I have written a c program. I want to pipe the program and I want to make it look meaningful. So instead of writing ./a.out each time, I want to name it changetext. To achieve that, I compiled my program following way: gcc -o changetext myprog.c. To the best of my knowledge, this should replace the use of ./a.out and changetext should do that instead. But I'm getting command not found. I am new to c and unix environment. Any suggestion appreciated.
As I said in a comment, you can either put a dot slash (./) in front of the executable to run it
./changetext
Or you put in in a directory that is referenced in the PATH environment variable. A nice explanation of this safety feature can be found here (thanks to rubenvb):
http://www.linfo.org/dot_slash.html
It says that this is more or less to distinguish built-in commands from user-written commands with the same name. I am not convinced though. The shell could simply prefer built-in names to user-supplied ones, and look in the current directory as well as in the PATH.
But this is the *nix way.
In order to compile and run a program such as your changetext with just the command chanhetext, you must put the binary in a directory listed in your PATH environment variable. It is recommended that you put programs that you made for your own use in the ~/bin/ directory. The command you would use to accomplish this would be the following, assuming ~/bin/ already exists:
gcc -o ~/bin/changetext myprog.c
If it does not exist, you can simply create it, then log out and back in.
If you are tired of doing the ./ before the program name you can always make an alias such as
alias a='./a.out' or alias changetext='./changetext'
this just basically look for everytime you type changetext or a and then replaces it to have the ./ infront of it
Related
Now I'm writing a makefile for my C project, and I want to implement a RELEASE&DEBUG compilation branch, it's like the script within the makefile could know which target name the user inputted in the command line, then it can decide what kind of GCC options it will use to compile the current program, is there any way to achieve this? Thank U.
I tried to define a variable in the makefile and pass its value through the command line like make type=release, but this is not what I wanted.
Assuming you're talking about GNU make then there is: look up the MAKECMDGOALS macro in the GNU make manual.
But, this is really not the right way to do things. I recommend that instead you look up target-specific variables in the manual and see if that gives any ideas.
I'm on a Mac and in terminal I'm compiling my program
gcc -Wall -g -o example example.c
it compiles (there are no errors), but when I try to provide command line arguments
example 5 hello how are you
terminal responds with "-bash: example: command not found"
how am supposed to provide the arguments I want to provide after compiling?
Run it like this with path:
./example 5 hello how are you
Unless the directory where the example binary is part of the PATH variable, what you have won't work even if the binary you are running is in the current directory.
It is not a compilation issue, but an issue with your shell. The current directory is not in your PATH (look with echo $PATH and use which to find out how the shell uses it for some particular program, e.g. which gcc).
I suggest testing your program with an explicit file path for the program like
./example 5 hello how are you
You could perhaps edit your ~/.bashrc to add . at the end of your PATH. There are pro and conses (in particular some possible security issues if your current directory happens to be sometimes a "malicious" one like perhaps /tmp might be : bad guys might put there a gcc which is a symlink to /bin/rm so you need to add . at the end of your PATH if you do).
Don't forget to learn how to use a debugger (like gdb). This skill is essential when coding in C (or in C++). Perhaps consider also upgrading your gcc (Apple don"t like much its current GPLv3 license so don't distribute the recent one; try just gcc -v and notice that the latest released GCC is today 4.8.1).
./example 5 Hello how are you is the syntax you're looking for.
This article lends a good explanation as to why this is important.
Basically, when you hit Enter, the shell checks to see if the first set of characters is an absolute path. If it's not, it checks the PATH variable to find executables with the name of the command you are trying to run. If it's found, it will be run, but otherwise it will crash and burn and you will become very sad.
I tried to use a make file in code::blocks but I am doing it wrong. I have the version installed with the compilers included. http://sourceforge.net/projects/codeblocks/files/Binaries/10.05/Windows/codeblocks-10.05mingw-setup.exe/download. What do I do with the make file? It starts with:
CC=gcc
best, US
You don't tend to execute the make file itself, rather you execute make, giving it the make file as an argument:
make -f pax.mk
If your make file is actually one of the standard names (like makefile or Makefile), you don't even need to specify it. It'll be picked up by default (if you have more than one of these standard names in your build directory, you better look up the make man page to see which takes precedence).
As paxdiablo said make -f pax.mk would execute the pax.mk makefile, if you directly execute it by typing ./pax.mk, then you would get syntax error.
Also you can just type make if your file name is makefile/Makefile.
Suppose you have two files named makefile and Makefile in the same directory then makefile is executed if make alone is given. You can even pass arguments to makefile.
Check out more about makefile at this Tutorial : Basic understanding of Makefile
Can someone explain to me why, in particular, we are using ./a.out to run a program?
Is there any meaning behind this?
Can someone please provide an explanation?
The name stands for "assembler output", and was (and still is) the default name for the executable generated by the compiler. The reason you need ./ in front of it is because the current directory (.) is not in $PATH therefore the path to the executable must be explicitly given.
If you mean the ./ part, it's for safety. Windows by default appends current directory to PATH, which is bad (there's a risk of DLL injection, and so on).
If you mean a.out part, it's just a name (which came from name of format a.out), which you can change by modifying gcc -o parameter.
When running an executable like a shell like bash the executable must be in your PATH environment variable for bash to locate and run the program.
The ./ prefix is a shorthand way of specifying the full path to the executable, so that bash does not need to the consult the PATH variable (which usually does not contain the current directory) to run it.
[For a.out (short for "assembler output"), it is the default executable output for a compiler like gcc if no output filename is specified.]
It'd be worth you looking a bit more into C and the way that C programs are compiled.
Essentially, your source code is sent to the preprocessor, where directives like #define and #include are loaded (e.g. into memory). So any libraries you want to use are loaded, e.g.
#include <math.h>
will basically 'paste' the contents of math.h into source code at the point at which it is defined.
Once all this stuff has been expanded out, the compiler turns your source code into object code, which is your source in binary code. a.out is the default name for output if you do not specify a build name.
gcc -o mynewprogram mynewprogram.c
a.out is the default name for the compiler. AFAIK it is because the linking process is skipped and it is not compiled as an object or library.
I'm just starting programming and going through K&R to try and learn C. I've gotten to the section on command line arguments (5.10) but now I'm stumped. Every time I try and open a program I've written with command line arguments I'm told that file X, X being the argument, doesn't exist.
`gcc -o find find.c
open find test
The file /Documents/Learning_C/test does not exist.`
Any suggestions? Thanks
What system are you on? In Unix/Linux you compile & run your executable via:
gcc -o find find.c
./find test
As others have noted, when you prefix your binary with "./", there wont be any naming conflicts. However, if you made find accessible in your $PATH, you might have some conflicts with find and test--standard programs with most *nix distributions... Maybe you could choose more specific names (i.e. ./myFind testArg)
Try giving your output executable a different name.
I suspect your executing the system find command which is looking for a directory called 'test'.
Or try forcing it by executing
./find toto
Edit: Prepending the ./ to the command is important because it tells the shell to execute the find in the current directory as opposed to the first 'find' that exists in your PATH. It is normally recommended that you don't have . (the current directory) in your PATH for security reasons.
HTH
P.S. Forgot to say good one for working through K&R. I just finished doing the same after working in C for thirty years and it was good to get back and refresh the mind!
Instead of making us all individually guess what exactly you're doing wrong, perhaps you should paste the program you're using for the illustration mentioned ?