic [9:0] a; // can be anything, 9 & 8 were chosen as random logic [8:0] b; logic [?:0] c; // 17 wide ? logic [?:0] d; // 18 wide ? logic [?:0] e; // ?? always @(posedge clk) begin // Assume you can finish in 1 cycle c <= a*b; // I think c should be at least 8+9 bits wide ? d <= a*b + 1'b1; // d should be atleast 17 (^) + 1 bits wide ? e <= a*b + 3; // what does the 3 here mean ? //In my 23 years of RTL experience I only saw people using the syntax 2'b11 or something like that. //The interviewer was interested in what would verilog interpret the 3 as ? Is it hex by default ? 4'h3 ?? end
If I helped you, press the 'Helped Me' button.
It's only my opinion, and sometimes I'm wrong.

Advertisment

7th December 2018, 01:21 #2
 Join Date
 Sep 2013
 Location
 USA
 Posts
 6,965
 Helped
 1659 / 1659
 Points
 30,234
 Level
 42
Re: Verilog Interview Question
[9:0] is 10 bits wide, i.e. 9,8,7,6,5,4,3,2,1,0 which is 10 digits.
[8:0] is 9 bits wide
a*b is 19 bits wide not 17 (9+8)
a*b +1'b1 will be 19 bits wide not 20, a*b will never be all 1's so adding a 1'b1 will never increase the result's bit width.
a*b +3 will fit in 19 bits but the result is 32bits due to the 3 which is an integer (dependent on implementation, but is normally 32bit)
I've been using Verilog for ~20 years, so I've probably seen + used with practically every possible thing.
   Updated   
The integer math is the most annoying as it causes additional warnings in synthesis when it sees a 32bit result being assigned to a smaller bit width variable. As long as you're willing to ignore those (I usually do) and can get through any reviews with those warnings in your synthesis report they aren't a problem.
1 members found this post helpful.

Advertisment

7th December 2018, 10:50 #3
 Join Date
 Apr 2016
 Posts
 1,640
 Helped
 289 / 289
 Points
 7,795
 Level
 21
Re: Verilog Interview Question
adsee answer is correct. but damn, what a weird question to ask in an interview. knowing or not knowing this doesn't make you a good designer.
Really, I am not Sam.

Advertisment

7th December 2018, 20:45 #4
 Join Date
 Feb 2015
 Posts
 969
 Helped
 278 / 278
 Points
 5,899
 Level
 18
Re: Verilog Interview Question
This looks like a fine interview question. especially if the job is for an engineer that will work on porting dsp algorithms to HW. The OP just managed to hit every offbyone error possible.
The addition of an integer literal 3 could bring up the 32b (possible) conversion, but I would expect the smaller width answer to be accepted.
A weird question would be "y <= (x + 3) >> 1;" vs "y <= (x + 2'd3) >> 1;".

14th December 2018, 23:34 #5
 Join Date
 May 2013
 Location
 Mordor
 Posts
 421
 Helped
 96 / 96
 Points
 3,148
 Level
 13
Re: Verilog Interview Question
Sorry about this one, I posted this question in a hurry without checking the width of a & b in code. As it's a phone interview, the question was just multiply a 9 bit register & a 8 bit register  what's the outcome. So i just remembered my answer of 17. My apologies for writing a 9 bit reg as logic [9:0] a.
Shoot, I just answered that it will overflow to 20 bits. Now I realize that unless you add (19'h111FFFF  max possible a*b) only then we overflow
By implementation, are you talking about the environment setupIf I helped you, press the 'Helped Me' button.
It's only my opinion, and sometimes I'm wrong.

Advertisment

15th December 2018, 01:36 #6
 Join Date
 Sep 2013
 Location
 USA
 Posts
 6,965
 Helped
 1659 / 1659
 Points
 30,234
 Level
 42
Re: Verilog Interview Question
software implementation of the number of bits used by a long integer. Most software implementations consider it to be 32bits, I can't think of any simulators/synthesis tools that consider an integer in Verilog to be anything other than 32bits, e.g. 64bits or 128bits. If I remember correctly the LRM states the size of an integer has to be at least 32bits.
+ Post New Thread
Please login