I am using fortran 95. I have a question very similar to Accessing files in sub directory of main program
The additional problem that I am having is this: I am creating files in a loop using following commands:
write(fn,fmt='(a,i0,a)')"degseqA",filenumber,'.dat'
open(unit=filenumber,file=fn)
Hence I cannot use 'output/myfile.dat' to make myfile.dat go to the directory output. Is there any way to solve this?
Thanks
If the directory already exists, it is totally straightforward.
write(fn,fmt='(a,i0,a)') "output/degseqA",filenumber,'.dat'
open(unit=filenumber,file=fn)
or in general
write(fn,fmt='(a,i0,a)') trim(directory_name)//"degseqA",filenumber,'.dat'
where directory_name is a character variable with the name of the directory.
Make sure fn is large enough.
Related
I am working on an assignment where I have main.cpp,lego.cpp, and lego.h. My program currently runs fine and gives out the desired output. My issue is that to submit, I must have everything in a single cpp file. Would appreciate if someone could give a simple example. (I want everything in main.cpp)
Merge C++ files into a single source file
i found this link that is similar but I cant grasp the example
if anyone has this problem I just fixed it by doing this
class lego {things in class};
lego::functionA(){}
lego::functionB(){}
int main(){};
make sure that you delete any other cpp files that are in your source folder and remove any header files from main. Was what was causing my issue.
Combining two cpp files into one cpp file
On a POSIX system, you can use the following command:
cat main.cpp lego.cpp > combined.cpp
On windows, use type instead of cat.
This trivial approach works for most programs, and very likely for a simple assignment. But this can have problems with more complicated programs in particular those that rely heavily on pre-processor.
using the command prompt by combining the files. you need to put the files that you need to combine in a single folder and using the command line (windows + r, on windows 10) run commands that access the directory in which the files are.
Enter the command copy/b *.cpp combined.cpp That will save the two cpp files into one file combined.cpp.
Then access the combined file combined.cpp in the same directory.
This might help you too https://www.youtube.com/watch?v=0atEx7EngoE
I'm trying to build a simple C++ project (an executable) that calls a Haskell function,
using Shake for the build script and calling Stack from within the script to build the Haskell library.
Let's say the Haskell library is called haskell-simple-lib.
The shake script calls stack install haskell-simple-lib which outputs an .so file: libHShaskell-simple-lib-*version*-*unique identifier*.so
My Shake rules depends on filenames, and so I can't use the aforementioned name as I don't know in advance what the unique identifier will be. And so, the Shake script runs a cp on the file to _build/libHShaskell-simple-lib.so
The link options for the C++ executable has -L_build and -lhaskell-simple-lib.
When I try to run the executable I get an error saying:
error while loading shared libraries: libHShaskell-simple-lib-0.1.0.0-8DkaSm3F3d44RUd03fOuDx-ghc7.10.2.so: cannot open shared object file: No such file or directory
But, if I rename the file I copied to _build, to the original name that stack install outputted (the one with the unique identifier), the executable runs correctly.
One would think that all I need to do is to simply cp the file to _build without erasing the unique identifier from the name, however I need to know the name of the .so file in advance for the shake script.
I don't understand why when the executable is run the original .so filename is searched for. The link flag doesn't mention the fullname of the .so that stack install outputted, only libHShaskell-simple-lib.
Could it be that the original name is embedded in the .so file? If so, how does one go about solving this issue?
EDIT:
I'm aware this could be solved using a dummy file, but I'd like to know if there's a better way to do this.
The original identifier is embedded in the .so. I don't remember all the details, but I do know that I've solved such problems using rpath twiddling in the past.
My C++ project folder structure is as shown below.
I am trying to debug the program using gdb in my linux machine. My main function is in g2o.cpp which is inside g2o_cli folder. I am able to put breakpoints in the files in this folder by
break g2o.cpp:<line_number>
But I am not able to put breakpoints in the files in other folders, for example, a file optimizer.cpp in the 'core' folder. I tried giving the absolute path from my home directory
break ~/HOME/g2o/core/optimizer.cpp:<line_number>
but it is giving an error
No source file named ~/HOME/g2o/core/optimizer.cpp
I also tried ../../core/optimizer.cpp instead of the absolute path. Still it did not work.
I saw a similar question here. But none of the possible reasons mentioned in the answer is applicable in my case. The file optimizer.cpp is not a shared library and the entire project was compiled using cmake.
How does gdb take folder paths? How can I give the relative folder path?
A dirty hack you can use on x86 is to use int3. Just use the statement asm volatile ("int 3"); in the code where you want the breakpoint.
I'm quite confused about this weird behaviour of my .cpp project. I've got the following folder structure:
include/mylib.h
myproject/src/eval.cpp
myproject/data/file.csv
myproject/Makefile
In eval.cpp I include mylib.h as follows:
#include "../../include/mylib.h"
and compile it through Makefile:
all:
g++ -I include ../include/mylib.h src/eval.cpp -o eval.out
Now in my eval.cpp I'm reading the file.csv from data directory and if I refer to it like this
../data/file.csv
it doesn't find it (gets empty lines all the time), but this
data/file.csv
works fine.
So, to include mylib.h it goes two directories up (from src folder) which seems right. But it doesn't make sense to me that to refer to another file from the same piece of code it assumes we are in project directory. I suppose it is connected with Makefile somehow, but I'm not sure.
Why is it so?
EDIT: After a few thing I tried it seems that the path which is used is not the path from binary location to the data location, but depends on where from I run the binary as well. I.e., if I have binary in bin directory and run it like:
./bin/eval.out
It works with data/file.csv.
This:
cd bin
./eval.out
works with ../data/file.csv.
Now it seems very confusing to me as depending on where I run the program from it will give different output. Can anyone please elaborate on the reasons for this behaviour and if it is normal or I'm making some mistake?
It is so because (as explained here ) the compiler will search for #included files with quotes (not with brackets) with the current working directory being the location of the source file.
Then, when you try to open your .csv file, it's now your program that looks for a file. But your program runs with the current working directory being myproject/ which explains why you must specify data/file.csv as your file path, and not ../data/file.csv. Your program does not run in your src folder, it will run in the directory the binary ends up being invoked from.
You could have noticed that in your Makefile, your -I options specify a different path for your header file than your .cpp file.
EDIT Answer: It's quite simple actually and completely normal. When you invoke your binary, the directory which you're in is the current working directory. That is, if you run it with the command ./myproject/bin/eval.out, the current working directory is . (e.g. /home/the_user/cpp_projects). My post was a bit misleading about that, I corrected it.
Note: You can use the command pwd in a command prompt to know which is the current working directory of this prompt (pwd stands for "print working directory").
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Closed 10 years ago.
Possible Duplicate:
What should Linux/Unix 'make install' consist of?
I'm making a program that can be invoked from the command line, like ./prog arg1 arg2. I was wondering, how can I make it so that I can run it from anywhere on the system? I know that I could put prog into /usr/bin/, but what if my program needs resources from its install directory (that can be wherever the user downloaded it)?
put the directory in which your program resides into the path environment variable or move your program into one of the directories already in path (usually requires superuser permission, which I gather you don't have for then you wouldn't ask this question).
to add a directory to the front of the search path and have the system refresh its database on tcsh, say
setenv "my/directory:"$PATH
rehash
on bash, I think, it's
PATH=/my/directory:$PATH
export PATH
(no need to rehash). Note that the above commands put your directory at the top of the search path, i.e. these will be searched before any other. Thus, if your program is called "gcc", then your program will be executed rather than the GNU C compiler. Alternatively, you can add your directory to the end of the search path, in which case your program will only be picked up if no other program of the same name is found in any of the other directories in the search path.
You probably also want to become familiar with the Linux Filesystem Hierarchy: the standard definition for "what goes where". Here's more information:
https://superuser.com/questions/90479/what-is-the-conventional-install-location-for-applications-in-linux
Environment variables can be defined globally ("for everybody", e.g. /etc/profile), or locally ("per user", e.g. ~/.bashrc). Here's a good summary of some of your options:
https://wiki.archlinux.org/index.php/Environment_Variables
When you execute a programme using prog arg1 arg2, it's thanks to your shell, which search in the $PATH environement variable for folders where programs are. (Try env | grep PATH to see those folder).
You need eather to add a new directory in this variable (export PATH="/new/directory/path/:$PATH" if under bash, setenv PATH "/new/directory/path/:$PATH" if with tcsh) or copy your program and all the files it need to execute in one of the PATH folder.
There are two ways of dealing with this (and Makefiles have nothing to do with them)
Your installer could just put the files where it wants them, so your program doesn't have to search -- it can use hardcoded paths. Or you could put the path to the data directory into yet another file, which would be hardcoded (like /etc/programname.config).
You put all your stuff into one directory (often something like /opt/programname). You can hardcode that too, of course, or your program can readlink() the /proc/pid/exe file for a good chance (no guarantee, though. In particular, it works if for example a symlink is used to point from /usr/bin/programname to your /opt/programname/bin/programname or whatever, but it won't work if that's a hardlink)
to get the path to the executable. From there you should be able to reach your data files.
If prefer the second solution, but that's just me. The first solution works well with package managers, and it's less overkill if you don't really have a lot of data files.