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If I create new project and I call this string predefined function or extern variables all works fine. But when I continuously modified or did some operation with those string function or variables then those functions are not working properly and shows some junk characters at the UART output.
It may because of the memory issue. please check the buzzer size if it is string array, make sure if are you using string then at end of the buzzer put the null ('\0') terminate.
If you are modifying variable in different process or function , please try to use "volatile " key word for the update.
Thank you.
Ketan
Setup: I am debugging a simple C++ program compiled with option -fno-omit-frame-pointer, using libwarf for DWARF 5. Main work is to write a debugger using libdwarf.
For a particular local variable, dwarfdump shows:
DW_AT_location len 0x0002: 915c: DW_OP_fbreg -36
In the following I will refer to '-36' as 'op1', which I get from libdwarf.
Problem: Using op1 directly results in incorrect value for the variable.
(fbPointer is current value of frame base pointer).
int32_t data = (int32_t) ptrace(PTRACE_PEEKDATA, processPid, fbPointer + op1, 0);
I also tried decoding -36 as sleb128 and usleb128, and for both I got 220. Not a good value.
Trial/error shows that if I add 16 to op1, it will work for any number of int variables as parameters and local objects. However, it does not work for float/double.
Question: Is -36, as mentioned everywhere, offset of variable from frame-base pointer? If so, what am I doing wrong?
What are the preceding values in DW_AT_location: "len 0x0002: 915c:"? If they are important in evaluating op1, how do I get them via libdwart?
Thank you very much. It has been more than a week I am stuck at this point.
It seems that DW_OP_fbreg is a reference to DWARF register which in this case is 16 bytes off. That is, we need to add 16 to RBP, the real register, then add -36 to that. Finally, somehow in this case -36 is a plain number as opposed to encoded sleb128.
I have one column (im = 160648) and row (jm = 1). I want to transform that to a matrix with sizes (im = 344) and (jm=467)
my program code is
program matrix
parameter (im=160648, jm=1)
dimension h(im,jm)
integer::h
open (1,file="Hasil.txt", status='old')
open (2,file="HasilNN.txt", status='unknown')
do i=1,jm
read(1,*)(h(i,j)),j=1,jm)
end do
do i=1,im
write(2,33)(h(i,j),j=1,jm)
end do
33 format(1x, 344f10.6)
end program matrix
the error code that appears when read(1,*)(h(i,j)),j=1,jm)
the data type is floating data.
Your read loop is:
do i=1,jm
read(1,*)(h(i,j)),j=1,jm)
end do
Shouldn't do i=1,jm be do i=1,im ?
This would imply there are "im" records (lines) in the formatted text file Hasil.txt, which your question suggests.
read(1,*)(h(i,j)),j=1,jm) implies each record (line of text) has "jm" values, which is 1 value per line. Is this what the file looks like ? (An unknown number of blank lines will be skipped with this read (lu,*) ... statement.)
You appear to be wanting to write this information to another file; HasilNN.txt using 33 format (1x, 344f10.6) which suggests 3441 characters per line, although your write statement will write only 1 value per line (as jm=1). This would be a very long line for a text file and probably difficult to manage outside the program. If you did wish to do this, you could achieve this with an implied do loop, such as:
write(2,33) ((h(i,j),j=1,jm),I=1,im)
A few comments:
using jm = 1 implies each row has only one value, which could be equivalently represented as a 1d vector "dimension h(im)", negating the need for j
File unit numbers 1 and 2 are typically reserved unit numbers for screen/keyboard. You would be better using units 11 and 12.
When devising this code, you need to address the record structure in the 2 files, as a simple vector could be used. You can control the line length with the format. A format of (1x,8f10.6) would create a record of 81 characters, which would be much easier to manage.
Format descriptor f10.6 also limits the range of values you can manage in the files. Values >= 1000 or <= -100 will overflow this format, while values smaller than 1.e-6 will be zero.
As #francescalus has noted, you have declared "h" as integer, but use a real format descriptor. This will produce an "Error : format-data mismatch" and has to be changed to what is expected in the file.
You should consider what you wish to achieve and adjust the code.
I am getting a run time error '6' Over Flow in vb 6
The "Overflow" error means that you are trying to put a number into a variable (or property etc), and the data type of the variable doesn't allow numbers that large.
Make sure that numbers used in calculations that are coerced into integers do not have results larger than integers.
What is the type of the data in the database?
My guess is that ADO returns it as either a String or a Decimal, and Decimal values only "fit into" a Variant in VB6.
VB6 has no syntax for a Decimal literal, however you can use something like:
CDec(111010114289#)
... inline, or declare a Const as in:
Private Const BigVal As Currency = 111010114289#
I you have to put a large number in a small variable, like C, check Remove integer bound check in project properties (if you are not compiling as PCode)
I am attempting to use fortran to write out a comma-delimited file for import into another commercial package. The issue is that I have an unknown number of data columns. My output needs to look like this:
a_string,a_float,a_different_float,float_array_elem1,float_array_elem2,...,float_array_elemn
which would result in something that might look like this:
L1080,546876.23,4325678.21,300.2,150.125,...,0.125
L1090,563245.1,2356345.21,27.1245,...,0.00983
I have three issues. One, I would prefer the elements to be tightly grouped (variable column width), two, I do not know how to define a variable number of array elements in the format statement, and three, the array elements can span a large range--maybe 12 orders of magnitude. The following code conceptually does what I want, but the variable 'n' and the lack of column-width definition throws an error (of course):
WRITE(50,900) linenames(ii),loc(ii,1:2),recon(ii,1:n)
900 FORMAT(A,',',F,',',F,n(',',F))
(I should note that n is fixed at run-time.) The write statement does what I want it to when I do WRITE(50,*), except that it's width-delimited.
I think this thread almost answered my question, but I got quite confused: SO. Right now I have a shell script with awk fixing the issue, but that solution is...inelegant. I could do some manipulation to make the output a string, and then just write it, but I would rather like to avoid that option if at all possible.
I'm doing this in Fortran 90 but I like to try to keep my code as backwards-compatible as possible.
the format close to what you want is f0.3, this will give no spaces and a fixed number of decimal places. I think if you want to also lop off trailing zeros you'll need to do a good bit of work.
The 'n' in your write statement can be larger than the number of data values, so one (old school) approach is to put a big number there, eg 100000. Modern fortran does have some syntax to specify indefinite repeat, i'm sure someone will offer that up.
----edit
the unlimited repeat is as you might guess an asterisk..and is evideltly "brand new" in f2008
In order to make sure that no space occurs between the entries in your line, you can write them separately in character variables and then print them out using theadjustl() function in fortran:
program csv
implicit none
integer, parameter :: dp = kind(1.0d0)
integer, parameter :: nn = 3
real(dp), parameter :: floatarray(nn) = [ -1.0_dp, -2.0_dp, -3.0_dp ]
integer :: ii
character(30) :: buffer(nn+2), myformat
! Create format string with appropriate number of fields.
write(myformat, "(A,I0,A)") "(A,", nn + 2, "(',',A))"
! You should execute the following lines in a loop for every line you want to output
write(buffer(1), "(F20.2)") 1.0_dp ! a_float
write(buffer(2), "(F20.2)") 2.0_dp ! a_different_float
do ii = 1, nn
write(buffer(2+ii), "(F20.3)") floatarray(ii)
end do
write(*, myformat) "a_string", (trim(adjustl(buffer(ii))), ii = 1, nn + 2)
end program csv
The demonstration above is only for one output line, but you can easily write a loop around the appropriate block to execute it for all your output lines. Also, you can choose different numerical format for the different entries, if you wish.