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I have 8085 assembly code for dividing 2 data 8 bit
:
MVI C,FFH
LXI H,1900H
MOV A,M (A=08)
INX H
MOV B,M (B=3)
REPEAT: INR C
SUB B
JNC REPEAT
ADD B
INX H
MOV M,C
INX H
MOV M,A
HLT
If you don't use the special opcodes RIM and SIM that only the 8085 has, the resulting machine code will run in almost all cases on the Z80 without changes. This is the case with your program.
However, if your task is to translate the mnemonics, just do a search-and-replace session. Start with the first one, MVI, and change it to LD. And so on.
You will need to change operands like M to (HL), too, because that is the syntax of the Z80 assembler.
Anyway, you need both instruction sets to do this.
I am writing a code in Fortran that involve computation of linear least squares (A^(-1)*b). For this I have used the subroutine "DGELSD". I am able to get the correct answer from my code. I have cross checked my solution with that of the data available from MATLAB.
Now when it comes to the computation time, MATLAB is taking much less time than that of my .f90 code. The main motivation for me to write the .f90 code was to do heavy computations, as MATLAB was unable to handle this problem (as the matrix size is increased, it takes more and more time). I am talking of the order of matrix around (10^6 x 10^6).
I know that I may be lacking somewhere around the vectorization or parallelization of code (as I am new to it). But would it make any difference? As the subroutine "DGELSD" is already highly optimized. Also I am using Intel ifort compiler and visual studio as an editor.
I have attached the part of the main code (.f90) below for reference. Please suggest what can be done to decrease the computation time. I am having good hardware configuration to run heavy computation.
Workstation specification: Intel Xeon 32 core - 64 bit processor, 64 GB RAM.
Code information:
a) I have used 'DGELSD' example available on Intel MKL example site..
b) I am using x64 architecture.
c) This is the code I am using in MATLAB for comparison.
function min_norm_check(m,n)
tic
a=15*m*n;
b=15*m*n;
c=6;
A=rand(a,b);
B=rand(b,c);
C=A\B;
toc
end
This is the Fotran code given below:
! Program to check the time required to
! calculate linear least square solution
! using DGELSD subroutine
PROGRAM MAIN
IMPLICIT NONE
INTEGER :: M,N,LDA,LDB,NRHS,NB,NG
REAL :: T1,T2,START,FINISH
DOUBLE PRECISION, DIMENSION (:,:), ALLOCATABLE :: D_CAP,A,B,TG,DG
INTEGER :: I=0
NB=10
NG=10
M = 15*NB*NG
N = 15*NB*NG
NRHS = 6
!!
LDA = MAX(M,N)
LDB = MAX(M,NRHS)
!
ALLOCATE (A(LDA,N))
ALLOCATE (B(LDB,NRHS))
A = 0
B = 0
CALL RANDOM_NUMBER(A)
CALL RANDOM_NUMBER(B)
CALL CPU_TIME(START)
DO I=1,1
WRITE(*,*) I
CALL CALC_MIN_NORM_D(M,N,LDA,LDB,NRHS,A,B,D_CAP)
ENDDO
CALL CPU_TIME(FINISH)
WRITE(*,*)'("TIME =',FINISH-START,'SECONDS.")'
END
SUBROUTINE CALC_MIN_NORM_D(M,N,LDA,LDB,NRHS,A,B,D_CAP)
!
! SUBROUTINE DEFINITION TO CALCULATE THE
! LINEAR LEAST SQUARE SOLUTION OF ([A]^-1*B)
!
IMPLICIT NONE
INTEGER :: M,N,NRHS,LDA,LDB,LWMAX,INFO,LWORK,RANK
DOUBLE PRECISION RCOND
INTEGER, ALLOCATABLE, DIMENSION(:) :: IWORK
DOUBLE PRECISION :: A(LDA,N),B(LDB,NRHS),D_CAP(LDB,NRHS)
DOUBLE PRECISION, ALLOCATABLE, DIMENSION(:) :: S,WORK
!
WRITE(*,*)'IN CALC MIN NORM BEGINING'
WRITE(*,*)'DGELSD EXAMPLE PROGRAM RESULTS'
LWMAX = 1E8
ALLOCATE(S(M))
ALLOCATE(IWORK(3*M*0+11*M))
ALLOCATE(WORK(LWMAX))
! NEGATIVE RCOND MEANS USING DEFAULT (MACHINE PRECISION) VALUE
RCOND = -1.0
!
! QUERY THE OPTIMAL WORKSPACE.
!
LWORK = -1
CALL DGELSD( M, N, NRHS, A, LDA, B, LDB, S, RCOND, RANK, WORK,LWORK, IWORK, INFO )
LWORK = MIN( LWMAX, INT( WORK( 1 ) ) )
!WRITE(*,*) 'AFTER FIRST DGELSD'
!
! SOLVE THE EQUATIONS A*X = B.
!!!
CALL DGELSD( M, N, NRHS, A, LDA, B, LDB, S, RCOND, RANK, WORK,LWORK, IWORK, INFO )
!
! CHECK FOR CONVERGENCE.
!
IF( INFO.GT.0 ) THEN
WRITE(*,*)'THE ALGORITHM COMPUTING SVD FAILED TO CONVERGE;'
WRITE(*,*)'THE LEAST SQUARES SOLUTION COULD NOT BE COMPUTED.'
STOP
END IF
!
!
WRITE(*,*)' EFFECTIVE RANK = ', RANK
!D_CAP = B
END
This is the build log for successful compilation of the file. As I am using Visual studio with intel visual fortran, I can compile my program using Compile option available in the editor. I mean to say I don't have to use command line interface to run my program.
Compiling with Intel(R) Visual Fortran Compiler 17.0.2.187 [IA-32]...
ifort /nologo /debug:full /Od /warn:interfaces /module:"Debug\\" /object:"Debug\\" /Fd"Debug\vc110.pdb" /traceback /check:bounds /check:stack /libs:dll /threads /dbglibs /c /Qlocation,link,"C:\Program Files (x86)\Microsoft Visual Studio 11.0\VC\\bin" /Qm32 "D:\Google Drive\Friday_May_27_2016\Mtech Thesis Job\All new work\Fortran coding\fortran hfgmc\Console6\Console6\Console6.f90"
Linking...
Link /OUT:"Debug\Console6.exe" /INCREMENTAL:NO /NOLOGO /MANIFEST /MANIFESTFILE:"Debug\Console6.exe.intermediate.manifest" /MANIFESTUAC:"level='asInvoker' uiAccess='false'" /DEBUG /PDB:"D:\Google Drive\Friday_May_27_2016\Mtech Thesis Job\All new work\Fortran coding\Console6\Console6\Debug\Console6.pdb" /SUBSYSTEM:CONSOLE /IMPLIB:"D:\Google Drive\Friday_May_27_2016\Mtech Thesis Job\All new work\Fortran coding\fortran hfgmc\Console6\Console6\Debug\Console6.lib" -qm32 "Debug\Console6.obj"
Embedding manifest...
mt.exe /nologo /outputresource:"D:\Google Drive\Friday_May_27_2016\Mtech Thesis Job\All new work\Fortran coding\fortran hfgmc\Console6\Console6\Debug\Console6.exe;#1" /manifest "Debug\Console6.exe.intermediate.manifest"
Console6 - 0 error(s), 0 warning(s)
Also I have included the intel mkl library, shown in the figure below:
I have compiled the program and the output has been written in file 'OUTPUT_FILE.TXT'.
IN CALC MIN NORM BEGINING
DGELSD EXAMPLE PROGRAM RESULTS
EFFECTIVE RANK = 1500
IN CALC MIN NORM ENDING
("TIME TAKEN TO SOLVE DGELSD= 4.290028 SECONDS.")
On the other hand MATLAB gives the following result in the output command window:
min_norm_check(10,10)
Elapsed time is 0.224172 seconds.
Also I don't want to outperform MATLAB, its easy and simple. But with increase in the size of the problem, MATLAB stops responding. I have left my program to run on MATLAB for more than two days. It still haven't produced any results.
If I have both alwas_comb and always_ff in a single module, which one executed first?.
for example, I have seen this code in a book but I am confused about the functionality. for example, if WE=0 what will be the value of Qout?
module SyncRAM #(parameter M = 4, N = 8)(output logic [N-1:0] Qout,
input logic [M-1:0] Address, input logic [N-1:0] Data, input logic clk, WE);
logic [N-1:0] mem [0:(1<<M)-1];
always_comb
Qout = mem[Address];
always_ff #(posedge clk)
if (~WE)
mem[Address] <= Data;
endmodule
Any help about the truth table of this code is appreciated,
regards
The specific answer to your question is that Qout will just track the value of mem[Address]. In other words, on the rising edge of the clock, if WE is 0, Qout will be driven with the value written to the memory. This is because the memory will behave like a bank of flip-flops, while the Qout output will behave as if it is directly connected to the Q output of a bank of flip-flops.
The order of the execution of the two always blocks is deterministic, because Qout is driven using a blocking assignment (=), whereas the memory is written to using a non-blocking assignment (<=). See the answer here for much more detail.
EDIT: I am Making this just a question about the Fortran and will start a new question about converting to MATLAB.
ORIGINAL:
I am working on a project and am trying to port some old Fortran code into Matlab. I have almost no Fortran experience so I am not quite sure what is going on in the following code. The Fortran code is simply for interpreting data from a binary file and I have made some decent progress at porting stuff into MATLAB but have got stuck at the following part:
IMPLICIT NONE
DOUBLE PRECISION SCALE
CHARACTER*72 BLINE
CHARACTER*72 DATA_FILE_FULL_NAME
CHARACTER*6 DATA_FILE_NAME
CHARACTER*4 CH4, CH4F
REAL*4 RL4
EQUIVALENCE (RL4,CH4)
CHARACTER*2 C2
LOGICAL LFLAG
c2='69'
LFLAG=.TRUE.
DATA_FILE_FULL_NAME='./'//DATA_FILE_NAME//'.DAT'
OPEN(UNIT=20, FILE=DATA_FILE_FULL_NAME, ACCESS='DIRECT',
. RECL=72, status='OLD')
READ(20,REC=1) BLINE
CH4f=BLINE(7:10)
call flip(4,lflag,ch4f,ch4)
SCALE=RL4
RETURN
END
c ..................................................
subroutine flip(n,lflag,ch1,ch2)
c ..................................................
integer*4 n, i
character*(*) ch1, ch2
logical lflag
if(lflag) then
do i=1,n
ch2(i:i)=ch1(n-i+1:n-i+1)
enddo
else
ch2=ch1
endif
return
end
What I am stuck with is getting the correct value for SCALE because I am not exactly sure what is happening with the RL4. Can someone explain to me why RL4 changes value and what process changes it so that I can put this process into MATLAB?
The code probably does change of Endianness - the byte swap. The equivalence means that the equivalenced variables share the memory and you can treat the 4 bytes read from the file as a number after the byte swap.
The four bytes are therefore interpreted as a four byte real RL4 and converted to a double precision value scale after that.
The byte swap is done only when the logical lflag is true. That will probably be when the byte order of the file is not the same as the machine native byte order.
In Matlab you can probably use http://www.mathworks.com/help/matlab/ref/swapbytes.html .
Just read the 4 byte floating point number and swap the bytes if necessary.
I am writing a test bench for Ripple counter using d flip flop. My program is compiling without errors, however, I get undefined result. How can I solve this problem?
Here is the code:
module RCounter;
reg d,d2,d3,d4,clk;
wire q,q2,q3,q4;
DFlipFlop a(d,q,clk);
DFlipFlop a1(d2,q2,q);
DFlipFlop a2(d3,q3,q2);
DFlipFlop a3(d4,q4,q3);
initial
begin
clk =1;
d=0;d2=0;d3=0;d4=0;
#2 d=1;d2=~q2; d3=~q3; d4=~q4;
#2 d=0;d2=~q2; d3=~q3; d4=~q4;
#2 d=1;d2=~q2; d3=~q3; d4=~q4;
#2 d=0;d2=~q2; d3=~q3; d4=~q4;
#2 d=1;d2=~q2; d3=~q3; d4=~q4;
#2 d=0;d2=~q2; d3=~q3; d4=~q4;
#2 d=1;d2=~q2; d3=~q3; d4=~q4;
end
always
begin
#2 assign clk = ~ clk;
end
endmodule
What am I doing wrong and how can I solve this?
What you have there is not a ripple counter, and it seems that you don't really understand the boundary between your testbench and your DUT (design under test, or in your case, the 'ripple counter').
What you have is a testbench that is simulating four independent flip flops. If you were simulating a ripple counter, you should have a module called something like 'RCounter', which is instanced inside something else called 'RCounter_TB'. The testbench should only drive the inputs (for a counter, clock and reset), it should not drive the d pins of the individual flops, as the connections between these flops is what you want to test.
Inside the ripple counter module, you define the wired connections between your flip flops. There should not be any # time delays inside this module, because a module does not have a concept of fixed time delays. If you want the d2 pin to be driven from ~q2, then you just assign it as such:
assign d2 = ~q2
Because in hardware, this is just a wire looping back from the output of ~q2 back to d2. It always exists, and it has no concept of time.
As to specifically why you're getting X's on your output, I'll guess that it comes from the flip flop design you posted in your last question.
module DFlipFlop(d,q,clk);
input d,clk;
output q;
assign q = clk?( (d==1)? 1:0) : q;
endmodule
This is not a flip flop, as there is no state retention here, it's just an assign statement with an infinite feedback loop, (you essentially just have a wire driving itself).
If you want to model a flip flop, you need to use an always #(posedge clk) block, to imply that you want some state retention. I'll leave it to you to look up how to use an always block to model a flip flop.