I understand Verilog 2001 added the ">>>" operator for sign extension right shift. But how does it work? I was implementing a shifter unit as part of an ALU. I had done the following: assign #1 rarithmetic = b >>> shiftamount; where b and shiftamount were both input [31:0] vectors. However, the sign extension does not occur. I had b = 0xd1200000 and the result was 0x0000d120 instead of 0xffffd120. I know there is also the "signed" keyword but it didn't seem to have any effect when I tried "input signed [31:0] ...". What am I doing wrong? Would synthesis results be different if "signed" keyword is used? In the end, I just end up coding up a variable 32 bit sign extension shifter as shown below. Is there a better/more compact way to do the same thing? I'd like to keep it synthesizable if possible. Thanks for any help. Kind regards. /* * 32-bit sign extend shifter */ module signextshifter32(/*AUTOARG*/ // Outputs y, // Inputs a, shiftamt ); input [31:0] a; // input input [4:0] shiftamt; // shift amount output [31:0] y; // output assign #1 y = (shiftamt == 5'b00000) ? {a[31:0]} : (shiftamt == 5'b00001) ? {{1{a[31]}}, a[31:1]} : (shiftamt == 5'b00010) ? {{2{a[31]}}, a[31:2]} : (shiftamt == 5'b00011) ? {{3{a[31]}}, a[31:3]} : (shiftamt == 5'b00100) ? {{4{a[31]}}, a[31:4]} : (shiftamt == 5'b00101) ? {{5{a[31]}}, a[31:5]} : (shiftamt == 5'b00110) ? {{6{a[31]}}, a[31:6]} : (shiftamt == 5'b00111) ? {{7{a[31]}}, a[31:7]} : (shiftamt == 5'b01000) ? {{8{a[31]}}, a[31:8]} : (shiftamt == 5'b01001) ? {{9{a[31]}}, a[31:9]} : (shiftamt == 5'b01010) ? {{10{a[31]}}, a[31:10]} : (shiftamt == 5'b01011) ? {{11{a[31]}}, a[31:11]} : (shiftamt == 5'b01100) ? {{12{a[31]}}, a[31:12]} : (shiftamt == 5'b01101) ? {{13{a[31]}}, a[31:13]} : (shiftamt == 5'b01110) ? {{14{a[31]}}, a[31:14]} : (shiftamt == 5'b01111) ? {{15{a[31]}}, a[31:15]} : (shiftamt == 5'b10000) ? {{16{a[31]}}, a[31:16]} : (shiftamt == 5'b10001) ? {{17{a[31]}}, a[31:17]} : (shiftamt == 5'b10010) ? {{18{a[31]}}, a[31:18]} : (shiftamt == 5'b10011) ? {{19{a[31]}}, a[31:19]} : (shiftamt == 5'b10100) ? {{20{a[31]}}, a[31:20]} : (shiftamt == 5'b10101) ? {{21{a[31]}}, a[31:21]} : (shiftamt == 5'b10110) ? {{22{a[31]}}, a[31:22]} : (shiftamt == 5'b10111) ? {{23{a[31]}}, a[31:23]} : (shiftamt == 5'b11000) ? {{24{a[31]}}, a[31:24]} : (shiftamt == 5'b11001) ? {{25{a[31]}}, a[31:25]} : (shiftamt == 5'b11010) ? {{26{a[31]}}, a[31:26]} : (shiftamt == 5'b11011) ? {{27{a[31]}}, a[31:27]} : (shiftamt == 5'b11100) ? {{28{a[31]}}, a[31:28]} : (shiftamt == 5'b11101) ? {{29{a[31]}}, a[31:29]} : (shiftamt == 5'b11110) ? {{30{a[31]}}, a[31:30]} : {31 {a[31]}}; endmodule

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4/30/2009 4:12:55 AM

On Wed, 29 Apr 2009 21:12:55 -0700 (PDT), pallav <pallavgupta@gmail.com> wrote: >where b and shiftamount were both input [31:0] vectors. However, the >sign extension does not occur. I had b = 0xd1200000 and the result was >0x0000d120 instead of 0xffffd120. I know there is also the "signed" >keyword but it didn't seem to have any effect when I tried "input >signed [31:0] ...". What am I doing wrong? Would synthesis results be >different if "signed" keyword is used? > >In the end, I just end up coding up a variable 32 bit sign extension >shifter as shown below. Is there a better/more compact way to do the >same thing? I'd like to keep it synthesizable if possible. Thanks for >any help. > >Kind regards. module signextshifter32( input signed [31:0] a, input [4:0] shiftamt; // shift amount output signed [31:0] y) assign y = a >>> shiftamnt; endmodule I am not sure what you're doing wrong because you don't post your code with the signed keyword but the above should work for both simulation and synthesis the same way. Muzaffer Kal DSPIA INC. ASIC/FPGA Design Services http://www.dspia.com

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4/30/2009 4:57:22 AM

On Apr 30, 12:57=A0am, Muzaffer Kal <k...@dspia.com> wrote: > On Wed, 29 Apr 2009 21:12:55 -0700 (PDT), pallav > > <pallavgu...@gmail.com> wrote: > >where b and shiftamount were both input [31:0] vectors. However, the > >sign extension does not occur. I had b =3D 0xd1200000 and the result was > >0x0000d120 instead of 0xffffd120. I know there is also the "signed" > >keyword but it didn't seem to have any effect when I tried "input > >signed [31:0] ...". What am I doing wrong? Would synthesis results be > >different if "signed" keyword is used? > > >In the end, I just end up coding up a variable 32 bit sign extension > >shifter as shown below. Is there a better/more compact way to do the > >same thing? I'd like to keep it synthesizable if possible. Thanks for > >any help. > > >Kind regards. > > module signextshifter32( > input signed [31:0] a, > input [4:0] =A0shiftamt; =A0// shift amount > output signed [31:0] y) > > assign y =3D a >>> shiftamnt; > > endmodule > > I am not sure what you're doing wrong because you don't post your code > with the signed keyword but the above should work for both simulation > and synthesis the same way. > > Muzaffer Kal > > DSPIA INC. > ASIC/FPGA Design Serviceshttp://www.dspia.com Hi Muzaffer, Thank you for your response. I tried the code that you posted on two different simulators: Icarus Verilog and Synopsys VCS. On Icarus Verilog I got the wrong answer (zero extension) while Synopsys VCS gave the correct answer (sign extension). So it looks like there is a bug on Icarus Verilog and I was trying all my experiments on that! I'll file a bug report for that. So it looks like my problem is solved. Thanks. Kind regards.

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4/30/2009 5:16:40 AM

On Apr 29, 10:16=A0pm, pallav <pallavgu...@gmail.com> wrote: > On Apr 30, 12:57=A0am, Muzaffer Kal <k...@dspia.com> wrote: > > > > > On Wed, 29 Apr 2009 21:12:55 -0700 (PDT), pallav > > > <pallavgu...@gmail.com> wrote: > > >where b and shiftamount were both input [31:0] vectors. However, the > > >sign extension does not occur. I had b =3D 0xd1200000 and the result w= as > > >0x0000d120 instead of 0xffffd120. I know there is also the "signed" > > >keyword but it didn't seem to have any effect when I tried "input > > >signed [31:0] ...". What am I doing wrong? Would synthesis results be > > >different if "signed" keyword is used? > > > >In the end, I just end up coding up a variable 32 bit sign extension > > >shifter as shown below. Is there a better/more compact way to do the > > >same thing? I'd like to keep it synthesizable if possible. Thanks for > > >any help. > > > >Kind regards. > > > module signextshifter32( > > input signed [31:0] a, > > input [4:0] =A0shiftamt; =A0// shift amount > > output signed [31:0] y) > > > assign y =3D a >>> shiftamnt; > > > endmodule > > > I am not sure what you're doing wrong because you don't post your code > > with the signed keyword but the above should work for both simulation > > and synthesis the same way. > > > Muzaffer Kal > > > DSPIA INC. > > ASIC/FPGA Design Serviceshttp://www.dspia.com > > Hi Muzaffer, > > Thank you for your response. I tried the code that you posted on > two different simulators: Icarus Verilog and Synopsys VCS. > > On Icarus Verilog I got the wrong answer (zero extension) while > Synopsys VCS gave the > correct answer (sign extension). So it looks like there is a bug on > Icarus Verilog > and I was trying all my experiments on that! I'll file a bug report > for that. > > So it looks like my problem is solved. Thanks. > > Kind regards. I tried the code using Veritak and, after fixing a couple of typos in the suggested module, got the expected results, ie, signed shifting works as expected. John Providenza

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4/30/2009 3:10:21 PM

On Apr 30, 12:12=A0am, pallav <pallavgu...@gmail.com> wrote: > I understand Verilog 2001 added the ">>>" operator for sign extension > right shift. But how does it work? The ">>>" operator is not specifically an arithmetic or sign-extension shift. It is an arithmetic shift in a signed expression, and a logical shift in an unsigned expression. It is basically what the ">>" should have been, except that ">>" had to continue always being a logical shift for backward compatibility reasons. So if you want to use ">>>" as a signed shift, you will need to make sure that its operand is signed. > I was implementing a shifter unit > as part of an ALU. I had done the following: > > assign #1 rarithmetic =3D b >>> shiftamount; > > where b and shiftamount were both input [31:0] vectors. You will have to declare b to be signed, or convert it to signed with the $signed() function. The signedness of shiftamount doesn't matter. > However, the > sign extension does not occur. I had b =3D 0xd1200000 and the result was > 0x0000d120 instead of 0xffffd120. I know there is also the "signed" > keyword but it didn't seem to have any effect when I tried "input > signed [31:0] ...". That should work. > What am I doing wrong? I don't know, without a more specific example. > Would synthesis results be > different if "signed" keyword is used? Yes, if the synthesis tool is correct. > In the end, I just end up coding up a variable 32 bit sign extension > shifter as shown below. Is there a better/more compact way to do the > same thing? I'd like to keep it synthesizable if possible. Thanks for > any help. There are tricky ways to do sign extension without hardcoding all the cases, even without an arithmetic right shift. Make a word that is 32 copies of the sign bit, then shift it left by 32-shiftamount and OR it with your logically shifted value.

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5/5/2009 10:57:50 PM