I have a common C function that I want to call from C, Fortran and COBOL. It fetches x bytes of data from a database and places it in a char pointer supplied to it. My example below fetches 1024 bytes, but in the real situation I want to be able to fetch much larger chunks of data than 1024 bytes as well, hence the dynamic memory allocation.
void fetch_data(char *fetched)
{
static struct {unsigned long data_length; char some_data[1024];} a_struct;
// Fetch data into a_struct.
memcpy(fetched, &(a_struct.some_data), 1024);
}
I was able to call this function successfully from C.
char *mydata;
mydata = malloc(1024);
fetch_data(mydata);
// Do something with the data.
free(mydata);
I was also able to call this function successfully from Fortran.
INTEGER*4, ALLOCATABLE :: MYDATA(:)
ALLOCATE(MYDATA(1024))
CALL FETCH_DATA(MYDATA)
// Do something with the data.
DEALLOCATE(MYDATA)
But how do I allocate and deallocate dynamic memory in COBOL? I have been unable to find built-in functions/procedures for this purpose.
I also don't see an alternative where C could handle the allocation and deallocation for Fortran and COBOL, as they need to access the data outside C.
As you've only talked about "COBOL" without specifying any actual implementation I assume you mean "standard COBOL".
This could mean COBOL85 - which doesn't have this feature but allows you to just define DATA-FOR-C PIC X(1024) and pass this as reference (COBOL85 actually doesn't specify anything about calling into C space but this should work with most if not all COBOL implementations). Note: This is actually more a detail of Acorns answer.
If you want to use real dynamic memory allocation and you mean standard COBOL - no problem with COBOL 2002 as it introduced the statements ALLOCATE and FREE (Note: this is actually the detail of the comments from roygvib and Rick):
77 pointer-variable USAGE POINTER.
77 address-holder PIC X BASED.
ALLOCATE variable-number CHARACTERS RETURNING pointer-variable
SET ADDRESS OF address-holder TO pointer-variable
CALL "fetch_data" USING address-holder
PERFORM stuff
FREE pointer-variable
If you don't use a COBOL implementation that support these statements you'd have to use the implementor specific routines (normally via CALL) to get/release the memory.
MicroFocus/NetCOBOL (see answer of Rick): CBL_ALLOC_MEM/CBL_FREE_MEM[2]
ACUCOBOL: M$ALLOC/M$FREE
IBM: CEEGTST
any COBOL compiler and runtime that allows to directly call C functions (which may adds additional needs as specifying the appropriate CALL-CONVENTION for those): malloc/free
... see your implementor's manual ...
One example with a very old compiler (Micro Focus COBOL v3.2.50). Much of this is taken directly from the supplemental materials. And since I didn't have an equally old C compiler available, I included a COBOL program as a subtitute.
program-id. dynam.
data division.
working-storage section.
1 ptr pointer.
1 mem-size pic x(4) comp-5 value 1024.
1 flags pic x(4) comp-5 value 1.
1 status-code pic x(2) comp-5.
linkage section.
1 mem pic x(1024).
procedure division.
call "CBL_ALLOC_MEM" using ptr
by value mem-size flags
returning status-code
if status-code not = 0
display "memory allocation failed"
stop run
else
set address of mem to ptr
end-if
call "fetch_data" using mem
display mem
call "CBL_FREE_MEM" using mem
returning status-code
if status-code not = 0
display "memory deallocation failed"
stop run
else
set address of mem to null
end-if
stop run
.
end program dynam.
program-id. "fetch_data".
data division.
working-storage section.
1 some-struct pic x(1024) value all "abcd".
linkage section.
1 mem pic x(1024).
procedure division using mem.
move some-struct to mem
exit program
.
end program "fetch_data".
The display (trimmed) is:
abcdabcdabcdabcd...(for 1024 characters total)
Maybe that will be of some help.
If you do not need the entire data in memory, then consider working chunk-by-chunk: allocate fixed-size storage in COBOL, fetch a chunk into it using the C function, work with it and loop to continue with the next chunk. This way you can avoid allocating dynamic memory altogether.
If you are on Z or using Gnu Cobol you can simply call malloc():
CALL "malloc" USING BY VALUE MEM-SIZE
RETURNING MEM-PTR.
CALL "free" USING BY VALUE MEM-PTR.
Related
I am facing a problem while debugging a COBOL program, When using animator and when a call is made to another COBOL program with a CALL ... USING statement.
After entering the sub program and pressing P (for Perform) E (for Exit). The animator asks "Perform Level = 1 Zoom to calling program ?", and on pressing Y(for Yes) Animator shows the message "Perform Level 1", and doesn't proceed to the the next line in the calling program (i.e, MAIN.COB) .
The only option available then is to Zoom all along which is not helping much during debugging.
Is there a runtime Switch or a Compiler option to make animator pass the control back to the calling program
Example:
This is the main program that iam debugging
IDENTIFICATION DIVISION.
PROGRAM-ID. MAIN.
DATA DIVISION.
WORKING-STORAGE SECTION.
01 WS-STUDENT-ID PIC 9(4) VALUE 1000.
01 WS-STUDENT-NAME PIC A(15) VALUE 'Tim'.
PROCEDURE DIVISION.
CALL 'UTIL' USING WS-STUDENT-ID, WS-STUDENT-NAME.
DISPLAY 'Student Id : ' WS-STUDENT-ID
DISPLAY 'Student Name : ' WS-STUDENT-NAME
STOP RUN.
This is the called program, UTIL,
IDENTIFICATION DIVISION.
PROGRAM-ID. UTIL.
DATA DIVISION.
LINKAGE SECTION.
01 LS-STUDENT-ID PIC 9(4).
01 LS-STUDENT-NAME PIC A(15).
PROCEDURE DIVISION USING LS-STUDENT-ID, LS-STUDENT-NAME.
DISPLAY 'In Called Program'.
MOVE 1111 TO LS-STUDENT-ID.
EXIT PROGRAM.
So when iam debugging MAIN program, and pass control to UTIL subroutine (by Perform and Step ). when i try to return back to calling program MAIN, animator gives the message "Perform Level = 1 Zoom to calling program ? Y/N" - when i press Y is shows a message at lower left saying ""Perform Level = 1" and doesnt skip to the program MAIN
When we look at the Perform / Call stack ( by pressing P and V ) animator displays the below Stack view
Perform/Call Stack View-----------------------------------
Program Name Section Name
----------------------------------------------------------
UTIL.int No Section or Paragraph
---------------------------------------------------------
Zoon Scroll <so on... >
---------------------------------------------------------
As you can see, only the Sub module is visible in the stack. Whereas it should have been.
Perform/Call Stack View-----------------------------------
Program Name Section Name
----------------------------------------------------------
UTIL.int No Section or Paragraph
MAIN.int No Section or Paragraph
---------------------------------------------------------
Zoon Scroll <so on... >
---------------------------------------------------------
(i.e., The calling point from the MAIN.int should also be visible)
This is why i suspect it has something to do with some missing compilation option or Environment settings
I have the following simple COBOL program - written for the PC. It simply reads a file from the computer and writes to the file:
ENVIRONMENT DIVISION.
INPUT-OUTPUT SECTION.
FILE-CONTROL.
SELECT CUSTOMER-FILE ASSIGN TO
"C:Customers.dat"
ORGANIZATION IS LINE SEQUENTIAL.
DATA DIVISION.
FILE SECTION.
FD CUSTOMER-FILE.
01 CUSTOMER-RECORD.
05 FIRST-NAME PIC X(20).
05 LAST-NAME PIC X(20).
WORKING-STORAGE SECTION.
01 WS-CUSTOMER-RECORD.
05 WS-FIRST-NAME PIC X(20).
05 WS-LAST-NAME PIC X(20).
01 WS-EOF PIC X.
PROCEDURE DIVISION.
OPEN OUTPUT CUSTOMER-FILE
PERFORM UNTIL CUSTOMER-RECORD = SPACES
DISPLAY "Enter the first and last name for the customer"
ACCEPT CUSTOMER-RECORD
WRITE CUSTOMER-RECORD
END-PERFORM
CLOSE CUSTOMER-FILE
DISPLAY "Output from the Customer File:"
OPEN INPUT CUSTOMER-FILE.
PERFORM UNTIL WS-EOF = 'Y'
READ CUSTOMER-FILE INTO WS-CUSTOMER-RECORD
AT END MOVE 'Y' TO WS-EOF
NOT AT END DISPLAY WS-CUSTOMER-RECORD
END-READ
END-PERFORM.
CLOSE CUSTOMER-FILE.
GOBACK.
My question: I'm not too familiar with JCL. So if I were to put this program on a mainframe, what would I do for the JCL?
I presume your Identification Division got lost in a cut & paste incident on its way to Stack Overflow; you'll need that.
The current incarnation of IBM Enterprise COBOL does not allow free format source so in order to get your code to compile you would have to reformat and follow the traditional fixed format.
Rather than referring to your data file by name, your Assign clause must refer to a name (limited to 8 characters) which corresponds to a DD name in your JCL. Pick something meaningful, to the extent you can in 8 characters, maybe CUSTOMER.
Since you're running with JCL, your Accept statement will work a bit differently. Probably data will come from a SYSIN DD.
Your JCL will look something like this...
[job card, which is shop-specific]
//TOMSPGM EXEC PGM=yourProgramName
//STEPLIB DD DISP=SHR,DSN=mainframe.dataset.where.you.bound.your.program
//SYSIN DD *
[your customer records]
//CUSTOMER DD DISP=(NEW,CATLG,DELETE),
// DSN=mainframe.dataset.where.your.data.should.end.up,
// LRECL=40,
// AVGREC=U,
// RECFM=FB,
// SPACE=(40,(10,10),RLSE) Adjust to your needs
//SYSOUT SYSOUT=*
//CEEDUMP SYSOUT=*
I'm not sure how this will work with your creating the customer file and then reading it in the same program. In 30 years of mainframe work I've never seen that.
Adding to answer from #cschneid.
Great to see AVGREC is being used on the DD statement to allocate space for the data set. This is much better than using the old-fashioned CYL, or TRK units.
Unfortunately, IMHO, the IBM z/OS architects missed to implement a more modern was to specify space: KiB, or MiB. (ISPF supports KB, and MB as space unit, JCL doesn't.)
With AVGREC you tell the system that the SPACE= primary and secondary space values are number of records, instead of physical units such as tracks, or cylinders.
//CUSTOMER DD ...
// AVGREC=U,
// SPACE=(40,(10,20),RLSE)
Above statement tells the system that the records written will have an average length of 40 bytes (this completely is independent of RECFM=, or LRECL=!). With AVGREC=U (U means units), this further tells the system to allocate initial (primary) space for 10 records, and to add additional space for 20 records each time more space is needed (with an upper limit).
Physical allocations are still in tracks, or cylinders under the hood. The system calculates tracks, or cylinders needed from
"average record length" * "number of records" * avgrec-unit
For the primary allocation, this is
40 * 10 * 1 = 400 bytes
Good. But how can we specify our space needs in KiB or MiB using these keywords?
Remember that the average record length in the SPACE= parameter is completely unreleated to the actual record length specified via LRECL=. Great, so we can freely choose the average record length, and set it to, say, 1. And let us also change the wording "number of records* in above forumla to "number of units". The formula becomes:
1 * "number of units" * avgrec-unit
or
"number of units" * avgrec-unit
AVGREC= supports the units U (1), K (1024), and M (1024*1024). So, to allocate space in megabytes (MiB), we simply code:
//CUSTOMER DD ...
// AVGREC=M,
// SPACE=(1,(10,20),RLSE)
This will allocate 10 MiB primary space, and 20 MiB secondary space. Each allocation is rounded up to the next integral number of tracks, or cylinders, depending on physical disk structures. You simply don't have to care anymore. Neat, isn't it?
123456*
IDENTIFICATION DIVISION.
PROGRAM-ID. "EVEN-OR-ODD".
DATA DIVISION.
WORKING-STORAGE SECTION.
01 Num-1 PIC 9(02).
02 Answer PIC XXXX.
PROCEDURE DIVISION.
GOBACK.
EVEN-OR-ODD.
IF FUNCTION REM(NUM-1, 2) = 0
COMPUTE ANSWER = "Even"
ELSE
COMPUTE ANSWER = "Odd"
END-IF
END PROGRAM EVEN-OR-ODD.
Its a simple even odd function. It should check if number is even return "even" else return "odd"
Can someone explain what's wrong ?
So much things a COBOL compiler would have told you...
GOBACK as first statement, so the rest would not be executed
the program misses a final period and a necessary/reasonable (that depends on the compiler) statement to end the program (END PROGRAM is only parsed for the compilation phase) - you likely want to move your GOBACK. to the end
COMPUTE does not set anything to alphanumeric, you likely want MOVE
there is no way to know what the program would have done, so possibly want DISPLAY instead of MOVE
NUM-1 is never set and has no initial VALUE - so it could theoretically even abend
I'm trying to obtain how much free memory I have on the device. To do this I call the cuda function cuMemGetInfo from a fortran code, but it returns negative values for the free amount of memory, so there's clearly something wrong.
Does anyone know how I can do that?
Thanks
EDIT:
Sorry, in fact my question was not very clear. I'm using OpenACC in Fortran and I call the C++ cuda function cudaMemGetInfo. Finally I could fix the code, the problem was effectively the kind of variables that I was using. Switching to size_ fixed everything. This is the interface in fortran that I'm using:
interface
subroutine get_dev_mem(total,free) bind(C,name="get_dev_mem")
use iso_c_binding
integer(kind=c_size_t)::total,free
end subroutine get_dev_mem
end interface
and this the cuda code
#include <cuda.h>
#include <cuda_runtime.h>
extern "C" {
void get_dev_mem(size_t& total, size_t& free)
{
cuMemGetInfo(&free, &total);
}
}
There's one last question: I pushed an array on the gpu and I checked its size using cuMemGetInfo, then I computed it's size counting the number of bytes, but I don't have the same answer, why? In the first case it is 3052mb large, in the latter 3051mb. This difference of 1mb could be the size of the array descriptor? Here there's the code that I used:
integer, parameter:: long = selected_int_kind(12)
integer(kind=c_size_t) :: total, free1,free2
real(8), dimension(:),allocatable::a
integer(kind=long)::N, eight, four
allocate(a(four*N))
!some OpenACC stuff in order to init the gpu
call get_dev_mem(total,free1)
!$acc data copy(a)
call get_dev_mem(total,free2)
print *,"size a in the gpu = ",(free1-free2)/1024/1024, " mb"
print *,"size a in theory = ", (eight*four*N)/1024/1024, " mb"
!$acc end data
deallocate(a)
Right, so, like commenters have suggested, we're not sure exactly what you're running, but filling in the missing details by guessing, here's a shot:
Most CUDA API calls return a status code (or error code if you will); this is true both in C/C++ and in Fortran, as we can see in the Portland Group's CUDA Fortran Manual:
Most of the runtime API routines are integer functions that return an error code; they return a value of zero if the call was successful, and a nonzero value if there was an error. To interpret the error codes, refer to “Error Handling,” on page 48.
This is the case for cudaMemGetInfo() specifically:
integer function cudaMemGetInfo( free, total )
integer(kind=cuda_count_kind) :: free, total
The two integers for free and total are cuda_count_kind, which if I am not mistaken are effectively unsigned... anyway, I would guess that what you're getting is an error code. Have a look at the Error Handling section on page 48 of the manual.
Segments of memory - BSS, Stack, Heap, Data, Code/Text (Are there any more?).
Say I have a 128MB RAM, Can someone tell me:
How much memory is allocated for each of these memory segments?
Where do they start? Please specify the address range or something like that for better clarity.
What factors influence which should start where?
That question depends on the number of variables used. Since you did not specify what compiler or language or even operating system, that is a difficult one to pin down on! It all rests with the operating system who is responsible for the memory management of the applications. In short, there is no definite answer to this question, think about this, the compiler/linker at runtime, requests the operating system to allocate a block of memory, that allocation is dependent on how many variables there are, how big are they, the scope and usage of the variables. For instance, this simple C program, in a file called simpletest.c:
#include <stdio.h>
int main(int argc, char **argv){
int num = 42;
printf("The number is %d!\n", num);
return 0;
}
Supposing the environment was Unix/Linux based and was compiled like this:
gcc -o simpletest simpletest.c
If you were to issue a objdump or nm on the binary image simpletest, you will see the sections of the executable, in this instance, 'bss', 'text'. Make note of the sizes of these sections, now add a int var[100]; to the above code, recompile and reissue the objdump or nm, you will find that the data section has appeared - why? because we added a variable of an array type of int, with 100 elements.
This simple exercise will prove that the sections grows, and hence the binary gets bigger, and it will also prove that you cannot pre-determine how much memory will be allocated as the runtime implementation varies from compiler to compiler and from operating system to operating system.
In short, the OS calls the shot on the memory management!
you can get all this information compiling your program
# gcc -o hello hello.c // you might compile with -static for simplicity
and then readelf:
# readelf -l hello
Elf file type is EXEC (Executable file)
Entry point 0x80480e0
There are 3 program headers, starting at offset 52
Program Headers:
Type Offset VirtAddr PhysAddr FileSiz MemSiz Flg Align
LOAD 0x000000 0x08048000 0x08048000 0x55dac 0x55dac R E 0x1000
LOAD 0x055dc0 0x0809edc0 0x0809edc0 0x01df4 0x03240 RW 0x1000
NOTE 0x000094 0x08048094 0x08048094 0x00020 0x00020 R 0x4
Section to Segment mapping:
Segment Sections...
00 .init .text .fini .rodata __libc_atexit __libc_subfreeres .note.ABI-tag
01 .data .eh_frame .got .bss
02 .note.ABI-tag
The output shows the overall structure of hello. The first program header corresponds to the process' code segment, which will be loaded from file at offset 0x000000 into a memory region that will be mapped into the process' address space at address 0x08048000. The code segment will be 0x55dac bytes large and must be page-aligned (0x1000). This segment will comprise the .text and .rodata ELF segments discussed earlier, plus additional segments generated during the linking procedure. As expected, it's flagged read-only (R) and executable (X), but not writable (W).
The second program header corresponds to the process' data segment. Loading this segment follows the same steps mentioned above. However, note that the segment size is 0x01df4 on file and 0x03240 in memory. This is due to the .bss section, which is to be zeroed and therefore doesn't need to be present in the file. The data segment will also be page-aligned (0x1000) and will contain the .data and .bss ELF segments. It will be flagged readable and writable (RW). The third program header results from the linking procedure and is irrelevant for this discussion.
If you have a proc file system, you can check this, as long as you get "Hello World" to run long enough (hint: gdb), with the following command:
# cat /proc/`ps -C hello -o pid=`/maps
08048000-0809e000 r-xp 00000000 03:06 479202 .../hello
0809e000-080a1000 rw-p 00055000 03:06 479202 .../hello
080a1000-080a3000 rwxp 00000000 00:00 0
bffff000-c0000000 rwxp 00000000 00:00 0
The first mapped region is the process' code segment, the second and third build up the data segment (data + bss + heap), and the fourth, which has no correspondence in the ELF file, is the stack. Additional information about the running hello process can be obtained with GNU time, ps, and /proc/pid/stat.
example taken from:
http://www.lisha.ufsc.br/teaching/os/exercise/hello.html
memory depend on the global variable and local variable